How to pass the grep result into a variable in Unix

Question

I have the following statements in my unix batch script:
JOBNAME=`grep "$1" rssc_plsbatch.txt|awk -F'~' '{print $1}'`;
PROCNAME=`grep "$1" rssc_plsbatch.txt|awk -F'~' '{print $2}'`;
JOBDESC=`grep "$1" rssc_plsbatch.txt|awk -F'~' '{print $3}'`;
PARMS=`grep "$1" rssc_plsbatch.txt|awk -F'~' '{print $4}'`;
I want to grep the the first line in the text file and pass it to a variable and then using awk programming, I want to print for each variable (JOBNAME, PROCNAME, JOBDESC and PARMS), instead of using grep 4 times.
Can somebody please help.
Each line of my text file is in the the following format:
00001JOB1~PROCNAME1~This is a procedure~10,'A','B'~
And also I want to check whether $1 passed (part of JONAME) is not fount, then it should exit with error message, if $1 (input parameter) does not match, I want to exit the program, instead of processing further statements.

Answer 1

If you want to return error codes, you can do that with the return command.
If the process name isn't found, grep will return a specific error code... from man grep
Normally, exit status is 0 if selected lines are found and 1 otherwise.
But the exit status is 2 if an error occurred, unless the -q or quiet or silent option is used and a selected line is found.
So you can use the exit code of grep in an if statement to decide what return code you want to return with your own script.
Mike