Catch bash exit code in if statement [duplicate]

Question

If I do a simple test of exit codes directly with single commands:

[user@localhost ~]$ date Tue Sep 8 14:00:11 CEST 2020 [user@localhost ~]$ echo $? 0 [user@localhost ~]$ dat bash: dat: command not found... [user@localhost ~]$ echo $? 127

But if I want to get them into an if sentence, I see that for the exit code 0 the output is OK, however, for the "command not found" code (it should be 127) it returns "1".

[user@localhost ~]$ date Tue Sep 8 14:02:18 CEST 2020 [user@localhost ~]$ if [ $? -eq 0 ]; then echo "Success: $?"; else echo "Failure: $?" >&2; fi Success: 0 [user@localhost ~]$ dat bash: dat: command not found... [user@localhost ~]$ if [ $? -eq 0 ]; then echo "Success: $?"; else echo "Failure: $?" >&2; fi Failure: 1

What am I missing?

The [ command modifies $? too. Use: dat; status=$?; if [ $status = 0 ]; then echo "Success: $status"; else echo "Failure: $status"; fi. — Jonathan Leffler
– Jonathan Leffler, Commented Sep 8, 2020 at 12:11
The $? in the else clause is the exit status of [ $? -eq 0 ] in the if clause. — M. Nejat Aydin
– M. Nejat Aydin, Commented Sep 8, 2020 at 12:14
@ZumodeVidrio : Looks like a XY-problem to me. What your use case? What do you want to achieve? — user1934428
– user1934428, Commented Sep 8, 2020 at 13:19
@user1934428 No no, it's not an XY-problem at all, I was just trying to add some debug into a script and wanted to take into account the error code in case it happens. — Zumo de Vidrio
– Zumo de Vidrio, Commented Sep 8, 2020 at 14:14
Then, why only 0 and 127? Codes larger 127 are virtually always errors occuring when starting a process, and most programs use codes between 1 (or 2) and 126 for errors too. And why do you want to continue on failure? Wouldn't a set -e better fulfil what you want? — user1934428
– user1934428, Commented Sep 8, 2020 at 14:38

user1934428 · Accepted Answer · 2020-09-08 15:37:40Z

7

If you want to catch and print the exit code for each command, I would not bother saving it in a variable. This would be necessary only if you need this exit code agains at a later time, or if there are some non-zero exit codes which you don't regard as error.

Instead, I would do it like this:

Replace your existing

command_to_verify

by

if command_to_verify then echo Success else echo Failure: $? fi

answered Sep 8, 2020 at 15:37

user1934428

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3 Comments

tripleee Over a year ago

See also Why is testing “$?” to see if a command succeeded or not, an anti-pattern?

Thick_propheT Over a year ago

I had code very similar to this that was not working that led me to this question. if ! command_to_verify; then echo Failure: $?; fi was not working.

user1934428 Over a year ago

If you write, ! command_to_verify, the ! changes $?, and this is the reason why you loose the actual status value of your command.

0stone0 · Accepted Answer · 2020-09-08 12:19:39Z

Every command has it own exit state, so the command [ ¹ will override $?. To prevent this, save the exit code to a variable, and use that in the if statement;

le=$? if [ "$le" -eq 0 ]; then echo "Success: $le"; else echo "Failure: $le" >&2; fi

Try it online!

1: The single bracket [ is actually an alias for the test command, it's not syntax. For more info, check this comment on unix.stackexchange

I think printing $le in the success case is redundandt; we know that it must be 0 here.

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