delavnog delavnog - 21 days ago 6
Linux Question

Do not interrupt script if command fails, but collect exit code

I need to execute a command in a script with

set -e
set. This command is an exception to the general script flow, in that it could fail and that would not be critical: the script can continue. I want that command to not interrupt the script; instead I want to keep the exit code for later evaluation.

I have come up with the following script:

#!/bin/bash

set -e
false && res1=$? || res1=$?
true && res2=$? || res2=$?
echo $res1
echo $res2


(My command would be in place of
false
or
true
)

Is this the right approach?

EDIT



Incidentally, this very similar construct does not work at all:

#!/bin/bash

set -e
false || res1=$? && res1=$?
true || res2=$? && res2=$?
echo $res1
echo $res2


EDIT



Testing one of the suggestions:

#!/bin/bash

set -e
false || { res1=$?; true; }
true || { res2=$?; true; }
echo $res1
echo $res2


This does not work. Result is:

1
(empty line)

Answer

Don't try to use && and || to form a ternary expression; it's too fragile. Just use an if statement.

if cmd; then res=$?; else res=$?; fi

Better(?) yet, you can put the assignment in the condition itself, reducing repetition at the cost of having a seemingly vacuous if statement:

if cmd; res=$?; then :; fi

That might be more clearly written with a single || though:

{ cmd; res=$?; } || true

(Notice I've used true and : interchangeably; : is guaranteed by POSIX to be a shell built-in, but true is probably more readable.)

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