Bash – How to set a variable to the output of a command in Bash

bashcommand-lineshell

I have a pretty simple script that is something like the following:

#!/bin/bash

VAR1="$1"
MOREF='sudo run command against $VAR1 | grep name | cut -c7-'

echo $MOREF

When I run this script from the command line and pass it the arguments, I am not getting any output. However, when I run the commands contained within the $MOREF variable, I am able to get output.

How can one take the results of a command that needs to be run within a script, save it to a variable, and then output that variable on the screen?

Best Solution

In addition to backticks `command`, command substitution can be done with $(command) or "$(command)", which I find easier to read, and allows for nesting.

OUTPUT=$(ls -1)
echo "${OUTPUT}"

MULTILINE=$(ls \
   -1)
echo "${MULTILINE}"

Quoting (") does matter to preserve multi-line variable values; it is optional on the right-hand side of an assignment, as word splitting is not performed, so OUTPUT=$(ls -1) would work fine.