Debugging my cron application

Question

I have a code like this :

set -e
set -x
folderName=$(echo `date +%Y/%m/%d/`);
fileName=x.x.x.x.x.x.x.log 
cp x.x.x.x.x.x/$1 $fileName  
gzip $fileName  
s3cmd put $fileName.gz s3://x.x.x.x.x/$folderName    
rm $fileName.gz

This is working fine if I run like this :

./shell logfilelocation

And when I added into the crontab like this :

* * * * * /home/x.x.x/testing/s3 -f x.x.x.log >> /tmp/mys3Log

And I waited! The file mys3Log gets created. But there is no content in it! I expect the result of command execution ( as I have used set -e ; set -x in my code ) should go into the mys3Log file as I'm doing a redirect there.

But something is going wrong. I'm very new to bash programming and cron.

Where I'm making the mistake?

Thanks in advance.

Answer 1

cron don't have the same environment like in interactive shell, so at the beginning of the script, after the shebang, add :

source ~/.bashrc || source /etc/profile

And remove set -e to see what's going on.

In your crontab, to log errors & output (STDERR, STDOUT), you need to do :

* * * * * /home/x.x.x/testing/s3 -f x.x.x.log >> /tmp/mys3Log 2>&1

Moreover, on line 4 of your script, you are using the variable $name thats is never declared.

Last but not least, like Janauary said, add the #!/bin/bash shegang on the first line and ensure that your script have executable rights. : chmod +x script.sh

Answer 2

As mentioned in the previous answer, the cron works in different environment than your shell. Which means that the environment variables would differ in shell. For eg:

In shell:

$ echo $PATH
/usr/local/bin:/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/sbin:/opt/aws/bin

Whereas in if you run the following in cron:

* * * * * echo $PATH > path.out

you'll get /usr/bin:/bin.

Find out where is your s3cmd executable using the command which s3cmd and then modify the cron accordingly. Eg:

$ which s3cmd
/usr/local/bin/s3cmd

Then, cron should look like:

PATH=/usr/bin:/bin:/usr/local/bin
* * * * * echo $PATH > path.out