problem in a shell command - unix

Question

i am trying the following command on the command line
ps -u `id | cut -f2 -d"=" | cut -f1 -d"("` -f | grep ppLSN | awk '{print $9}' | awk '{FS="=";print $2}' | grep KLMN | wc -l
the value of teh command is returned as 7.
but when i am putting the same command inside a script abc_sh like below
ps -u `id | cut -f2 -d"=" | cut -f1 -d"("` -f | grep ppLSN | awk '{print $9}' | awk '{FS="=";print $2}' | grep $XYZ | wc -l
and i am calling the script on the command line as abc_sh XYZ=KLMN and it does not work and returns 0
the problem is with the grep in the command grep $XYZ
could anybody please tell why this is not working?

Answer 1

Because your $1 variable (first argument to the script) is set to XYZ=KLMN.
Just use abc_sh KLMN and grep $1 instead of grep $XYZ.
(Assuming we are talking about bash here)
The other alternative is defining a temporary environment variable in which case you would have to call it like this: XYZ=KLMN abc_sh
EDIT:
Found what you were using, you have to use set -k (see SHELL BUILTIN COMMANDS in the BASH manual)
-k All arguments in the form of assignment statements are
placed in the environment for a command, not just those
that precede the command name.
So
vinko#parrot:~$ more abc
#!/bin/bash
echo $XYZ
vinko#parrot:~$ set -k
vinko#parrot:~$ ./abc XYZ=KLMN
KLMN
vinko#parrot:~$ set +k
vinko#parrot:~$ ./abc XYZ=KLMN
vinko#parrot:~$
So, the place where this was working probably has set -k in one of the startup scripts (bashrc or profile.)

Answer 2

Try any of these to set a temporary environment variable:
XYZ=KLMN abc_sh
env XYZ=KLMN abc_sh
(export XYZ=KLMN; abc_sh)

Answer 3

you are using so many commands chained together....
ps -u `id -u` -f | awk -v x="$XYZ" -v p="ppLSN" '$0~p{
m=split($9,a,"=")
if(a[2]~x){count++}
}
END{print count}'

Answer 4

Call this script:
#!/bin/ksh
ps -u $(id -u) -o args | grep $XYZ | cut -f2- -d " "
Like this:
XYZ=KLMN abc_sh