Shell Programming and Scripting

I am running on SunOS, KSH

For my script , I want to test if a file contain

'PLS' OR 'ORA' OR 'ERROR'

Can we do it in one line using grep ???

currently i am using something like below.

ERROR_CHK () {
ERR_DATA="$ABC"

if $(echo "$ABC" | grep ERROR) ;then
echo "Failed: Can't execute procedure ${PRG_NAME} ERROR found "
return 1
elsif $(echo $"ERR_DATA"| grep ORA) then
echo "Failed: Can't execute procedure ${PRG_NAME} ORA Error found "
elsif $(echo $"ERR_DATA"| grep PLS) then
echo "Failed: Can't execute procedure ${PRG_NAME} PLS Error found"
else
echo "Info : ${PRG_NAME} done"
fi

}


I will appriciate if you can come with a more streamlined version of above function

Code:

grep -e "PLS
ORA
ERROR" filelist....

I did like

ERROR_CHK () {

ERR_DATA="$ABC"
if `echo "$ABC" | egrep 'PLS|ERROR|ORA'` ; then
echo "Failed: Can't execute procedure ${PRG_NAME} ERROR found "
return 1
else
echo "Info : ${PRG_NAME} done"
fi

}

but it fails with

ERROR: not found message

;; any help

I thought I had just suggested something.

Like the previous "suggestion" stated, you can check for multiple patterns with the -e option with grep, here's an example:

Code:

/usr/xpg4/bin/grep -e root -e  daemon -e sys # or something like
echo $ERR_DATA | /usr/xpg4/bin/grep -e PLS -e ERROR -e ORA 
[ $? -eq 0 ] && echo Pattern match found, oracle prob occured!