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I found Terminal command to change desktop wallpaper:
gsettings set org.gnome.desktop.background picture-uri file:///path/to/your/image.png
but this command not working in cron and other desktops like Mate. pgrep gnome-session approach shows nothing for me.
You can use dconf to change background. Here is example of simple bash script:
#!/bin/bash
WP="$(find ~+ -type f -exec mimetype {} + 2>/dev/null | awk -F': +' '{ if ($2 ~ /^image\//) print $1 }' | sort -R | tail -30 | shuf -n 1)"
dconf write /org/mate/desktop/background/picture-filename "'${WP}'"
You can find distro-specific key using GUI app - dconf-editor
But to use this script in CRON you need to set session environment variables. Command pgrep gnome-session doesn't work in Mint and other not Gnome desktops. To solve this problem you need to save environment variables of specific user by running command at system startup:
env > ~/cronenv && sed -i '/%s/d' ~/cronenv
now you have cronenv file (without substitutional vars - %s) in users home dir. Just restore them back in cron before running dconf:
*/1 7-21 * * * cd ~/Pictures && env $(cat ~/cronenv | xargs) /path/to/first/script
Use crontab -e to edit cron jobs for current user. All works fine!
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We want to extend our backup system to include a monthly end backup. It will be performed the last Sunday in the month but code below is so that I can see it works today on a smaller scale.
Started with (which works)
0 12 * * 0 sudo tar -cpzf /media/BackupDisk/wwwJUNEbackup.tar.gz /var/www
Have trawled the internet and come up with this code, tested in script file
if [ $(date +%d -d '+7 days') -le '8' ] ; then
echo "Yes"
else
echo "No"
fi
(For reference this says - if today's date + 7 days is less then or equal to 8 then YES else NO)
But when I try to include into the Sudo's crontrab
26 17 * 6 5 [ $(date +%d -d '+7 days') -lt '8' ] && sudo tar -cpzf /media/BackupDisk/wwwJUNEbackup.tar.gz /var/www
I get a nothing.
What am I doing wrong?
Create trivial script last_weekday_in_month.sh and use it in your crontab entry.
You use syntax far beyond basic shell => IMHO it is better to move it to trivial script with specific shell enforced via #!/...
12 * * 0 /path/last_weekday_in.month.sh && sudo tar -cpzf /media/BackupDisk/wwwJUNEbackup.tar.gz /var/www
last_weekday_in.month.sh
#!/bin/bash
if [ $(date +%d -d '+7 days') -lt '8' ] ; then
exit 0
else
exit 1
fi
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on the Linux (CentOS 6), i want to kill process containing "pkgacc" in its command (so no exact command but just partial match) if it is running more than 36 hours.
There is related question: How do you kill all Linux processes that are older than a certain age? but any of the solutions provided do not work for me.
when executing:
if [[ "$(uname)" = "Linux" ]];then killall --older-than 1h someprocessname;fi
It just return killall usage page on how to use killall, in its manual page there is no mention about "--older-than" switch.
It is infinitely easier to invoke a program in a wrapper like timeout from GNU coreutils than to go hunting for them after the fact. In particular because timeout owns its process, there is no ambiguity that it kills the right process. Thus
timeout 36h pkgaccess --pkg_option --another_option package_name
where I made up the names and options for the pkgaccess command since you didn't give them. This process will run no longer than 36 hours.
I think you could do something like
ps -eo pid,cmd,etime
then you could parse the output with grep searching for you process,
something like that:
ps -eo pid,cmd,etime | grep pkgacc
you will have some output with one or more result, the last column from the output must be the time of running process, so one more little bash effort
and you could check if the time is greater than 36 hours.
#!/bin/bash
FOO=$(ps -eo pid,cmd,etime | grep -m 1 pkgacc | awk '{ print $1" "$3 }'| sed -e 's/\://g')
IFS=' ' read -r -a array <<< "$FOO"
if [ "${array[1]}" -gt "360000" ]; then
echo "kill the process: ${array[0]}"
else
echo "process was not found or time less than 36 hours"
fi
I think that could solve part of your problem,
see that I do not explicit kill the process but just indicate
what it is. You could improve the idea from that.
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I thought I could just use this related question: How Do I write a bash script to restart a process if it dies. #lhunath had a great answer and told me everything I might do about it was wrong, but I'm restarting a daemon process and if I'm hoping there's something I can do in a single script that works.
my process starts with a kick off script that shows the startup log, but then quits and leaves the process running off the shell:
>sudo ./start
R CMD Rserve --RS-conf /var/FastRWeb/code/rserve.conf --vanilla --no-save
...
Loading required package: FastRWeb
FastRWeb: TRUE
Loading data...
Rserv started in daemon mode.
>
The process is up and running,
ps -ale | grep Rserve
1 S 33 16534 1 0 80 0 - 60022 poll_s ? 00:00:00 Rserve
Is there a simple way to wrap or call the 'start' script from bash and restart when the process dies or is this a case where PID files are actually called for?
Dang - question got closed even after pointing to a very similar question that was not closed on stackoverflow. you guys suck
A very simple way to monitor the program is to use cron: check every minute (or so) if the program still is alive, ./start it otherwise.
As root, invoke crontab -e.
Append a line like this:
* * * * * if ! pidof Rserve 2>&1 >/dev/null; then /full/path/to/start; fi
This method will stay persistent, i.e., it will be executed after a reboot etc. If this is not what you want, move it to a shell script:
#! /bin/bash
# monitor.sh
while true; do
if ! pidof Rserve 2>&1 >/dev/null; then /full/path/to/start; fi
sleep 10
done
This script has to be started manually from the command line, and can be easily stopped with Ctrl-C.
The easiest solution, if you can run the process is NON-daemon mode, is to wrap it in a script.
#!/bin/bash
while (true)
do
xmessage "This is your process. Click OK to kill and respawn"
done
Edit
Many deamons leave a lock file, usually in /var/lock, that contains their PID. This keeps multiple copies of the deamon from running.
Under Linux, it is fairly simple to look throgh /proc and see if that process is still around.
Under other platforms you may need to play games with ps to check for the processes existence.
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I have a #!/bin/bash file in cron.week directory.
Is there a way to test if it works? Can't wait 1 week
I am on Debian 6 with root
Just do what cron does, run the following as root:
run-parts -v /etc/cron.weekly
... or the next one if you receive the "Not a directory: -v" error:
run-parts /etc/cron.weekly -v
Option -v prints the script names before they are run.
A wee bit beyond the scope of your question... but here's what I do.
The "how do I test a cron job?" question is closely connected to "how do I test scripts that run in non-interactive contexts launched by other programs?" In cron, the trigger is some time condition, but lots of other *nix facilities launch scripts or script fragments in non-interactive ways, and often the conditions in which those scripts run contain something unexpected and cause breakage until the bugs are sorted out. (See also: https://stackoverflow.com/a/17805088/237059 )
A general approach to this problem is helpful to have.
One of my favorite techniques is to use a script I wrote called 'crontest'. It launches the target command inside a GNU screen session from within cron, so that you can attach with a separate terminal to see what's going on, interact with the script, even use a debugger.
To set this up, you would use "all stars" in your crontab entry, and specify crontest as the first command on the command line, e.g.:
* * * * * crontest /command/to/be/tested --param1 --param2
So now cron will run your command every minute, but crontest will ensure that only one instance runs at a time. If the command takes time to run, you can do a "screen -x" to attach and watch it run. If the command is a script, you can put a "read" command at the top to make it stop and wait for the screen attachment to complete (hit enter after attaching)
If your command is a bash script, you can do this instead:
* * * * * crontest --bashdb /command/to/be/tested --param1 --param2
Now, if you attach with "screen -x", you'll be facing an interactive bashdb session, and you can step through the code, examine variables, etc.
#!/bin/bash
# crontest
# See https://github.com/Stabledog/crontest for canonical source.
# Test wrapper for cron tasks. The suggested use is:
#
# 1. When adding your cron job, use all 5 stars to make it run every minute
# 2. Wrap the command in crontest
#
#
# Example:
#
# $ crontab -e
# * * * * * /usr/local/bin/crontest $HOME/bin/my-new-script --myparams
#
# Now, cron will run your job every minute, but crontest will only allow one
# instance to run at a time.
#
# crontest always wraps the command in "screen -d -m" if possible, so you can
# use "screen -x" to attach and interact with the job.
#
# If --bashdb is used, the command line will be passed to bashdb. Thus you
# can attach with "screen -x" and debug the remaining command in context.
#
# NOTES:
# - crontest can be used in other contexts, it doesn't have to be a cron job.
# Any place where commands are invoked without an interactive terminal and
# may need to be debugged.
#
# - crontest writes its own stuff to /tmp/crontest.log
#
# - If GNU screen isn't available, neither is --bashdb
#
crontestLog=/tmp/crontest.log
lockfile=$(if [[ -d /var/lock ]]; then echo /var/lock/crontest.lock; else echo /tmp/crontest.lock; fi )
useBashdb=false
useScreen=$( if which screen &>/dev/null; then echo true; else echo false; fi )
innerArgs="$#"
screenBin=$(which screen 2>/dev/null)
function errExit {
echo "[-err-] $#" | tee -a $crontestLog >&2
}
function log {
echo "[-stat-] $#" >> $crontestLog
}
function parseArgs {
while [[ ! -z $1 ]]; do
case $1 in
--bashdb)
if ! $useScreen; then
errExit "--bashdb invalid in crontest because GNU screen not installed"
fi
if ! which bashdb &>/dev/null; then
errExit "--bashdb invalid in crontest: no bashdb on the PATH"
fi
useBashdb=true
;;
--)
shift
innerArgs="$#"
return 0
;;
*)
innerArgs="$#"
return 0
;;
esac
shift
done
}
if [[ -z $sourceMe ]]; then
# Lock the lockfile (no, we do not wish to follow the standard
# advice of wrapping this in a subshell!)
exec 9>$lockfile
flock -n 9 || exit 1
# Zap any old log data:
[[ -f $crontestLog ]] && rm -f $crontestLog
parseArgs "$#"
log "crontest starting at $(date)"
log "Raw command line: $#"
log "Inner args: $#"
log "screenBin: $screenBin"
log "useBashdb: $( if $useBashdb; then echo YES; else echo no; fi )"
log "useScreen: $( if $useScreen; then echo YES; else echo no; fi )"
# Were building a command line.
cmdline=""
# If screen is available, put the task inside a pseudo-terminal
# owned by screen. That allows the developer to do a "screen -x" to
# interact with the running command:
if $useScreen; then
cmdline="$screenBin -D -m "
fi
# If bashdb is installed and --bashdb is specified on the command line,
# pass the command to bashdb. This allows the developer to do a "screen -x" to
# interactively debug a bash shell script:
if $useBashdb; then
cmdline="$cmdline $(which bashdb) "
fi
# Finally, append the target command and params:
cmdline="$cmdline $innerArgs"
log "cmdline: $cmdline"
# And run the whole schlock:
$cmdline
res=$?
log "Command result: $res"
echo "[-result-] $(if [[ $res -eq 0 ]]; then echo ok; else echo fail; fi)" >> $crontestLog
# Release the lock:
9<&-
fi
After messing about with some stuff in cron which wasn't instantly compatible I found that the following approach was nice for debugging:
crontab -e
* * * * * /path/to/prog var1 var2 &>>/tmp/cron_debug_log.log
This will run the task once a minute and you can simply look in the /tmp/cron_debug_log.log file to figure out what is going on.
It is not exactly the "fire job" you might be looking for, but this helped me a lot when debugging a script that didn't work in cron at first.
I'd use a lock file and then set the cron job to run every minute. (use crontab -e and * * * * * /path/to/job) That way you can just keep editing the files and each minute they'll be tested out. Additionally, you can stop the cronjob by just touching the lock file.
#!/bin/sh
if [ -e /tmp/cronlock ]
then
echo "cronjob locked"
exit 1
fi
touch /tmp/cronlock
<...do your regular cron here ....>
rm -f /tmp/cronlock
What about putting it into cron.hourly, waiting until the next run of hourly cron jobs, then removing it? That would run it once within an hour, and in the cron environment. You can also run ./your_script, but that won't have the same environment as under cron.
Aside from that you can also use:
http://pypi.python.org/pypi/cronwrap
to wrap up your cron to send you an email upon success or failure.
None of these answers fit my specific situation, which was that I wanted to run one specific cron job, just once, and run it immediately.
I'm on a Ubuntu server, and I use cPanel to setup my cron jobs.
I simply wrote down my current settings, and then edited them to be one minute from now. When I fixed another bug, I just edited it again to one minute from now. And when I was all done, I just reset the settings back to how they were before.
Example: It's 4:34pm right now, so I put 35 16 * * *, for it to run at 16:35.
It worked like a charm, and the most I ever had to wait was a little less than one minute.
I thought this was a better option than some of the other answers because I didn't want to run all of my weekly crons, and I didn't want the job to run every minute. It takes me a few minutes to fix whatever the issues were before I'm ready to test it again. Hopefully this helps someone.
The solution I am using is as follows:
Edit crontab(use command :crontab -e) to run the job as frequently
as needed (every 1 minute or 5 minutes)
Modify the shell script which should be executed using cron to prints the output into some file (e.g: echo "Working fine" >>
output.txt)
Check the output.txt file using the command : tail -f output.txt, which will print the latest additions into this file, and thus you can track the execution of the script
I normally test by running the job i created like this:
It is easier to use two terminals to do this.
run job:
#./jobname.sh
go to:
#/var/log and run
run the following:
#tailf /var/log/cron
This allows me to see the cron logs update in real time. You can also review the log after you run it, I prefer watching in real time.
Here is an example of a simple cron job. Running a yum update...
#!/bin/bash
YUM=/usr/bin/yum
$YUM -y -R 120 -d 0 -e 0 update yum
$YUM -y -R 10 -e 0 -d 0 update
Here is the breakdown:
First command will update yum itself and next will apply system updates.
-R 120 : Sets the maximum amount of time yum will wait before performing a command
-e 0 : Sets the error level to 0 (range 0 - 10). 0 means print only critical errors about which you must be told.
-d 0 : Sets the debugging level to 0 - turns up or down the amount of things that are printed. (range: 0 - 10).
-y : Assume yes; assume that the answer to any question which would be asked is yes
After I built the cron job I ran the below command to make my job executable.
#chmod +x /etc/cron.daily/jobname.sh
Hope this helps,
Dorlack
sudo run-parts --test /var/spool/cron/crontabs/
files in that crontabs/ directory needs to be executable by owner - octal 700
source: man cron and NNRooth's
I'm using Webmin because its a productivity gem for someone who finds command line administration a bit daunting and impenetrable.
There is a "Save and Run Now" button in the "System > Scheduled Cron Jobs > Edit Cron Job" web interface.
It displays the output of the command and is exactly what I needed.