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Shell Script gives error when run through cron job.

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Operating Systems Solaris Shell Script gives error when run through cron job.
# 1  
Old 04-20-2011
Shell Script gives error when run through cron job.
Hi,

The following shell script runs without any problem when executed manulally.
Code:
USED=$(df -h /arch | tail -1 | awk '{print $5}' | cut -d '%' -f 1)
if [ ${USED} -gt 90 ]
then
  find /arch/AUBUAT/ -type f -mtime +0 | xargs rm
  find /arch/AUBMIG/ -type f -mtime +0 | xargs rm
fi



But the same gives below error when called from a cron job:
/backup/arch_test.sh: syntax error at line 1: `USED=$' unexpected

We are running following version:
Code:
bash-3.00$ uname -a
SunOS uaeaubdbdr01.aub.af 5.10 Generic_141444-09 sun4u sparc SUNW,SPARC-Enterprise

Kindly assist.

Thanks
Last edited by Franklin52; 04-20-2011 at 08:45 AM.. Reason: Please use code tags and indentation
# 2  
Old 04-20-2011
please read our fine FAQ:
https://www.unix.com/answers-frequent...n-crontab.html
# 3  
Old 04-20-2011
Hi,

Thx for the update. However, I tried different options but still the problem persists.

Thanks
# 4  
Old 04-20-2011
try backticks instead of $()

Code:
USED=`df -h /arch | tail -1 | awk '{print $5}' | cut -d '%' -f 1`

or define a posix conform shell in your script like "#!/bin/ksh".
# 5  
Old 04-20-2011
have you tried, in crontab:

Code:
* * * * * /usr/bin/bash -c your_script

# 6  
Old 04-21-2011
Check the shell in which you r running the scripts
# 7  
Old 04-22-2011
try cut -d% -f1
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