How to force a service (bash script) to read from stdin (ask me to input something) when the script is executing? - linux

I created a service with systemd that uses wendy (inotify replacement tool) to listen in a directory and run a bash script when something changes.
however, my script relies on stdin at a certain time to read a variable. but when the service runs it skips asking me in a terminal for the input entirely and proceeds with the rest of the bash script.
i'm new to systemd and services, is there anyway I can force it to ask me for input?
this is what happened from /var/log/syslog
Oct 7 21:52:09 server wendy.sh[13062]: was added to scripts.
Oct 7 21:52:09 server wendy.sh[13062]: enter scriptname: (/home/user/scripts/blah.sh)
Oct 7 21:52:09 server wendy.sh[13062]: chmod: missing operand after +x
Oct 7 21:52:09 server wendy.sh[13062]: Try 'chmod --help' for more information.
it was supposed to ask me for a scriptname to pass into chmod.
How to accomplish this?
Thank you

Unfortunately, it is very difficult to ask for user inputs in a script running as a background task, because they have no terminal attached. I would advise you to try to find an alternative to reading inputs from stdin.
If you really want to achieve that, you could for example run a program somewhere that will listen to a UNIX socket, and your automated script could communicate to this client to ask for an input (c.f. this Stack Exchange post).
With netcat-openbsd, there is a -U option. If you don't have it, you probably have netcat-traditional installed instead; I'd suggest switching.
Example command: nc -U /var/run/socket

Related

Cronjob does not create Logfiles [duplicate]

I have set up a cronjob for root user in ubuntu environment as follows by typing crontab -e
34 11 * * * sh /srv/www/live/CronJobs/daily.sh
0 08 * * 2 sh /srv/www/live/CronJobs/weekly.sh
0 08 1 * * sh /srv/www/live/CronJobs/monthly.sh
But the cronjob does not run. I have tried checking if the cronjob is running using pgrep cron and that gives process id 3033. The shell script calls a python file and is used to send an email. Running the python file is ok. There's no error in it but the cron doesn't run. The daily.sh file has the following code in it.
python /srv/www/live/CronJobs/daily.py
python /srv/www/live/CronJobs/notification_email.py
python /srv/www/live/CronJobs/log_kpi.py
WTF?! My cronjob doesn't run?!
Here's a checklist guide to debug not running cronjobs:
Is the Cron daemon running?
Run ps ax | grep cron and look for cron.
Debian: service cron start or service cron restart
Is cron working?
* * * * * /bin/echo "cron works" >> /tmp/file
Syntax correct? See below.
You obviously need to have write access to the file you are redirecting the output to. A unique file name in /tmp which does not currently exist should always be writable.
Probably also add 2>&1 to include standard error as well as standard output, or separately output standard error to another file with 2>>/tmp/errors
Is the command working standalone?
Check if the script has an error, by doing a dry run on the CLI
When testing your command, test as the user whose crontab you are editing, which might not be your login or root
Can cron run your job?
Check /var/log/cron.log or /var/log/messages for errors.
Ubuntu: grep CRON /var/log/syslog
Redhat: /var/log/cron
Check permissions
Set executable flag on the command: chmod +x /var/www/app/cron/do-stuff.php
If you redirect the output of your command to a file, verify you have permission to write to that file/directory
Check paths
Check she-bangs / hashbangs line
Do not rely on environment variables like PATH, as their value will likely not be the same under cron as under an interactive session. See How to get CRON to call in the correct PATHs
Don't suppress output while debugging
Commonly used is this suppression: 30 1 * * * command > /dev/null 2>&1
Re-enable the standard output or standard error message output by removing >/dev/null 2>&1 altogether; or perhaps redirect to a file in a location where you have write access: >>cron.out 2>&1 will append standard output and standard error to cron.out in the invoking user's home directory.
If you don't redirect output from a cron job, the daemon will try to send you any output or error messages by email. Check your inbox (maybe simply more $MAIL if you don't have a mail client). If mail is not available, maybe check for a file named dead.letter in your home directory, or system log entries saying that the output was discarded. Especially in the latter case, probably edit the job to add redirection to a file, then wait for the job to run, and examine the log file for error messages or other useful feedback.
If you are trying to figure out why something failed, the error messages will be visible in this file. Read it and understand it.
Still not working? Yikes!
Raise the cron debug level
Debian
in /etc/default/cron
set EXTRA_OPTS="-L 2"
service cron restart
tail -f /var/log/syslog to see the scripts executed
Ubuntu
in /etc/rsyslog.d/50-default.conf
add or comment out line cron.* /var/log/cron.log
reload logger sudo /etc/init.d/rsyslog restart
re-run cron
open /var/log/cron.log and look for detailed error output
Reminder: deactivate log level, when you are done with debugging
Run cron and check log files again
Cronjob Syntax
# Minute Hour Day of Month Month Day of Week User Command
# (0-59) (0-23) (1-31) (1-12 or Jan-Dec) (0-6 or Sun-Sat)
0 2 * * * root /usr/bin/find
This syntax is only correct for the root user. Regular user crontab syntax doesn't have the User field (regular users aren't allowed to run code as any other user);
# Minute Hour Day of Month Month Day of Week Command
# (0-59) (0-23) (1-31) (1-12 or Jan-Dec) (0-6 or Sun-Sat)
0 2 * * * /usr/bin/find
Crontab Commands
crontab -l
Lists all the user's cron tasks.
crontab -e, for a specific user: crontab -e -u agentsmith
Starts edit session of your crontab file.
When you exit the editor, the modified crontab is installed automatically.
crontab -r
Removes your crontab entry from the cron spooler, but not from crontab file.
Another reason crontab will fail: Special handling of the % character.
From the manual file:
The entire command portion of the line, up to a newline or a
"%" character, will be executed by /bin/sh or by the shell specified
in the SHELL variable of the cronfile. A "%" character in the
command, unless escaped with a backslash (\), will be changed into
newline characters, and all data after the first % will be sent to
the command as standard input.
In my particular case, I was using date --date="7 days ago" "+%Y-%m-%d" to produce parameters to my script, and it was failing silently. I finally found out what was going on when I checked syslog and saw my command was truncated at the % symbol. You need to escape it like this:
date --date="7 days ago" "+\%Y-\%m-\%d"
See here for more details:
http://www.ducea.com/2008/11/12/using-the-character-in-crontab-entries/
Finally I found the solution. Following is the solution:-
Never use relative path in python scripts to be executed via crontab.
I did something like this instead:-
import os
import sys
import time, datetime
CLASS_PATH = '/srv/www/live/mainapp/classes'
SETTINGS_PATH = '/srv/www/live/foodtrade'
sys.path.insert(0, CLASS_PATH)
sys.path.insert(1,SETTINGS_PATH)
import other_py_files
Never supress the crontab code instead use mailserver and check the mail for the user. That gives clearer insights of what is going.
I want to add 2 points that I learned:
Cron config files put in /etc/cron.d/ should not contain a dot (.). Otherwise, it won't be read by cron.
If the user running your command is not in /etc/shadow. It won't be allowed to schedule cron.
Refs:
http://manpages.ubuntu.com/manpages/xenial/en/man8/cron.8.html
https://help.ubuntu.com/community/CronHowto
To add another point, a file in /etc/cron.d must contain an empty new line at the end. This is likely related to the response by Luciano which specifies that:
The entire command portion of the line, up to a newline or a "%"
character, will be executed
I found useful debugging information on an Ubuntu 16.04 server by running:
systemctl status cron.service
In my case I was kindly informed I had left a comment '#' off of a remark line:
Aug 18 19:12:01 is-feb19 cron[14307]: Error: bad minute; while reading /etc/crontab
Aug 18 19:12:01 is-feb19 cron[14307]: (*system*) ERROR (Syntax error, this crontab file will be ignored)
It might also be a timezone problem.
Cron uses the local time.
Run the command timedatectl to see the machine time and make sure that your crontab is in this same timezone.
https://askubuntu.com/a/536489/1043751
I had a similar problem to the link below.
similar to my problem
my original post
My Issue
My issue was that cron / crontab wouldn't execute my bash script. that bash script executed a python script.
original bash file
#!/bin/bash
python /home/frosty/code/test_scripts/test.py
python file (test.py)
from datetime import datetime
def main():
dt_now = datetime.now()
string_now = dt_now.strftime('%Y-%m-%d %H:%M:%S.%f')
with open('./text_file.txt', 'a') as f:
f.write(f'wrote at {string_now}\n')
return None
if __name__ == '__main__':
main()
the error I was getting
File "/home/frosty/code/test_scripts/test.py", line 7
string_to_write = f'wrote at {string_now}\n'
^
SyntaxError: invalid syntax
this error didn't make sense because the code executed without error from the bash file and the python file.
** Note -> ensure in the crontab -e file you don't suppress the output. I sent the output to a file by adding >>/path/to/cron/output/file.log 2>&1 after the command. below is my crontab -e entry
*/5 * * * * /home/frosty/code/test_scripts/echo_message_sh >>/home/frosty/code/test_scripts/cron_out.log 2>&1
the issue
cron was using the wrong python interpreter, probably python 2 from the syntax error.
how I solved the problem
I changed my bash file to the following
#!/bin/bash
conda_shell=/home/frosty/anaconda3/etc/profile.d/conda.sh
conda_env=base
source ${conda_shell}
conda activate ${conda_env}
python /home/frosty/code/test_scripts/test.py
And I changed my python file to the following
from datetime import datetime
def main():
dt_now = datetime.now()
string_now = dt_now.strftime('%Y-%m-%d %H:%M:%S.%f')
string_file = '/home/frosty/code/test_scripts/text_file.txt'
string_to_write = 'wrote at {}\n'.format(string_now)
with open(string_file, 'a') as f:
f.write(string_to_write)
return None
if __name__ == '__main__':
main()
No MTA installed, discarding output
I had a similar problem with a PHP file executed as a CRON job.
When I manually execute the file it works, but not with CRON tab.
I got the output message: "No MTA installed, discarding output"
Postfix is the default Mail Transfer Agent (MTA) in Ubuntu and can be installed it using
sudo apt-get install postfix
But this same message can be also output when you add a log file as below and it does not have proper write permission to /path/to/logfile.log
/path/to/php -f /path/to/script.php >> /path/to/logfile.log
The permission issue can occur if you create the cron-log file manually using a command like touch while you are logged in as a different user and you add CRONs in the tab of another user(group) like www-data using: sudo crontab -u www-data -e. Then CRON daemon tries to write to the log file and fail, then tries to send the output as an email using Ubuntu's MTA and when it's not found, outputs "No MTA installed, discarding output".
To prevent this:
Create the file with proper permission.
Avoid creating the relevant CRON log file manually, add the log in CRON tab and let the log file get created automatically when the cron is run.
I've found another reason for user's crontab not running: the hostname is not present on the hosts file:
user#ubuntu:~$ cat /etc/hostname
ubuntu
Now the hosts file:
user#ubuntu:~$ cat /etc/hosts
127.0.0.1 localhost
# The following lines are desirable for IPv6 capable hosts
::1 ip6-localhost ip6-loopback
fe00::0 ip6-localnet
ff00::0 ip6-mcastprefix
ff02::1 ip6-allnodes
ff02::2 ip6-allrouters
ff02::3 ip6-allhosts
This is on a Ubuntu 14.04.3 LTS, the way to fix it is adding the hostname to the hosts file so it resembles something like this:
user#ubuntu:~$ cat /etc/hosts
127.0.0.1 ubuntu localhost
# The following lines are desirable for IPv6 capable hosts
::1 ip6-localhost ip6-loopback
fe00::0 ip6-localnet
ff00::0 ip6-mcastprefix
ff02::1 ip6-allnodes
ff02::2 ip6-allrouters
ff02::3 ip6-allhosts
For me, the solution was that the file cron was trying to run was in an encrypted directory, more specifcically a user diretory on /home/. Although the crontab was configured as root, because the script being run exisited in an encrypted user directory in /home/ cron could only read this directory when the user was actually logged in. To see if the directory is encrypted check if this directory exists:
/home/.ecryptfs/<yourusername>
if so then you have an encrypted home directory.
The fix for me was to move the script in to a non=encrypted directory and everythig worked fine.
As this is becoming a canonical for troubleshooting cron issues, allow me to add one specific but rather complex issue: If you are attempting to run a GUI program from cron, you are probably Doing It Wrong.
A common symptom is receiving error messages about DISPLAY being unset, or the cron job's process being unable to access the display.
In brief, this means that the program you are trying to run is attempting to render something on an X11 (or Wayland etc) display, and failing, because cron is not attached to a graphical environment, or in fact any kind of input/output facility at all, beyond being able to read and write files, and send email if the system is configured to allow that.
For the purposes of "I'm unable to run my graphical cron job", let's just point out in broad strokes three common scenarios for this problem.
Probably identify the case you are trying to implement, and search for related questions about that particular scenario to learn more, and find actual solutions with actual code.
If you are trying to develop an interactive program which communicates with a user, you want to rethink your approach. A common, but nontrivial, arrangement is to split the program in two: A back-end service which can run from cron, but which does not have any user-visible interactive facilities, and a front-end client which the user runs from their GUI when they want to communicate with the back-end service.
Probably your user client should simply be added to the user(s)' GUI startup script if it needs to be, or they want to, run automatically when they log in.
I suppose the back-end service could be started from cron, but if it requires a GUI to be useful, maybe start it from the X11 server's startup scripts instead; and if not, probably run it from a regular startup script (systemd these days, or /etc/rc.local or a similar system startup directory more traditionally).1
If you are trying to run a GUI program without interacting with a real user 2, you may be able to set up a "headless" X11 server 3 and run a cron job which starts up that server, runs your job, and quits.
Probably your job should simply run a suitable X11 server from cron (separate from any interactive X11 server which manages the actual physical display(s) and attached graphics card(s) and keyboard(s) available to the system), and pass it a configuration which runs the client(s) you want to run once it's up and running. (See also the next point for some practical considerations.)
You are running a computer for the sole purpose of displaying a specific application in a GUI, and you want to start that application when the computer is booted.
Probably your startup scripts should simply run the GUI (X11 or whatever) and hook into its startup script to also run the client program once the GUI is up and running. In other words, you don't need cron here; just configure the startup scripts to run the desktop GUI, and configure the desktop GUI to run your application as part of the (presumably automatic, guest?) login sequence.4
There are ways to run X11 programs on the system's primary display (DISPLAY=:0.0) but doing that from a cron job is often problematic, as that display is usually reserved for actual interactive use by the first user who logs in and starts a graphical desktop. On a single-user system, you might be able to live with the side effects if that user is also you, but this tends to have inconvenient consequences and scale very poorly.
An additional complication is deciding which user to run the cron job as. A shared system resource like a back-end service can and probably should be run by root (though ideally have a dedicated system account which it switches into once it has acquired access to any privileged resources it needs) but anything involving a GUI should definitely not be run as root at any point.
A related, but distinct problem is to interact in any meaningful way with the user. If you can identify the user's active session (to the extent that this is even well-defined in the first place), how do you grab their attention without interfering with whatever else they are in the middle of? But more fundamentally, how do you even find them? If they are not logged in at all, what do you do then? If they are, how do you determine that they are active and available? If they are logged in more than once, which terminal are they using, and is it safe to interrupt that session? Similarly, if they are logged in to the GUI, they might miss a window you spring up on the local console, if they are actually logged in remotely via VNC or a remote X11 server.
As a further aside: On dedicated servers (web hosting services, supercomputing clusters, etc) you might even be breaking the terms of service of the hosting company or institution if you install an interactive graphical desktop you can connect to from the outside world, or even at all.
1
The #reboot hook in cron is a convenience for regular users who don't have any other facility for running something when the system comes up, but it's just inconvenient and obscure to hide something there if you are root anyway and have complete control over the system. Use the system facilities to launch system services.
2
A common use case is running a web browser which needs to run a full GUI client, but which is being controlled programmatically and which doesn't really need to display anything anywhere, for example to scrape sites which use Javascript and thus require a full graphical browser to render the information you want to extract.
Another is poorly designed scientific or office software which was not written for batch use, and thus requires a GUI even when you just want to run a batch job and then immediately quit without any actual need to display anything anywhere.
(In the latter case, probably review the documentation to check if there isn't a --batch or --noninteractive or --headless or --script or --eval option or similar to run the tool without the GUI, or perhaps a separate utility for noninteractive use.)
3
Xvfb is the de facto standard solution; it runs a "virtual framebuffer" where the computer can spit out pixels as if to a display, but which isn't actually connected to any display hardware.
4
There are several options here.
The absolutely simplest is to set up the system to automatically log in a specific user at startup without a password prompt, and configure that user's desktop environment (Gnome or KDE or XFCE or what have you) to run your script from its "Startup Items" or "Login Actions" or "Autostart" or whatever the facility might be called. If you need more control over the environment, maybe run bare X11 without a desktop environment or window manager at all, and just run your script instead. Or in some cases, maybe replace the X11 login manager ("greeter") with something custom built.
The X11 stack is quite modular, and there are several hooks in various layers where you could run a script either as part of a standard startup process, or one which completely replaces a standard layer. These things tend to differ somewhat between distros and implementations, and over time, so this answer is necessarily vague and incomplete around these matters. Again, probably try to find an existing question about how to do things for your specific platform (Ubuntu, Raspbian, Gnome, KDE, what?) and scenario. For simple scenarios, perhaps see Ubuntu - run bash script on startup with visible terminal
I experienced same problem where crons are not running.
We fixed by changing permissions and owner by
Crons made root owner as we had mentioned in crontab AND
Cronjobs 644 permission given
There is already a lot of answers, but none of them helped me so I'll add mine here in case it's useful for somebody else.
In my situation, my cronjobs were working find until there was a power shortage that cut the power to my Raspberry Pi. Cron got corrupted. I think it was running a long python script exactly when the shortage happened. Nothing in the main answer above worked for me. The solution was however quite simple. I just had to force reinstallation of cron with:
sudo apt-get --reinstall install cron
It work right away after this.
Copying my answer for a duplicated question here.
cron may not know where to find the Python interpreter because it doesn't share your user account's environment variables.
There are 3 solutions to this:
If Python is at /usr/bin/python, you can change the cron job to use an absolute path: /usr/bin/python /srv/www/live/CronJobs/daily.py
Alternatively you can also add a PATH value to the crontab with PATH=/usr/bin.
Another solution would be to specify an interpreter in the script file, make it executable, and call the script itself in your crontab:
a. Put shebang at the top of your python file: #!/usr/bin/python.
b. Set it to executable: $ chmod +x /srv/www/live/CronJobs/daily.py
c. Put it in crontab: /srv/www/live/CronJobs/daily.py
Adjust the path to the Python interpreter if it's different on your system.
Reference
CRON uses a different TIMEZONE
A very common issue is: cron time settings may is different than your. In particular, the timezone could be not be the same:
# Notice that tasks will be started based on the cron's system
# daemon's notion of time and timezones.
You can run:
* * * * * echo $(date) >> /tmp/test.txt
This should generate a file like:
# cat test.txt
Sun 03 Apr 2022 09:02:01 AM UTC
Sun 03 Apr 2022 09:03:01 AM UTC
Sun 03 Apr 2022 09:04:01 AM UTC
Sun 03 Apr 2022 09:05:01 AM UTC
Sun 03 Apr 2022 09:06:01 AM UTC
If you are using a TZ other than UTC, you can try:
timedatectl set-timezone America/Sao_Paulo
replace America/Sao_Paulo according to you settings.
I'm not sure if it is actually necessary, but you can run:
sudo systemctl restart cron.service
After that, cron works as I expected:
# cat test.txt
Sun 03 Apr 2022 09:02:01 AM UTC
Sun 03 Apr 2022 09:03:01 AM UTC
Sun 03 Apr 2022 09:04:01 AM UTC
Sun 03 Apr 2022 09:05:01 AM UTC
Sun 03 Apr 2022 09:06:01 AM UTC
Sun 03 Apr 2022 09:07:01 AM UTC
Sun 03 Apr 2022 09:08:01 AM UTC
Sun 03 Apr 2022 09:09:01 AM UTC
Sun 03 Apr 2022 09:10:01 AM UTC
Sun 03 Apr 2022 06:11:01 AM -03
Sun 03 Apr 2022 06:12:01 AM -03
Sun 03 Apr 2022 06:13:01 AM -03
Sun 03 Apr 2022 06:14:01 AM -03
Try
service cron start
or
systemctl start cron
In my case I was trying to run cron locally.
I checked status:
service cron status
It showed me:
* cron is not running
Then I simply started the service:
service cron start
Sometimes the command that cron needs to run is in a directory where cron has no access, typically on systems where users' home directories' permissions are 700 and the command is in that directory.
Although answer has been accepted for this question, I will like to add what worked for me.
it's a good idea to quote the URL, if it contains a query it may not work without everything being quoted.
DONT FORGET TO PUT YOUR URL WHICH CONTAINS "?, =, #, %" IN A QUOTE.
Example.
https://paystack.com/indexphp?docs/api/#transaction-charge-authorization&date=today
should be in a quote like so
"https://paystack.com/indexphp?docs/api/#transaction-charge-authorization&date=today"

Cronjob is running but not executing python script [duplicate]

I have set up a cronjob for root user in ubuntu environment as follows by typing crontab -e
34 11 * * * sh /srv/www/live/CronJobs/daily.sh
0 08 * * 2 sh /srv/www/live/CronJobs/weekly.sh
0 08 1 * * sh /srv/www/live/CronJobs/monthly.sh
But the cronjob does not run. I have tried checking if the cronjob is running using pgrep cron and that gives process id 3033. The shell script calls a python file and is used to send an email. Running the python file is ok. There's no error in it but the cron doesn't run. The daily.sh file has the following code in it.
python /srv/www/live/CronJobs/daily.py
python /srv/www/live/CronJobs/notification_email.py
python /srv/www/live/CronJobs/log_kpi.py
WTF?! My cronjob doesn't run?!
Here's a checklist guide to debug not running cronjobs:
Is the Cron daemon running?
Run ps ax | grep cron and look for cron.
Debian: service cron start or service cron restart
Is cron working?
* * * * * /bin/echo "cron works" >> /tmp/file
Syntax correct? See below.
You obviously need to have write access to the file you are redirecting the output to. A unique file name in /tmp which does not currently exist should always be writable.
Probably also add 2>&1 to include standard error as well as standard output, or separately output standard error to another file with 2>>/tmp/errors
Is the command working standalone?
Check if the script has an error, by doing a dry run on the CLI
When testing your command, test as the user whose crontab you are editing, which might not be your login or root
Can cron run your job?
Check /var/log/cron.log or /var/log/messages for errors.
Ubuntu: grep CRON /var/log/syslog
Redhat: /var/log/cron
Check permissions
Set executable flag on the command: chmod +x /var/www/app/cron/do-stuff.php
If you redirect the output of your command to a file, verify you have permission to write to that file/directory
Check paths
Check she-bangs / hashbangs line
Do not rely on environment variables like PATH, as their value will likely not be the same under cron as under an interactive session. See How to get CRON to call in the correct PATHs
Don't suppress output while debugging
Commonly used is this suppression: 30 1 * * * command > /dev/null 2>&1
Re-enable the standard output or standard error message output by removing >/dev/null 2>&1 altogether; or perhaps redirect to a file in a location where you have write access: >>cron.out 2>&1 will append standard output and standard error to cron.out in the invoking user's home directory.
If you don't redirect output from a cron job, the daemon will try to send you any output or error messages by email. Check your inbox (maybe simply more $MAIL if you don't have a mail client). If mail is not available, maybe check for a file named dead.letter in your home directory, or system log entries saying that the output was discarded. Especially in the latter case, probably edit the job to add redirection to a file, then wait for the job to run, and examine the log file for error messages or other useful feedback.
If you are trying to figure out why something failed, the error messages will be visible in this file. Read it and understand it.
Still not working? Yikes!
Raise the cron debug level
Debian
in /etc/default/cron
set EXTRA_OPTS="-L 2"
service cron restart
tail -f /var/log/syslog to see the scripts executed
Ubuntu
in /etc/rsyslog.d/50-default.conf
add or comment out line cron.* /var/log/cron.log
reload logger sudo /etc/init.d/rsyslog restart
re-run cron
open /var/log/cron.log and look for detailed error output
Reminder: deactivate log level, when you are done with debugging
Run cron and check log files again
Cronjob Syntax
# Minute Hour Day of Month Month Day of Week User Command
# (0-59) (0-23) (1-31) (1-12 or Jan-Dec) (0-6 or Sun-Sat)
0 2 * * * root /usr/bin/find
This syntax is only correct for the root user. Regular user crontab syntax doesn't have the User field (regular users aren't allowed to run code as any other user);
# Minute Hour Day of Month Month Day of Week Command
# (0-59) (0-23) (1-31) (1-12 or Jan-Dec) (0-6 or Sun-Sat)
0 2 * * * /usr/bin/find
Crontab Commands
crontab -l
Lists all the user's cron tasks.
crontab -e, for a specific user: crontab -e -u agentsmith
Starts edit session of your crontab file.
When you exit the editor, the modified crontab is installed automatically.
crontab -r
Removes your crontab entry from the cron spooler, but not from crontab file.
Another reason crontab will fail: Special handling of the % character.
From the manual file:
The entire command portion of the line, up to a newline or a
"%" character, will be executed by /bin/sh or by the shell specified
in the SHELL variable of the cronfile. A "%" character in the
command, unless escaped with a backslash (\), will be changed into
newline characters, and all data after the first % will be sent to
the command as standard input.
In my particular case, I was using date --date="7 days ago" "+%Y-%m-%d" to produce parameters to my script, and it was failing silently. I finally found out what was going on when I checked syslog and saw my command was truncated at the % symbol. You need to escape it like this:
date --date="7 days ago" "+\%Y-\%m-\%d"
See here for more details:
http://www.ducea.com/2008/11/12/using-the-character-in-crontab-entries/
Finally I found the solution. Following is the solution:-
Never use relative path in python scripts to be executed via crontab.
I did something like this instead:-
import os
import sys
import time, datetime
CLASS_PATH = '/srv/www/live/mainapp/classes'
SETTINGS_PATH = '/srv/www/live/foodtrade'
sys.path.insert(0, CLASS_PATH)
sys.path.insert(1,SETTINGS_PATH)
import other_py_files
Never supress the crontab code instead use mailserver and check the mail for the user. That gives clearer insights of what is going.
I want to add 2 points that I learned:
Cron config files put in /etc/cron.d/ should not contain a dot (.). Otherwise, it won't be read by cron.
If the user running your command is not in /etc/shadow. It won't be allowed to schedule cron.
Refs:
http://manpages.ubuntu.com/manpages/xenial/en/man8/cron.8.html
https://help.ubuntu.com/community/CronHowto
To add another point, a file in /etc/cron.d must contain an empty new line at the end. This is likely related to the response by Luciano which specifies that:
The entire command portion of the line, up to a newline or a "%"
character, will be executed
I found useful debugging information on an Ubuntu 16.04 server by running:
systemctl status cron.service
In my case I was kindly informed I had left a comment '#' off of a remark line:
Aug 18 19:12:01 is-feb19 cron[14307]: Error: bad minute; while reading /etc/crontab
Aug 18 19:12:01 is-feb19 cron[14307]: (*system*) ERROR (Syntax error, this crontab file will be ignored)
It might also be a timezone problem.
Cron uses the local time.
Run the command timedatectl to see the machine time and make sure that your crontab is in this same timezone.
https://askubuntu.com/a/536489/1043751
I had a similar problem to the link below.
similar to my problem
my original post
My Issue
My issue was that cron / crontab wouldn't execute my bash script. that bash script executed a python script.
original bash file
#!/bin/bash
python /home/frosty/code/test_scripts/test.py
python file (test.py)
from datetime import datetime
def main():
dt_now = datetime.now()
string_now = dt_now.strftime('%Y-%m-%d %H:%M:%S.%f')
with open('./text_file.txt', 'a') as f:
f.write(f'wrote at {string_now}\n')
return None
if __name__ == '__main__':
main()
the error I was getting
File "/home/frosty/code/test_scripts/test.py", line 7
string_to_write = f'wrote at {string_now}\n'
^
SyntaxError: invalid syntax
this error didn't make sense because the code executed without error from the bash file and the python file.
** Note -> ensure in the crontab -e file you don't suppress the output. I sent the output to a file by adding >>/path/to/cron/output/file.log 2>&1 after the command. below is my crontab -e entry
*/5 * * * * /home/frosty/code/test_scripts/echo_message_sh >>/home/frosty/code/test_scripts/cron_out.log 2>&1
the issue
cron was using the wrong python interpreter, probably python 2 from the syntax error.
how I solved the problem
I changed my bash file to the following
#!/bin/bash
conda_shell=/home/frosty/anaconda3/etc/profile.d/conda.sh
conda_env=base
source ${conda_shell}
conda activate ${conda_env}
python /home/frosty/code/test_scripts/test.py
And I changed my python file to the following
from datetime import datetime
def main():
dt_now = datetime.now()
string_now = dt_now.strftime('%Y-%m-%d %H:%M:%S.%f')
string_file = '/home/frosty/code/test_scripts/text_file.txt'
string_to_write = 'wrote at {}\n'.format(string_now)
with open(string_file, 'a') as f:
f.write(string_to_write)
return None
if __name__ == '__main__':
main()
No MTA installed, discarding output
I had a similar problem with a PHP file executed as a CRON job.
When I manually execute the file it works, but not with CRON tab.
I got the output message: "No MTA installed, discarding output"
Postfix is the default Mail Transfer Agent (MTA) in Ubuntu and can be installed it using
sudo apt-get install postfix
But this same message can be also output when you add a log file as below and it does not have proper write permission to /path/to/logfile.log
/path/to/php -f /path/to/script.php >> /path/to/logfile.log
The permission issue can occur if you create the cron-log file manually using a command like touch while you are logged in as a different user and you add CRONs in the tab of another user(group) like www-data using: sudo crontab -u www-data -e. Then CRON daemon tries to write to the log file and fail, then tries to send the output as an email using Ubuntu's MTA and when it's not found, outputs "No MTA installed, discarding output".
To prevent this:
Create the file with proper permission.
Avoid creating the relevant CRON log file manually, add the log in CRON tab and let the log file get created automatically when the cron is run.
I've found another reason for user's crontab not running: the hostname is not present on the hosts file:
user#ubuntu:~$ cat /etc/hostname
ubuntu
Now the hosts file:
user#ubuntu:~$ cat /etc/hosts
127.0.0.1 localhost
# The following lines are desirable for IPv6 capable hosts
::1 ip6-localhost ip6-loopback
fe00::0 ip6-localnet
ff00::0 ip6-mcastprefix
ff02::1 ip6-allnodes
ff02::2 ip6-allrouters
ff02::3 ip6-allhosts
This is on a Ubuntu 14.04.3 LTS, the way to fix it is adding the hostname to the hosts file so it resembles something like this:
user#ubuntu:~$ cat /etc/hosts
127.0.0.1 ubuntu localhost
# The following lines are desirable for IPv6 capable hosts
::1 ip6-localhost ip6-loopback
fe00::0 ip6-localnet
ff00::0 ip6-mcastprefix
ff02::1 ip6-allnodes
ff02::2 ip6-allrouters
ff02::3 ip6-allhosts
For me, the solution was that the file cron was trying to run was in an encrypted directory, more specifcically a user diretory on /home/. Although the crontab was configured as root, because the script being run exisited in an encrypted user directory in /home/ cron could only read this directory when the user was actually logged in. To see if the directory is encrypted check if this directory exists:
/home/.ecryptfs/<yourusername>
if so then you have an encrypted home directory.
The fix for me was to move the script in to a non=encrypted directory and everythig worked fine.
As this is becoming a canonical for troubleshooting cron issues, allow me to add one specific but rather complex issue: If you are attempting to run a GUI program from cron, you are probably Doing It Wrong.
A common symptom is receiving error messages about DISPLAY being unset, or the cron job's process being unable to access the display.
In brief, this means that the program you are trying to run is attempting to render something on an X11 (or Wayland etc) display, and failing, because cron is not attached to a graphical environment, or in fact any kind of input/output facility at all, beyond being able to read and write files, and send email if the system is configured to allow that.
For the purposes of "I'm unable to run my graphical cron job", let's just point out in broad strokes three common scenarios for this problem.
Probably identify the case you are trying to implement, and search for related questions about that particular scenario to learn more, and find actual solutions with actual code.
If you are trying to develop an interactive program which communicates with a user, you want to rethink your approach. A common, but nontrivial, arrangement is to split the program in two: A back-end service which can run from cron, but which does not have any user-visible interactive facilities, and a front-end client which the user runs from their GUI when they want to communicate with the back-end service.
Probably your user client should simply be added to the user(s)' GUI startup script if it needs to be, or they want to, run automatically when they log in.
I suppose the back-end service could be started from cron, but if it requires a GUI to be useful, maybe start it from the X11 server's startup scripts instead; and if not, probably run it from a regular startup script (systemd these days, or /etc/rc.local or a similar system startup directory more traditionally).1
If you are trying to run a GUI program without interacting with a real user 2, you may be able to set up a "headless" X11 server 3 and run a cron job which starts up that server, runs your job, and quits.
Probably your job should simply run a suitable X11 server from cron (separate from any interactive X11 server which manages the actual physical display(s) and attached graphics card(s) and keyboard(s) available to the system), and pass it a configuration which runs the client(s) you want to run once it's up and running. (See also the next point for some practical considerations.)
You are running a computer for the sole purpose of displaying a specific application in a GUI, and you want to start that application when the computer is booted.
Probably your startup scripts should simply run the GUI (X11 or whatever) and hook into its startup script to also run the client program once the GUI is up and running. In other words, you don't need cron here; just configure the startup scripts to run the desktop GUI, and configure the desktop GUI to run your application as part of the (presumably automatic, guest?) login sequence.4
There are ways to run X11 programs on the system's primary display (DISPLAY=:0.0) but doing that from a cron job is often problematic, as that display is usually reserved for actual interactive use by the first user who logs in and starts a graphical desktop. On a single-user system, you might be able to live with the side effects if that user is also you, but this tends to have inconvenient consequences and scale very poorly.
An additional complication is deciding which user to run the cron job as. A shared system resource like a back-end service can and probably should be run by root (though ideally have a dedicated system account which it switches into once it has acquired access to any privileged resources it needs) but anything involving a GUI should definitely not be run as root at any point.
A related, but distinct problem is to interact in any meaningful way with the user. If you can identify the user's active session (to the extent that this is even well-defined in the first place), how do you grab their attention without interfering with whatever else they are in the middle of? But more fundamentally, how do you even find them? If they are not logged in at all, what do you do then? If they are, how do you determine that they are active and available? If they are logged in more than once, which terminal are they using, and is it safe to interrupt that session? Similarly, if they are logged in to the GUI, they might miss a window you spring up on the local console, if they are actually logged in remotely via VNC or a remote X11 server.
As a further aside: On dedicated servers (web hosting services, supercomputing clusters, etc) you might even be breaking the terms of service of the hosting company or institution if you install an interactive graphical desktop you can connect to from the outside world, or even at all.
1
The #reboot hook in cron is a convenience for regular users who don't have any other facility for running something when the system comes up, but it's just inconvenient and obscure to hide something there if you are root anyway and have complete control over the system. Use the system facilities to launch system services.
2
A common use case is running a web browser which needs to run a full GUI client, but which is being controlled programmatically and which doesn't really need to display anything anywhere, for example to scrape sites which use Javascript and thus require a full graphical browser to render the information you want to extract.
Another is poorly designed scientific or office software which was not written for batch use, and thus requires a GUI even when you just want to run a batch job and then immediately quit without any actual need to display anything anywhere.
(In the latter case, probably review the documentation to check if there isn't a --batch or --noninteractive or --headless or --script or --eval option or similar to run the tool without the GUI, or perhaps a separate utility for noninteractive use.)
3
Xvfb is the de facto standard solution; it runs a "virtual framebuffer" where the computer can spit out pixels as if to a display, but which isn't actually connected to any display hardware.
4
There are several options here.
The absolutely simplest is to set up the system to automatically log in a specific user at startup without a password prompt, and configure that user's desktop environment (Gnome or KDE or XFCE or what have you) to run your script from its "Startup Items" or "Login Actions" or "Autostart" or whatever the facility might be called. If you need more control over the environment, maybe run bare X11 without a desktop environment or window manager at all, and just run your script instead. Or in some cases, maybe replace the X11 login manager ("greeter") with something custom built.
The X11 stack is quite modular, and there are several hooks in various layers where you could run a script either as part of a standard startup process, or one which completely replaces a standard layer. These things tend to differ somewhat between distros and implementations, and over time, so this answer is necessarily vague and incomplete around these matters. Again, probably try to find an existing question about how to do things for your specific platform (Ubuntu, Raspbian, Gnome, KDE, what?) and scenario. For simple scenarios, perhaps see Ubuntu - run bash script on startup with visible terminal
I experienced same problem where crons are not running.
We fixed by changing permissions and owner by
Crons made root owner as we had mentioned in crontab AND
Cronjobs 644 permission given
There is already a lot of answers, but none of them helped me so I'll add mine here in case it's useful for somebody else.
In my situation, my cronjobs were working find until there was a power shortage that cut the power to my Raspberry Pi. Cron got corrupted. I think it was running a long python script exactly when the shortage happened. Nothing in the main answer above worked for me. The solution was however quite simple. I just had to force reinstallation of cron with:
sudo apt-get --reinstall install cron
It work right away after this.
Copying my answer for a duplicated question here.
cron may not know where to find the Python interpreter because it doesn't share your user account's environment variables.
There are 3 solutions to this:
If Python is at /usr/bin/python, you can change the cron job to use an absolute path: /usr/bin/python /srv/www/live/CronJobs/daily.py
Alternatively you can also add a PATH value to the crontab with PATH=/usr/bin.
Another solution would be to specify an interpreter in the script file, make it executable, and call the script itself in your crontab:
a. Put shebang at the top of your python file: #!/usr/bin/python.
b. Set it to executable: $ chmod +x /srv/www/live/CronJobs/daily.py
c. Put it in crontab: /srv/www/live/CronJobs/daily.py
Adjust the path to the Python interpreter if it's different on your system.
Reference
CRON uses a different TIMEZONE
A very common issue is: cron time settings may is different than your. In particular, the timezone could be not be the same:
# Notice that tasks will be started based on the cron's system
# daemon's notion of time and timezones.
You can run:
* * * * * echo $(date) >> /tmp/test.txt
This should generate a file like:
# cat test.txt
Sun 03 Apr 2022 09:02:01 AM UTC
Sun 03 Apr 2022 09:03:01 AM UTC
Sun 03 Apr 2022 09:04:01 AM UTC
Sun 03 Apr 2022 09:05:01 AM UTC
Sun 03 Apr 2022 09:06:01 AM UTC
If you are using a TZ other than UTC, you can try:
timedatectl set-timezone America/Sao_Paulo
replace America/Sao_Paulo according to you settings.
I'm not sure if it is actually necessary, but you can run:
sudo systemctl restart cron.service
After that, cron works as I expected:
# cat test.txt
Sun 03 Apr 2022 09:02:01 AM UTC
Sun 03 Apr 2022 09:03:01 AM UTC
Sun 03 Apr 2022 09:04:01 AM UTC
Sun 03 Apr 2022 09:05:01 AM UTC
Sun 03 Apr 2022 09:06:01 AM UTC
Sun 03 Apr 2022 09:07:01 AM UTC
Sun 03 Apr 2022 09:08:01 AM UTC
Sun 03 Apr 2022 09:09:01 AM UTC
Sun 03 Apr 2022 09:10:01 AM UTC
Sun 03 Apr 2022 06:11:01 AM -03
Sun 03 Apr 2022 06:12:01 AM -03
Sun 03 Apr 2022 06:13:01 AM -03
Sun 03 Apr 2022 06:14:01 AM -03
Try
service cron start
or
systemctl start cron
In my case I was trying to run cron locally.
I checked status:
service cron status
It showed me:
* cron is not running
Then I simply started the service:
service cron start
Sometimes the command that cron needs to run is in a directory where cron has no access, typically on systems where users' home directories' permissions are 700 and the command is in that directory.
Although answer has been accepted for this question, I will like to add what worked for me.
it's a good idea to quote the URL, if it contains a query it may not work without everything being quoted.
DONT FORGET TO PUT YOUR URL WHICH CONTAINS "?, =, #, %" IN A QUOTE.
Example.
https://paystack.com/indexphp?docs/api/#transaction-charge-authorization&date=today
should be in a quote like so
"https://paystack.com/indexphp?docs/api/#transaction-charge-authorization&date=today"

User input during a systemctl/service call for CentOS?

Lets say I have a service (like rsyslog) that when I stop it, I want to log the reasoning behind someone shutting it down. Expressing this simply would be
systemctl start rsyslog
systemctl stop rsyslog
(Begin the prompt as to why a user is doing this after shutting down rsyslog)
#!/bin/bash
echo "you are shutting down the syslog service. Please state your name and reason."
read -p "[name?]" name && read -p "[reason?]" reason
logger $name:$reason
Modifying the Unit Files (located in /usr/lib/systemd/system/rsyslog.service), to include an ExecStop of the path to the script, I am able to run the above script. I know that the script is working as checking the log messages shows a :, the nonvariable portion that was passed to logger.
I need to be able to have this script operate like said shell script the instant someone attempts to shutdown the logging service. This means that echo commands are shown on the terminal and variables can be recorded using the read command.
This may be similar to this persons question, but I can not understand it.
Thanks to Mark Stosberg for sharing information about the systemd-ask-password command that takes user input during a systemctl call.
For those unaware, the systemd-ask-password command is a password prompt that's available for all machines sponsoring the systemd service. A unique feature of this command is the fact it can allow for user input during a systemctl/service call. Knowing this, one can prompt for as much data as they like which integrates perfectly into standard bash scripts, allowing for the interaction that may or may not be needed during a call.
Here is an example script:
#!/bin/bash
date=$(/bin/date)
echo "Rsyslog has been turned off at $date. Below is the name and reason"
name=`systemd-ask-password --echo --no-tty "name:"`
reason=`systemd-ask-password --echo --no-tty "reason for shutting down rsyslog:"`
echo LOG:$name:$reason:ENDLOG >>/var/log/messages
You must make sure that when initializing any of your services you make changes to the unit files located in /usr/lib/systemd/system/ with an ExecStart, ExecStop, and so forth under the [Service] tag to equal the path to your scripts. You can find what other options you can here as well as some syntax, and tie in with the unit file as needed.

How do I run a bash script (that resides on a remote server) in windows task scheduler?

SOLVED
Scenario: I am a beginner in bash script, windows task scheduler and such. I am able to run a local bash script in my Windows Task Scheduler successfully.
Problem: I need to do this on many computers, thus I think storing just 1 copy of the bash script on a remote server may be of help. What my Task Scheduler needs to do is just to run the script and output a log. However, I can't get the correct syntax for the argument.
The below is what I have currently:
Program/Script: C:\cygwin64\bin\bash.exe
Argument (works successfully):
-l -c "ssh -p 222 ME#ME.com "httpdocs/bashscript.sh" >> /cygdrive/c/Users/ME/Desktop/`date +%Y%m%d`.log 2>&1"
Start in: C:\cygwin64\bin
Also had to make sure that the user account under Properties in Task Scheduler is correct, as mine was incorrect before. And need key authentication for ME#ME.com too.
For the password issue, you really should use ssh keys. I think your command would simply be ssh -p 222 ME#ME.com:.... I.e., just get rid of the --rsh stuff. – chrisaycock

Best practice to run Linux service as a different user

Services default to starting as root at boot time on my RHEL box. If I recall correctly, the same is true for other Linux distros which use the init scripts in /etc/init.d.
What do you think is the best way to instead have the processes run as a (static) user of my choosing?
The only method I'd arrived at was to use something like:
su my_user -c 'daemon my_cmd &>/dev/null &'
But this seems a bit untidy...
Is there some bit of magic tucked away that provides an easy mechanism to automatically start services as other, non-root users?
EDIT: I should have said that the processes I'm starting in this instance are either Python scripts or Java programs. I'd rather not write a native wrapper around them, so unfortunately I'm unable to call setuid() as Black suggests.
On Debian we use the start-stop-daemon utility, which handles pid-files, changing the user, putting the daemon into background and much more.
I'm not familiar with RedHat, but the daemon utility that you are already using (which is defined in /etc/init.d/functions, btw.) is mentioned everywhere as the equivalent to start-stop-daemon, so either it can also change the uid of your program, or the way you do it is already the correct one.
If you look around the net, there are several ready-made wrappers that you can use. Some may even be already packaged in RedHat. Have a look at daemonize, for example.
After looking at all the suggestions here, I've discovered a few things which I hope will be useful to others in my position:
hop is right to point me back
at /etc/init.d/functions: the
daemon function already allows you
to set an alternate user:
daemon --user=my_user my_cmd &>/dev/null &
This is implemented by wrapping the
process invocation with runuser -
more on this later.
Jonathan Leffler is right:
there is setuid in Python:
import os
os.setuid(501) # UID of my_user is 501
I still don't think you can setuid
from inside a JVM, however.
Neither su nor runuser
gracefully handle the case where you
ask to run a command as the user you
already are. E.g.:
[my_user#my_host]$ id
uid=500(my_user) gid=500(my_user) groups=500(my_user)
[my_user#my_host]$ su my_user -c "id"
Password: # don't want to be prompted!
uid=500(my_user) gid=500(my_user) groups=500(my_user)
To workaround that behaviour of su and runuser, I've changed my init script to something like:
if [[ "$USER" == "my_user" ]]
then
daemon my_cmd &>/dev/null &
else
daemon --user=my_user my_cmd &>/dev/null &
fi
Thanks all for your help!
Some daemons (e.g. apache) do this by themselves by calling setuid()
You could use the setuid-file flag to run the process as a different user.
Of course, the solution you mentioned works as well.
If you intend to write your own daemon, then I recommend calling setuid().
This way, your process can
Make use of its root privileges (e.g. open log files, create pid files).
Drop its root privileges at a certain point during startup.
Just to add some other things to watch out for:
Sudo in a init.d script is no good since it needs a tty ("sudo: sorry, you must have a tty to run sudo")
If you are daemonizing a java application, you might want to consider Java Service Wrapper (which provides a mechanism for setting the user id)
Another alternative could be su --session-command=[cmd] [user]
on a CENTOS (Red Hat) virtual machine for svn server:
edited /etc/init.d/svnserver
to change the pid to something that svn can write:
pidfile=${PIDFILE-/home/svn/run/svnserve.pid}
and added option --user=svn:
daemon --pidfile=${pidfile} --user=svn $exec $args
The original pidfile was /var/run/svnserve.pid. The daemon did not start becaseu only root could write there.
These all work:
/etc/init.d/svnserve start
/etc/init.d/svnserve stop
/etc/init.d/svnserve restart
Some things to watch out for:
As you mentioned, su will prompt for a password if you are already the target user
Similarly, setuid(2) will fail if you are already the target user (on some OSs)
setuid(2) does not install privileges or resource controls defined in /etc/limits.conf (Linux) or /etc/user_attr (Solaris)
If you go the setgid(2)/setuid(2) route, don't forget to call initgroups(3) -- more on this here
I generally use /sbin/su to switch to the appropriate user before starting daemons.
Why not try the following in the init script:
setuid $USER application_name
It worked for me.
I needed to run a Spring .jar application as a service, and found a simple way to run this as a specific user:
I changed the owner and group of my jar file to the user I wanted to run as.
Then symlinked this jar in init.d and started the service.
So:
#chown myuser:myuser /var/lib/jenkins/workspace/springApp/target/springApp-1.0.jar
#ln -s /var/lib/jenkins/workspace/springApp/target/springApp-1.0.jar /etc/init.d/springApp
#service springApp start
#ps aux | grep java
myuser 9970 5.0 9.9 4071348 386132 ? Sl 09:38 0:21 /bin/java -Dsun.misc.URLClassPath.disableJarChecking=true -jar /var/lib/jenkins/workspace/springApp/target/springApp-1.0.jar

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