Execute multiple commands in a find

Shell Programming and Scripting
1

I am checking that a file is older than a reference file that I build with a touch command before processing it. If it is not old enough, I want to sleep for an hour and check again.

My problem is if it is old enough to process, I want to exit when I am done, but I cannot find a way to exit after doing a successful find. It keeps looping and processing until the count is fulfilled. I have tried doing an exit as a -exec on the find as shown below and I have tried checking the status of the find with $?, but it is always "0" whether the file is old or new.

Any ideas?

typeset -i  count=1

while ((${count} <= 3))
do
 find  /export/home/MYACCT/ -type f -name \ar5.log ! -newer /WORKDIR/REF -exec /export/home/MYACCT/datetest.ksh \; -exec exit 0 \;
   count=${count}+1
   sleep 3600
done

echo "No file to process"
exit 5
2

You can try putting the output of the find command into a variable and then checking to see if the variable value is empty or not.

3

Thanks Padow, that worked great.

abc=`find  /export/home/MYACCT/ -type f -name \ar5.log ! -newer /WORKDIR/REF`

if [ -z "$abc" ]; then
   count=${count}+1
   sleep 3600
else
  /export/home/MYACCT/datetest.ksh
  exit 0
fi
done

echo "No file to process"
exit 5