Use bash script $1 argument in awk command

Question

I want to create a bash script that takes an argument and using awk to print if

it's a orphan process :

#find_orphan.sh :

ps -elf | awk '/$1/&&!/awk/ {if ($5 == 1){print $15}}'

Unfortunately I'm having problems in order to render the $1 as a bash argument..

if I run this without using $1 my program will work :

ps -elf | awk '/some_program/&&!/awk/ {if ($5 == 1){print $15}}'

I really need to be able to specify the awk search pattern in the bash script argument.

Any help ?

Answer 1

Use awk -v var=val to pass a command line value to awk and use $0 ~ val syntax to compare it using regex:

ps -elf | awk -v arg="$1" '$0 ~ arg && !/awk/ && $5 == 1{print $15}'

Without regex:

ps -elf | awk -v arg="$1" 'index($0, arg) && !/awk/ && $5 == 1{print $15}'