How do i store the output of a bash command in a variable? [duplicate]

Answer 1

You need to use echo (or printf) to actually put the value of $LINE onto the standard input of the awk command.

LINE=$(ps aux | grep "$1")
PROCESS=$(echo "$LINE" | awk '{print $2}')
echo $PROCESS
kill -9 $PROCESS

There's no need use LINE; you can set PROCESS with a single line

PROCESS=$(ps aux | grep "$1" | awk '{print $2}')

or better, skip the grep:

PROCESS=$(ps aux | awk -v pname="$1" '$1 ~ pname {print $2}')

Finally, don't use kill -9; that's a last resort for debugging faulty programs. For any program that you didn't write yourself, kill "$PROCESS" should be sufficient.

Answer 2

PROCESS=$(echo "$LINE" | awk '{print $2}')

or

PROCESS=$(ps aux | grep "$1" | awk '{print $2}')

I don't know why you're getting the error you quoted. I can't reproduce it. When you say this:

PROCESS=$LINE | awk '{print $2}'

the shell expands it to something like this:

PROCESS='mayoff  10732 ...' | awk '{print $2}'

(I've shortened the value of $LINE to make the example readable.)

The first subcommand of the pipeline sets variable PROCESS; this variable-setting command has no output so awk reads EOF immediately and prints nothing. And since each subcommand of the pipeline runs in a subshell, the setting of PROCESS takes place only in a subshell, not in the parent shell running the script, so PROCESS is still not set for later commands in your script.

(Note that some versions of bash can run the last subcommand of the pipeline in the current shell instead of in a subshell, but that doesn't affect this example.)

Instead of setting PROCESS in a subshell and feeding nothing to awk on standard input, you want to feed the value of LINE to awk and store the result in PROCESS in the current shell. So you need to run a command that writes the value of LINE to its standard output, and connects that standard output to the standard input of awk. The echo command can do this (or the printf command, as chepner pointed out in his answer).