linux - [Solved-5 Solutions] How to set a variable to the output from a command in Bash ? - ubuntu - red hat - debian - linux server - linux pc
Linux - Problem :
How to set a variable to the output from a command in Bash ?
Linux - Solution 1:
In addition to the backticks, you can use $().
OUTPUT="$(ls -1)"
echo "${OUTPUT}"click below button to copy the code. By - Linux tutorial - team
Quoting (") does matter to preserve multi-line values.
Linux - Solution 2:
You're using the wrong kind of apostrophe. You need `, not '. This character is called "backticks" (or "grave accent").
#!/bin/bash
VAR1="$1"
VAR2="$2"
MOREF=`sudo run command against $VAR1 | grep name | cut -c7-`
echo $MOREFclick below button to copy the code. By - Linux tutorial - team
Linux - Solution 3:
The $(command) works as well, and it also easier to read, but note that it is valid only with bash or korn shells (and shells derived from those), so if your scripts have to be really portable on various Unix systems, you should prefer the old backticks notation.
Linux - Solution 4:
There are three ways to do:
1) Functions:
func (){
ls -l
}click below button to copy the code. By - Linux tutorial - team
2) Another suitable solution could be eval:
var="ls -l"
eval $varclick below button to copy the code. By - Linux tutorial - team
3) Using variables directly:
var=$(ls -l)
OR
var=`ls -l`click below button to copy the code. By - Linux tutorial - team
You can get output of third solution in good way:
echo "$var"click below button to copy the code. By - Linux tutorial - team
and also in nasty way:
echo $varclick below button to copy the code. By - Linux tutorial - team
Linux - Solution 5:
MOREF=$(sudo run command against $VAR1 | grep name | cut -c7-)