linux - How to set a variable to the output from a command in Bash? -

i working on simple scripting project work involves use of bash. have pretty simple script following:

#!/bin/bash  var1="$1" var2="$2"  moref='sudo run command against $var1 | grep name | cut -c7-'  echo $moref 

when run script command line , pass arguments, not getting output. however, when run commands contained within $moref variable, able output. know how 1 can take results of command needs run within script, save variable, , output variable on screen?

in addition backticks, can use $(), find easier read, , allows nesting.

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

quoting (") matter preserve multi-line values.