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

Question

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)

@Jahid: no, because a failing command will interrupt script, which is what I want to avoid — blueFast
– blueFast, Commented Nov 11, 2016 at 11:56
@Jahid: Maybe false ; res1=$? would be better? Would that record the result of the previous command, without interrupting script? — blueFast
– blueFast, Commented Nov 11, 2016 at 11:57

Jahid · Accepted Answer · 2016-11-11 12:34:06Z

6

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.)

edited Nov 11, 2016 at 12:34

Jahid

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answered Nov 11, 2016 at 12:15

chepner

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3 Comments

blueFast Over a year ago

There you are talking!

chepner Over a year ago

@Jahid Thanks; no idea why I typed $! originally.

Jahid Over a year ago

This kind of thing sometimes happens :D. When I upvoted your answer, there were three $!, yet I didn't notice that. Even after three upvotes, no-one noticed... :D

Community · Accepted Answer · 2017-05-23 12:30:23Z

1

Putting the && after || doesn't have the effect that you want. For example,

false || echo true && echo false

will print both true and false, i.e whatever after && will be executed whether the first command was successful or not. You can do it like this:

command && res1=$? || res1=$?

In this case, if command succeeds, the third one doesn't get executed, if fails, the second one doesn't get executed.

EDIT

As chepnar mentioned, the use of && and || isn't full-proof. The behavior isn't guaranteed.

edited May 23, 2017 at 12:30

CommunityBot

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answered Nov 11, 2016 at 12:07

Jahid

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1 Comment

blueFast Over a year ago

Ok, so my original implementation is the right one. Thanks!

anubhava · Accepted Answer · 2016-11-11 12:17:16Z

0

Rather than capturing same exit code twice why not just turn off -e, capture the exit status and turn on -e again like this:

#!/bin/bash
set -e

set +e          # turn off exit on error
false; res1=$?  # run command1 and capture exit status1
true; res2=$?   # run command2 and capture exit status2
set -e          # turn on exit on error

echo $res1
echo $res2

answered Nov 11, 2016 at 12:17

anubhava

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4 Comments

blueFast Over a year ago

Thanks, but I want to avoid having set +e / set -e all around my code.

anubhava Over a year ago

Hmm ok , but it will be cleaner than capturing same $? twice.

blueFast Over a year ago

Solution by @chepner is very clean.

anubhava Over a year ago

Yes, totally agree with that.

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