One Line for loop in CRON

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# 1  
One Line for loop in CRON


I'm trying to run a one line for loop from the CRON. This is the exact command
 /bin/bash for file in $(/bin/find /opt/local/edw_extract/logs -name "*.log"); do /bin/cat /dev/null > $file; done

This will run from the command successfully.

When I schedule this to run it will error out and produce the following output

/bin/bash for file in $(/bin/find /opt/local/edw_extract/logs -name "*.log"); do /bin/cat /dev/null > $file; done

produced the following output:

sh: syntax error at line 1: `(' unexpected

Could someone explain the proper way to do this?
Is this even possible?

Thank You.


Last edited by jim mcnamara; 01-06-2012 at 05:41 PM.. Reason: please use code tags
# 2  
Try this:
/bin/bash -c 'for file in $(/bin/find /opt/local/edw_extract/logs -name "*.log"); do > $file;  done'

See if that helps. Generally, complex commands in cron cause way more trouble than just writing a simple shell script that you invoke from cron. And you save nothing worth worrying about in performance
This User Gave Thanks to jim mcnamara For This Post:
# 3  
I'd check the environment variables. it seems they need to be specified for cron
# 4  

Thanks to everyone for the quick responses. Jim's response worked. Thank you.

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