I have to write a cron job to execute a java class every after 8 hours. To test, I wrote a simple cron using :
crontab -e
*/1 * * * * echo $PATH
When I do, crontab -u <user> -l, it prints the entire cron file with above job listed. But I want to ascertain that my job is getting executed as intended. I don't see output of "echo $PATH" every minute. How can I see this output...?
Related
I'm currently running jobs on a computing cluster that uses the slurm workload manager. I can view all of the jobs I currently have running with:
$ squeue -u <username>
JOBID PARTITION NAME USER ST TIME NODES NODELIST(REASON)
3379570 sixhour job01 <username> PD 0:00 1 (Priority)
3409269 sixhour job02 <username> R 03:06:13 1 n387
So in this example, I have two jobs that have been submitted. job01 is pending, and job02 has been running for around 3 hours.
I want to set up a cron job that will only submit the job if it's not listed in the above view. Take the following cron file as an example:
MAILTO=""
* */1 * * * sbatch job01.sh
* */1 * * * sbatch job02.sh
* */1 * * * sbatch job03.sh
Using this cron file, each job would get submitted every hour. But because job01 and job02 are already listed under squeue -u <username>, I only want job03 to actually get submitted.
Is there a way I can add some conditional logic to the cron file?
You can try a script like this:
#!/bin/bash
jobnames=$(squeue -h --user <username> --format %j)
for jobname in job{01..03} ; do
grep "$jobname" <<< "$jobnames" >/dev/null || sbatch "$jobname.sh"
done
This script will collect all job names related to jobs submitted by <username> in $jobnames and then iterate over the ones it should expect. If one is not found, grep will return non-zero exit code and the || sbatch will be executed.
Replace the three lines in your cron file with a single one running the above script.
I know this question has been asked many, many times and I've done a lot of research but still I'm not able to run this extremely simple cron:
$ crontab -l
* * * * * /bin/date
This should, ideally, print the date every minute.
There is no cron.allow or cron.deny file, and the cron daemon is working:
ps -e | grep cron
1119 ? 00:00:00 cron
17646 ? 00:00:00 cron
Any idea what might be wrong?
Cron processes run in a separate subprocess of their own, so the output of a cron job will not be visible to you in your shell.
Instead, you will have to capture the output of your cron commands and save them. So, set your cronjob like:
* * * * * /bin/date >> /home/user/date.log
And now, if you will tail this log file, you will start seeing the result.
I wrote a linux bash shell script which works fine except the output when I run it manually is different than when I run it from a cronjob.
The particular command is lftp:
lftp -e "lcd $outgoingpathlocal;mput -O $incomingpathremote *.CSV;exit" -u $FTPUSERNAME,$FTPPASSWORD $FTPSERVER >> ${SCRIPTLOGFILE} 2>&1
When I run the script manually, the ${SCRIPTLOGFILE} contains a lot of info such as how many files/bytes/etc transferred. But when I run the same script from a cronjob there is no output unless there was an error (such as could not connect). I have tried various terminal output configurations but none work for this lftp command. Suggestions?
It's worth reading this:
crontab PATH and USER
In particular, cron won't set the same environment variables you're used to an interactive shell.
You might want to wrap your entire cron job up in a script, and then you can, for example, temporarily add some code like export >> scriptenvironment.txt and see what the difference is between the cron invoked script and the interactively invoked script.
Try man 5 crontab for details.
Once you know what envrionment variables you need for your script to run, you can set them in the crontab as necessary, or source at the start of your own script.
EXAMPLE CRON FILE
# use /bin/sh to run commands, overriding the default set by cron
SHELL=/bin/sh
# mail any output to `paul', no matter whose crontab this is
MAILTO=paul
#
# run five minutes after midnight, every day
5 0 * * * $HOME/bin/daily.job >> $HOME/tmp/out 2>&1
# run at 2:15pm on the first of every month -- output mailed to paul
15 14 1 * * $HOME/bin/monthly
# run at 10 pm on weekdays, annoy Joe
0 22 * * 1-5 mail -s "It's 10pm" joe%Joe,%%Where are your kids?%
23 0-23/2 * * * echo "run 23 minutes after midn, 2am, 4am ..., everyday"
5 4 * * sun echo "run at 5 after 4 every sunday"
I'm running RHEL and I'm trying to set up a cron job to run a shell script every 5 minutes.
Following the directions here: https://access.redhat.com/documentation/en-US/Red_Hat_Enterprise_Linux/6/html/Deployment_Guide/ch-Automating_System_Tasks.html#s2-configuring-cron-jobs
I have service crond start and chkconfig crond on. Then I edited /etc/crontab and added:
*/5 * * * * my-user /path/to/shell.sh
I did a chmod +x shell.sh. And I made sure to add a new line character at the end.
I'm expecting it to run every 5 minutes but it never executes.
What am I doing wrong?
Simply try to add the cronjob entry and check the script is working fine or not by taking the viewable output in the script.
echo "test time - $(date)" > script.sh
chmod +x script.sh
crontab -e
Then enter the cronjob as below,
*/5 * * * * sh /path/to/script.sh > /path/to/log.file
Check if the log is writing correctly. If its fine, better cross check the script that you are trying to execute via cron. Otherwise it will be a cron issue.
I want to schedule a cron job in Linux by running a shell script.
The scenario is, I'm taking the time in HH:MM format in the shell script from the user and want to schedule a cron job from the shell script. I even want the cron job to be executed only once.
Thanks in advance...
For the crontab, you can do something like this:
cur=$(crontab -l)
new="$mm $hh * * * your_command"
echo "$cur$new" | crontab -
For one-shot, crontab is not the good candidate. Use at (note: your_command must be a file, e.g., a bash script)
at -f your_command $hh:$mm