double quotation bash scripting linux [duplicate]

Question

I do not understand why the output is the username because in line 3 and 4 must print /usr/bin/whoami. please explantation simple to me

#!/bin/bash

WHEREWHOAMI="`which whoami`"
ROOTORNOT="`$WHEREWHOAMI`"
echo "$ROOTORNOT"

Run the script with set -x to see what the commands expand to. — Benjamin W.
– Benjamin W., Commented Mar 19, 2019 at 19:56
Possible duplicate of What is the difference between $(command) and `command` in shell programming? and Command substitution: backticks or dollar sign / paren enclosed? — jww
– jww, Commented Mar 20, 2019 at 1:01
Related to using which, you usually want to use command -v. Also see How to check if a program exists from a Bash script? and How to check if command exists in a shell script? — jww
– jww, Commented Mar 20, 2019 at 1:02

kvantour · Accepted Answer · 2019-03-19 20:00:35Z

1

The variable ROOTORNOT is set to the output of the execution of WHEREWHOAMI which in turn is the output of the command which whoami.

WHEREWHOAMI=`which whoami`  # <- /usr/bin/whoami
ROOTWHOAMI="`$WHEREWHOAMI`" # <- `/usr/bin/whoami`  # <- username

You can easily figure out what is going on if you add the set -x flag to your script. Example:

$ set -x
$ WHEREWHOAMI="`which whoami`"
++ alias
++ declare -f
++ /usr/bin/which --tty-only --read-alias --read-functions --show-tilde --show-dot whoami
+ WHEREWHOAMI=/usr/bin/whoami
$ ROOTORNOT="`$WHEREWHOAMI`"
++ /usr/bin/whoami
+ ROOTORNOT=kvantour
$ echo "$ROOTORNOT"
+ echo kvantour
kvantour
$

answered Mar 19, 2019 at 20:00

kvantour

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Comments

Paul Hodges · Accepted Answer · 2019-03-19 20:15:33Z

1

Backticks are evaluated even inside double quotes.
(Suggestion - don't use backticks. use $() instead.)

WHEREWHOAMI="`which whoami`"

This executes which whoami and assigns /usr/bin/whoami to WHEREWHOAMI.

ROOTORNOT="`$WHEREWHOAMI`"

This executes /usr/bin/whoami in backticks, and assigns the USERNAME result to ROOTORNOT.

It's doing exactly what it should.
Is that not what you indended?

Perhaps what you wanted was something like -

$: [[ $( $(which whoami) ) == root ]] && echo ROOT || echo not-root
not-root

Though I do suggest storing the value and comparing that. Is there a reason you can't just use

if [[ root == "$LOGNAME" ]] 
then : ...

?

answered Mar 19, 2019 at 20:15

Paul Hodges

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1 Comment

vahid peivandy Over a year ago

thanks and yes dear paul

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