I'm trying to run this script every 5 minutes. It seems the only way to run CRON jobs on OpenShift is to use their CRON plugin. And the CRON plugin only allows for minutely, hourly, and daily scripts (by placing the script in the corresponding folder).
I am trying to run this script every 5 minutes:
#!/bin/bash
php /var/lib/openshift/53434c795973ca1ddc000668/app-root/runtime/repo/scheduled.php > /dev/null 2>&1
But right now it runs every minute (because it's placed in the minutely folder).
How can I re-write it so that it runs every 5 minutes?
Modify the script so it checks the current time, and bails out if it's not a multiple of 5 minutes.
Something like this:
#!/bin/bash
minute=$(date +%M)
if [[ $minute =~ [05]$ ]]; then
php ...
fi
The right operand of the =~ operator is a regular expression; the above matches if the current minute ends in 0 or 5. Several other approaches are possible:
if [[ $minute =~ .[05] ]]; then
(check for any character followed by a 0 or 5; $minute is always exactly 2 characters).
(User theshadowmonkey suggests in a comment:
if [ $(($minute % 5)) -eq 0 ]; then
which checks arithmetically whether $minute is a multiple of 5, but there's a problem with that. In the expression in a $(( ... )) expression, constants with leading zeros are treated as octal; if it's currently 8 or 9 minutes after the hour, the constant 08 or 09 is an error. You could work around this with sed, but it's probably not worthwhile given that there are other solutions.)
I will extend Keith Thompson answer:
His solution works perfectly for every 5 minutes but won't work for, let's say, every 13 minutes; if we use $minutes % 13 we get this schedule:
5:13
5:26
5:30
5:52
6:00 because 0%13 is 0
6:13
...
I'm sure you notice the issue. We can achieve any frequency if we count the minutes(, hours, days, or weeks) since Epoch:
#!/bin/bash
minutesSinceEpoch=$(($(date +'%s / 60')))
if [[ $(($minutesSinceEpoch % 13)) -eq 0 ]]; then
php [...]
fi
date(1) returns current date, we format it as seconds since Epoch (%s) and then we do basic maths:
# .---------------------- bash command substitution
# |.--------------------- bash arithmetic expansion
# || .------------------- bash command substitution
# || | .---------------- date command
# || | | .------------ FORMAT argument
# || | | | .----- formula to calculate minutes/hours/days/etc is included into the format string passed to date command
# || | | | |
# ** * * * *
$(($(date +'%s / 60')))
# * * ---------------
# | | |
# | | ·----------- date should result in something like "1438390397 / 60"
# | ·-------------------- it gets evaluated as an expression. (the maths)
# ·---------------------- and we can store it
And you may use this approach with hourly, daily, or monthly cron jobs on OpenShift:
#!/bin/bash
# We can get the
minutes=$(($(date +'%s / 60')))
hours=$(($(date +'%s / 60 / 60')))
days=$(($(date +'%s / 60 / 60 / 24')))
weeks=$(($(date +'%s / 60 / 60 / 24 / 7')))
# or even
moons=$(($(date +'%s / 60 / 60 / 24 / 656')))
# passed since Epoch and define a frequency
# let's say, every 7 hours
if [[ $(($hours % 7)) -ne 0 ]]; then
exit 0
fi
# and your actual script starts here
Notice that I used the -ne (not equal) operator to exit the script instead of using the -eq (equal) operator to wrap the script into the IF construction; I find it handy.
And remember to use the proper .openshift/cron/{minutely,hourly,daily,weekly,monthly}/ folder for your frequency.
You can make use of minutely cron job available. The jobs are run at a specified frequency and you can instrument your job to inspect the date and/or time when your job runs.
Here is an example for running a job every 5 minutes. Place the below code snippet in an executable file at path .openshift/cron/minutely/awesome_job and give proper permissions chmod +x awesome_job then add it to your application repository, commit and push.
The snippet
#!/bin/bash
if [ ! -f $OPENSHIFT_DATA_DIR/last_run ]; then
touch $OPENSHIFT_DATA_DIR/last_run
fi
if [[ $(find $OPENSHIFT_DATA_DIR/last_run -mmin +4) ]]; then #run every 5 mins
rm -f $OPENSHIFT_DATA_DIR/last_run
touch $OPENSHIFT_DATA_DIR/last_run
# The 'awesome_command(s)' that you want to run every 5 minutes
fi
Explanation:
This snippet actually creates an empty file named 'last_run' just to
note down the last run status and then starts your job.
The next minute, your 'awesome_job' will be executed and now it tries to find file 'last_run' modified in last 4 minutes. But your
file 'last_run' was created just a minute ago and hence the condition fails and exits. This continues.
At fifth minute, while finding (with -mmin +4), your 'last_run' will be shown and now, the file 'last_run' will be deleted and recreated and your awesome command(s) will be executed.
Source (Official Docs): https://developers.openshift.com/managing-your-applications/background-jobs.html#_execution_timing
Related
Does crontab have an argument for creating cron jobs without using the editor (crontab -e)? If so, what would be the code to create a cron job from a Bash script?
You can add to the crontab as follows:
#write out current crontab
crontab -l > mycron
#echo new cron into cron file
echo "00 09 * * 1-5 echo hello" >> mycron
#install new cron file
crontab mycron
rm mycron
Cron line explaination
* * * * * "command to be executed"
- - - - -
| | | | |
| | | | ----- Day of week (0 - 7) (Sunday=0 or 7)
| | | ------- Month (1 - 12)
| | --------- Day of month (1 - 31)
| ----------- Hour (0 - 23)
------------- Minute (0 - 59)
Source nixCraft.
You may be able to do it on-the-fly
crontab -l | { cat; echo "0 0 0 0 0 some entry"; } | crontab -
crontab -l lists the current crontab jobs, cat prints it, echo prints the new command and crontab - adds all the printed stuff into the crontab file. You can see the effect by doing a new crontab -l.
This shorter one requires no temporary file, it is immune to multiple insertions, and it lets you change the schedule of an existing entry.
Say you have these:
croncmd="/home/me/myfunction myargs > /home/me/myfunction.log 2>&1"
cronjob="0 */15 * * * $croncmd"
To add it to the crontab, with no duplication:
( crontab -l | grep -v -F "$croncmd" ; echo "$cronjob" ) | crontab -
To remove it from the crontab whatever its current schedule:
( crontab -l | grep -v -F "$croncmd" ) | crontab -
Notes:
grep -F matches the string literally, as we do not want to interpret it as a regular expression
We also ignore the time scheduling and only look for the command. This way; the schedule can be changed without the risk of adding a new line to the crontab
Thanks everybody for your help. Piecing together what I found here and elsewhere I came up with this:
The Code
command="php $INSTALL/indefero/scripts/gitcron.php"
job="0 0 * * 0 $command"
cat <(fgrep -i -v "$command" <(crontab -l)) <(echo "$job") | crontab -
I couldn't figure out how to eliminate the need for the two variables without repeating myself.
command is obviously the command I want to schedule. job takes $command and adds the scheduling data. I needed both variables separately in the line of code that does the work.
Details
Credit to duckyflip, I use this little redirect thingy (<(*command*)) to turn the output of crontab -l into input for the fgrep command.
fgrep then filters out any matches of $command (-v option), case-insensitive (-i option).
Again, the little redirect thingy (<(*command*)) is used to turn the result back into input for the cat command.
The cat command also receives echo "$job" (self explanatory), again, through use of the redirect thingy (<(*command*)).
So the filtered output from crontab -l and the simple echo "$job", combined, are piped ('|') over to crontab - to finally be written.
And they all lived happily ever after!
In a nutshell:
This line of code filters out any cron jobs that match the command, then writes out the remaining cron jobs with the new one, effectively acting like an "add" or "update" function.
To use this, all you have to do is swap out the values for the command and job variables.
EDIT (fixed overwriting):
cat <(crontab -l) <(echo "1 2 3 4 5 scripty.sh") | crontab -
There have been a lot of good answers around the use of crontab, but no mention of a simpler method, such as using cron.
Using cron would take advantage of system files and directories located at /etc/crontab, /etc/cron.daily,weekly,hourly or /etc/cron.d/:
cat > /etc/cron.d/<job> << EOF
SHELL=/bin/bash
PATH=/sbin:/bin:/usr/sbin:/usr/bin
MAILTO=root HOME=/
01 * * * * <user> <command>
EOF
In this above example, we created a file in /etc/cron.d/, provided the environment variables for the command to execute successfully, and provided the user for the command, and the command itself. This file should not be executable and the name should only contain alpha-numeric and hyphens (more details below).
To give a thorough answer though, let's look at the differences between crontab vs cron/crond:
crontab -- maintain tables for driving cron for individual users
For those who want to run the job in the context of their user on the system, using crontab may make perfect sense.
cron -- daemon to execute scheduled commands
For those who use configuration management or want to manage jobs for other users, in which case we should use cron.
A quick excerpt from the manpages gives you a few examples of what to and not to do:
/etc/crontab and the files in /etc/cron.d must be owned by root, and must not be group- or other-writable. In contrast to the spool area, the files under /etc/cron.d or the files under /etc/cron.hourly, /etc/cron.daily, /etc/cron.weekly and /etc/cron.monthly may also be symlinks, provided that both the symlink and the file it points to are owned by root. The files under /etc/cron.d do not need to be executable, while the files under /etc/cron.hourly, /etc/cron.daily, /etc/cron.weekly and /etc/cron.monthly do, as they are run by run-parts (see run-parts(8) for more information).
Source: http://manpages.ubuntu.com/manpages/trusty/man8/cron.8.html
Managing crons in this manner is easier and more scalable from a system perspective, but will not always be the best solution.
So, in Debian, Ubuntu, and many similar Debian based distros...
There is a cron task concatenation mechanism that takes a config file, bundles them up and adds them to your cron service running.
You can put a file under the /etc/cron.d/somefilename where somefilename is whatever you want.
sudo echo "0,15,30,45 * * * * ntpdate -u time.nist.gov" >> /etc/cron.d/vmclocksync
Let's disassemble this:
sudo - because you need elevated privileges to change cron configs under the /etc directory
echo - a vehicle to create output on std out. printf, cat... would work as well
" - use a doublequote at the beginning of your string, you're a professional
0,15,30,45 * * * * - the standard cron run schedule, this one runs every 15 minutes
ntpdate -u time.nist.gov - the actual command I want to run
" - because my first double quotes needs a buddy to close the line being output
>> - the double redirect appends instead of overwrites*
/etc/cron.d/vmclocksync - vmclocksync is the filename I've chosen, it goes in /etc/cron.d/
* if we used the > redirect, we could guarantee we only had one task entry. But, we would be at risk of blowing away any other rules in an existing file. You can decide for yourself if possible destruction with > is right or possible duplicates with >> are for you. Alternatively, you could do something convoluted or involved to check if the file name exists, if there is anything in it, and whether you are adding any kind of duplicate-- but, I have stuff to do and I can't do that for you right now.
For a nice quick and dirty creation/replacement of a crontab from with a BASH script, I used this notation:
crontab <<EOF
00 09 * * 1-5 echo hello
EOF
Chances are you are automating this, and you don't want a single job added twice.
In that case use:
__cron="1 2 3 4 5 /root/bin/backup.sh"
cat <(crontab -l) |grep -v "${__cron}" <(echo "${__cron}")
This only works if you're using BASH. I'm not aware of the correct DASH (sh) syntax.
Update: This doesn't work if the user doesn't have a crontab yet. A more reliable way would be:
(crontab -l ; echo "1 2 3 4 5 /root/bin/backup.sh") | sort - | uniq - | crontab -
Alternatively, if your distro supports it, you could also use a separate file:
echo "1 2 3 4 5 /root/bin/backup.sh" |sudo tee /etc/crond.d/backup
Found those in another SO question.
echo "0 * * * * docker system prune --force >/dev/null 2>&1" | sudo tee /etc/cron.daily/dockerprune
A variant which only edits crontab if the desired string is not found there:
CMD="/sbin/modprobe fcpci"
JOB="#reboot $CMD"
TMPC="mycron"
grep "$CMD" -q <(crontab -l) || (crontab -l>"$TMPC"; echo "$JOB">>"$TMPC"; crontab "$TMPC")
(2>/dev/null crontab -l ; echo "0 3 * * * /usr/local/bin/certbot-auto renew") | crontab -
cat <(crontab -l 2>/dev/null) <(echo "0 3 * * * /usr/local/bin/certbot-auto renew") | crontab -
#write out current crontab
crontab -l > mycron 2>/dev/null
#echo new cron into cron file
echo "0 3 * * * /usr/local/bin/certbot-auto renew" >> mycron
#install new cron file
crontab mycron
rm mycron
If you're using the Vixie Cron, e.g. on most Linux distributions, you can just put a file in /etc/cron.d with the individual cronjob.
This only works for root of course. If your system supports this you should see several examples in there. (Note the username included in the line, in the same syntax as the old /etc/crontab)
It's a sad misfeature in cron that there is no way to handle this as a regular user, and that so many cron implementations have no way at all to handle this.
My preferred solution to this would be this:
(crontab -l | grep . ; echo -e "0 4 * * * myscript\n") | crontab -
This will make sure you are handling the blank new line at the bottom correctly. To avoid issues with crontab you should usually end the crontab file with a blank new line. And the script above makes sure it first removes any blank lines with the "grep ." part, and then add in a new blank line at the end with the "\n" in the end of the script. This will also prevent getting a blank line above your new command if your existing crontab file ends with a blank line.
Bash script for adding cron job without the interactive editor.
Below code helps to add a cronjob using linux files.
#!/bin/bash
cron_path=/var/spool/cron/crontabs/root
#cron job to run every 10 min.
echo "*/10 * * * * command to be executed" >> $cron_path
#cron job to run every 1 hour.
echo "0 */1 * * * command to be executed" >> $cron_path
Here is a bash function for adding a command to crontab without duplication
function addtocrontab () {
local frequency=$1
local command=$2
local job="$frequency $command"
cat <(fgrep -i -v "$command" <(crontab -l)) <(echo "$job") | crontab -
}
addtocrontab "0 0 1 * *" "echo hello"
CRON="1 2 3 4 5 /root/bin/backup.sh"
cat < (crontab -l) |grep -v "${CRON}" < (echo "${CRON}")
add -w parameter to grep exact command, without -w parameter adding the cronjob "testing" cause deletion of cron job "testing123"
script function to add/remove cronjobs. no duplication entries :
cronjob_editor () {
# usage: cronjob_editor '<interval>' '<command>' <add|remove>
if [[ -z "$1" ]] ;then printf " no interval specified\n" ;fi
if [[ -z "$2" ]] ;then printf " no command specified\n" ;fi
if [[ -z "$3" ]] ;then printf " no action specified\n" ;fi
if [[ "$3" == add ]] ;then
# add cronjob, no duplication:
( crontab -l | grep -v -F -w "$2" ; echo "$1 $2" ) | crontab -
elif [[ "$3" == remove ]] ;then
# remove cronjob:
( crontab -l | grep -v -F -w "$2" ) | crontab -
fi
}
cronjob_editor "$1" "$2" "$3"
tested :
$ ./cronjob_editor.sh '*/10 * * * *' 'echo "this is a test" > export_file' add
$ crontab -l
$ */10 * * * * echo "this is a test" > export_file
No, there is no option in crontab to modify the cron files.
You have to: take the current cron file (crontab -l > newfile), change it and put the new file in place (crontab newfile).
If you are familiar with perl, you can use this module Config::Crontab.
LLP, Andrea
script function to add cronjobs. check duplicate entries,useable expressions * > "
cronjob_creator () {
# usage: cronjob_creator '<interval>' '<command>'
if [[ -z $1 ]] ;then
printf " no interval specified\n"
elif [[ -z $2 ]] ;then
printf " no command specified\n"
else
CRONIN="/tmp/cti_tmp"
crontab -l | grep -vw "$1 $2" > "$CRONIN"
echo "$1 $2" >> $CRONIN
crontab "$CRONIN"
rm $CRONIN
fi
}
tested :
$ ./cronjob_creator.sh '*/10 * * * *' 'echo "this is a test" > export_file'
$ crontab -l
$ */10 * * * * echo "this is a test" > export_file
source : my brain ;)
Say you're logged in as the user "ubuntu", but you want to add a job to a different user's crontab, like "john", for example. You can do the following:
(sudo crontab -l -u john; echo "* * * * * command") | awk '!x[$0]++' | sudo crontab -u john -
Source for most of this solution: https://www.baeldung.com/linux/create-crontab-script
I was having tons of issues trying to add a job to another user's crontab. It kept duplicating crontabs, or just flat-out deleting them. After some testing, though, I'm confident this line of code will append a new job to a specified user's crontab, non-destructively, including not creating a job that already exists.
I wanted to find an example like this, so maybe it helps:
COMMAND="/var/lib/postgresql/backup.sh"
CRON="0 0 * * *"
USER="postgres"
CRON_FILE="postgres-backup"
# At CRON times, the USER will run the COMMAND
echo "$CRON $USER $COMMAND" | sudo tee /etc/cron.d/$CRON_FILE
echo "Cron job created. Remove /etc/cron.d/$CRON_FILE to stop it."
I'd like to configure cron to run script on the first day of every month if it is a working day. If the month starts on a weekend, I want the script to be run on the first following week day.
I tried do it this way
0 9 1-3 * * dom=$(date +\%d); dow=$(date +\%u); [[ $dow -lt 6 && (( $dom -eq 1 || $dow -eq 1 )) ]] && script.sh
This works in bash, but seems like cron can't execute comparisons.
By advice of Roadowl, adding
SHELL=/bin/bash
at the beginning of cron file made things work.
I currently have a script in crontab for rsync (and some other small stuff). Right now the script is executed every 5 minutes. I modified the script to look for a specific line from the rsync part (example from my machine, not the actual code):
#!/bin/bash
Number=`/usr/bin/rsync -n --stats -avz -e ssh 1/ root#127.0.0.1 | grep "Number of regular files transferred" | cut -d':' -f 2 | tr -d 040\054\012`
echo $Number
Let's say the number is 10. If the number is 10 or below I want the script executed through the crontab. But if the number is bigger I want to be executed ONLY manually.
Any ides?
Maybe you can use an argument to execute it manually, for example:
if [[ $Number -le 10 || $1 == true ]];then
echo "executing script..."
fi
This will execute if $Number is less or equal to 10 or if you execute it with true as the first positional argument, so if $Number is greater than 10 it won't execute in your crontab and you can execute your script manually with ./your_script true.
Does crontab have an argument for creating cron jobs without using the editor (crontab -e)? If so, what would be the code to create a cron job from a Bash script?
You can add to the crontab as follows:
#write out current crontab
crontab -l > mycron
#echo new cron into cron file
echo "00 09 * * 1-5 echo hello" >> mycron
#install new cron file
crontab mycron
rm mycron
Cron line explaination
* * * * * "command to be executed"
- - - - -
| | | | |
| | | | ----- Day of week (0 - 7) (Sunday=0 or 7)
| | | ------- Month (1 - 12)
| | --------- Day of month (1 - 31)
| ----------- Hour (0 - 23)
------------- Minute (0 - 59)
Source nixCraft.
You may be able to do it on-the-fly
crontab -l | { cat; echo "0 0 0 0 0 some entry"; } | crontab -
crontab -l lists the current crontab jobs, cat prints it, echo prints the new command and crontab - adds all the printed stuff into the crontab file. You can see the effect by doing a new crontab -l.
This shorter one requires no temporary file, it is immune to multiple insertions, and it lets you change the schedule of an existing entry.
Say you have these:
croncmd="/home/me/myfunction myargs > /home/me/myfunction.log 2>&1"
cronjob="0 */15 * * * $croncmd"
To add it to the crontab, with no duplication:
( crontab -l | grep -v -F "$croncmd" ; echo "$cronjob" ) | crontab -
To remove it from the crontab whatever its current schedule:
( crontab -l | grep -v -F "$croncmd" ) | crontab -
Notes:
grep -F matches the string literally, as we do not want to interpret it as a regular expression
We also ignore the time scheduling and only look for the command. This way; the schedule can be changed without the risk of adding a new line to the crontab
Thanks everybody for your help. Piecing together what I found here and elsewhere I came up with this:
The Code
command="php $INSTALL/indefero/scripts/gitcron.php"
job="0 0 * * 0 $command"
cat <(fgrep -i -v "$command" <(crontab -l)) <(echo "$job") | crontab -
I couldn't figure out how to eliminate the need for the two variables without repeating myself.
command is obviously the command I want to schedule. job takes $command and adds the scheduling data. I needed both variables separately in the line of code that does the work.
Details
Credit to duckyflip, I use this little redirect thingy (<(*command*)) to turn the output of crontab -l into input for the fgrep command.
fgrep then filters out any matches of $command (-v option), case-insensitive (-i option).
Again, the little redirect thingy (<(*command*)) is used to turn the result back into input for the cat command.
The cat command also receives echo "$job" (self explanatory), again, through use of the redirect thingy (<(*command*)).
So the filtered output from crontab -l and the simple echo "$job", combined, are piped ('|') over to crontab - to finally be written.
And they all lived happily ever after!
In a nutshell:
This line of code filters out any cron jobs that match the command, then writes out the remaining cron jobs with the new one, effectively acting like an "add" or "update" function.
To use this, all you have to do is swap out the values for the command and job variables.
EDIT (fixed overwriting):
cat <(crontab -l) <(echo "1 2 3 4 5 scripty.sh") | crontab -
There have been a lot of good answers around the use of crontab, but no mention of a simpler method, such as using cron.
Using cron would take advantage of system files and directories located at /etc/crontab, /etc/cron.daily,weekly,hourly or /etc/cron.d/:
cat > /etc/cron.d/<job> << EOF
SHELL=/bin/bash
PATH=/sbin:/bin:/usr/sbin:/usr/bin
MAILTO=root HOME=/
01 * * * * <user> <command>
EOF
In this above example, we created a file in /etc/cron.d/, provided the environment variables for the command to execute successfully, and provided the user for the command, and the command itself. This file should not be executable and the name should only contain alpha-numeric and hyphens (more details below).
To give a thorough answer though, let's look at the differences between crontab vs cron/crond:
crontab -- maintain tables for driving cron for individual users
For those who want to run the job in the context of their user on the system, using crontab may make perfect sense.
cron -- daemon to execute scheduled commands
For those who use configuration management or want to manage jobs for other users, in which case we should use cron.
A quick excerpt from the manpages gives you a few examples of what to and not to do:
/etc/crontab and the files in /etc/cron.d must be owned by root, and must not be group- or other-writable. In contrast to the spool area, the files under /etc/cron.d or the files under /etc/cron.hourly, /etc/cron.daily, /etc/cron.weekly and /etc/cron.monthly may also be symlinks, provided that both the symlink and the file it points to are owned by root. The files under /etc/cron.d do not need to be executable, while the files under /etc/cron.hourly, /etc/cron.daily, /etc/cron.weekly and /etc/cron.monthly do, as they are run by run-parts (see run-parts(8) for more information).
Source: http://manpages.ubuntu.com/manpages/trusty/man8/cron.8.html
Managing crons in this manner is easier and more scalable from a system perspective, but will not always be the best solution.
So, in Debian, Ubuntu, and many similar Debian based distros...
There is a cron task concatenation mechanism that takes a config file, bundles them up and adds them to your cron service running.
You can put a file under the /etc/cron.d/somefilename where somefilename is whatever you want.
sudo echo "0,15,30,45 * * * * ntpdate -u time.nist.gov" >> /etc/cron.d/vmclocksync
Let's disassemble this:
sudo - because you need elevated privileges to change cron configs under the /etc directory
echo - a vehicle to create output on std out. printf, cat... would work as well
" - use a doublequote at the beginning of your string, you're a professional
0,15,30,45 * * * * - the standard cron run schedule, this one runs every 15 minutes
ntpdate -u time.nist.gov - the actual command I want to run
" - because my first double quotes needs a buddy to close the line being output
>> - the double redirect appends instead of overwrites*
/etc/cron.d/vmclocksync - vmclocksync is the filename I've chosen, it goes in /etc/cron.d/
* if we used the > redirect, we could guarantee we only had one task entry. But, we would be at risk of blowing away any other rules in an existing file. You can decide for yourself if possible destruction with > is right or possible duplicates with >> are for you. Alternatively, you could do something convoluted or involved to check if the file name exists, if there is anything in it, and whether you are adding any kind of duplicate-- but, I have stuff to do and I can't do that for you right now.
For a nice quick and dirty creation/replacement of a crontab from with a BASH script, I used this notation:
crontab <<EOF
00 09 * * 1-5 echo hello
EOF
Chances are you are automating this, and you don't want a single job added twice.
In that case use:
__cron="1 2 3 4 5 /root/bin/backup.sh"
cat <(crontab -l) |grep -v "${__cron}" <(echo "${__cron}")
This only works if you're using BASH. I'm not aware of the correct DASH (sh) syntax.
Update: This doesn't work if the user doesn't have a crontab yet. A more reliable way would be:
(crontab -l ; echo "1 2 3 4 5 /root/bin/backup.sh") | sort - | uniq - | crontab -
Alternatively, if your distro supports it, you could also use a separate file:
echo "1 2 3 4 5 /root/bin/backup.sh" |sudo tee /etc/crond.d/backup
Found those in another SO question.
echo "0 * * * * docker system prune --force >/dev/null 2>&1" | sudo tee /etc/cron.daily/dockerprune
A variant which only edits crontab if the desired string is not found there:
CMD="/sbin/modprobe fcpci"
JOB="#reboot $CMD"
TMPC="mycron"
grep "$CMD" -q <(crontab -l) || (crontab -l>"$TMPC"; echo "$JOB">>"$TMPC"; crontab "$TMPC")
(2>/dev/null crontab -l ; echo "0 3 * * * /usr/local/bin/certbot-auto renew") | crontab -
cat <(crontab -l 2>/dev/null) <(echo "0 3 * * * /usr/local/bin/certbot-auto renew") | crontab -
#write out current crontab
crontab -l > mycron 2>/dev/null
#echo new cron into cron file
echo "0 3 * * * /usr/local/bin/certbot-auto renew" >> mycron
#install new cron file
crontab mycron
rm mycron
If you're using the Vixie Cron, e.g. on most Linux distributions, you can just put a file in /etc/cron.d with the individual cronjob.
This only works for root of course. If your system supports this you should see several examples in there. (Note the username included in the line, in the same syntax as the old /etc/crontab)
It's a sad misfeature in cron that there is no way to handle this as a regular user, and that so many cron implementations have no way at all to handle this.
My preferred solution to this would be this:
(crontab -l | grep . ; echo -e "0 4 * * * myscript\n") | crontab -
This will make sure you are handling the blank new line at the bottom correctly. To avoid issues with crontab you should usually end the crontab file with a blank new line. And the script above makes sure it first removes any blank lines with the "grep ." part, and then add in a new blank line at the end with the "\n" in the end of the script. This will also prevent getting a blank line above your new command if your existing crontab file ends with a blank line.
Bash script for adding cron job without the interactive editor.
Below code helps to add a cronjob using linux files.
#!/bin/bash
cron_path=/var/spool/cron/crontabs/root
#cron job to run every 10 min.
echo "*/10 * * * * command to be executed" >> $cron_path
#cron job to run every 1 hour.
echo "0 */1 * * * command to be executed" >> $cron_path
Here is a bash function for adding a command to crontab without duplication
function addtocrontab () {
local frequency=$1
local command=$2
local job="$frequency $command"
cat <(fgrep -i -v "$command" <(crontab -l)) <(echo "$job") | crontab -
}
addtocrontab "0 0 1 * *" "echo hello"
CRON="1 2 3 4 5 /root/bin/backup.sh"
cat < (crontab -l) |grep -v "${CRON}" < (echo "${CRON}")
add -w parameter to grep exact command, without -w parameter adding the cronjob "testing" cause deletion of cron job "testing123"
script function to add/remove cronjobs. no duplication entries :
cronjob_editor () {
# usage: cronjob_editor '<interval>' '<command>' <add|remove>
if [[ -z "$1" ]] ;then printf " no interval specified\n" ;fi
if [[ -z "$2" ]] ;then printf " no command specified\n" ;fi
if [[ -z "$3" ]] ;then printf " no action specified\n" ;fi
if [[ "$3" == add ]] ;then
# add cronjob, no duplication:
( crontab -l | grep -v -F -w "$2" ; echo "$1 $2" ) | crontab -
elif [[ "$3" == remove ]] ;then
# remove cronjob:
( crontab -l | grep -v -F -w "$2" ) | crontab -
fi
}
cronjob_editor "$1" "$2" "$3"
tested :
$ ./cronjob_editor.sh '*/10 * * * *' 'echo "this is a test" > export_file' add
$ crontab -l
$ */10 * * * * echo "this is a test" > export_file
No, there is no option in crontab to modify the cron files.
You have to: take the current cron file (crontab -l > newfile), change it and put the new file in place (crontab newfile).
If you are familiar with perl, you can use this module Config::Crontab.
LLP, Andrea
script function to add cronjobs. check duplicate entries,useable expressions * > "
cronjob_creator () {
# usage: cronjob_creator '<interval>' '<command>'
if [[ -z $1 ]] ;then
printf " no interval specified\n"
elif [[ -z $2 ]] ;then
printf " no command specified\n"
else
CRONIN="/tmp/cti_tmp"
crontab -l | grep -vw "$1 $2" > "$CRONIN"
echo "$1 $2" >> $CRONIN
crontab "$CRONIN"
rm $CRONIN
fi
}
tested :
$ ./cronjob_creator.sh '*/10 * * * *' 'echo "this is a test" > export_file'
$ crontab -l
$ */10 * * * * echo "this is a test" > export_file
source : my brain ;)
Say you're logged in as the user "ubuntu", but you want to add a job to a different user's crontab, like "john", for example. You can do the following:
(sudo crontab -l -u john; echo "* * * * * command") | awk '!x[$0]++' | sudo crontab -u john -
Source for most of this solution: https://www.baeldung.com/linux/create-crontab-script
I was having tons of issues trying to add a job to another user's crontab. It kept duplicating crontabs, or just flat-out deleting them. After some testing, though, I'm confident this line of code will append a new job to a specified user's crontab, non-destructively, including not creating a job that already exists.
I wanted to find an example like this, so maybe it helps:
COMMAND="/var/lib/postgresql/backup.sh"
CRON="0 0 * * *"
USER="postgres"
CRON_FILE="postgres-backup"
# At CRON times, the USER will run the COMMAND
echo "$CRON $USER $COMMAND" | sudo tee /etc/cron.d/$CRON_FILE
echo "Cron job created. Remove /etc/cron.d/$CRON_FILE to stop it."
I want to repeatedly run a program for a maximum of 5 seconds.
I know that timeout executes a command for the amount of time specified, e.g.:
timeout 5 ./a.out
But I want to keep executing the program until 5 seconds have passed so I can tell how
many times it was executed.
I figured that I need something like this:
timeout 5 `while true; do ./a.out; done`
but this is not working. I've already tried to create a shell script that calculates
the elapsed time of every loop iteration and subtracts it from the start time,
but that is inefficient.
Any help would be appreciated.
If you want to use timeout:
timeout 5s ./a.out
You can write a short script and easily set an end time with date -d "date string" +%s to get a future time in seconds. Then just compare current time to end time and break on true. This allows you to capture additional data during your execution time period. For example, the following code sets the end time 5 seconds in the future and then loops until current time equals end.
#!/bin/bash
end=$(date -d "+ 5 seconds" +%s) # set end time with "+ 5 seconds"
declare -i count=0
while [ $(date +%s) -lt $end ]; do # compare current time to end until true
((count++))
printf "working... %s\n" "$count" # do stuff
sleep .5
done
output:
$ bash timeexec.sh
working... 1
working... 2
working... 3
working... 4
working... 5
working... 6
working... 7
working... 8
working... 9
In your case you would do something like
./a.out & # start your application in background
apid=$(pidof a.out) # save PID of a.out
while [ $(date +%s) -lt $end ]; do
# do stuff, count, etc.
sleep .5 # something to prevent continual looping
done
kill $apid # kill process after time test true