bash variable capture stderr and stdout separately or get exit value

user_263328 0
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I need to capture the output and error of a command in my bash script and know whether the command succeeded or not.

At the moment, I am capturing both like this:

output=$(mycommand 2>&1)

I then need to check the exit value of mycommand. If it failed, I need to do some stuff with the output, if the command succeeded, I don't need to touch the output.

Since I am capturing the output, checking $? is always a 0 since bash succeeded at capturing the output into the variable.

This is a very time sensitive script, so we are trying to avoid any slower solutions like outputting to a file and re-reading it in.

If I could capture stdout to one variable and stderr to another, that would solve my problem because I could just check if the error variable was empty or not.

Thanks.

stderr capture stdout variables bash
2 Answers

The problem only seems to manifest when the output is captured to a local variable within a function:

$ echo $BASH_VERSION
3.2.48(1)-release
$ false; echo $?
1
$ echo $?
0
$ x=$(false 2>&1) ; echo $?
1
$ function f {
> local x=$(false 2>&1) ; echo $?
> }
$ f
0
$ function g {
> x=$(false 2>&1) ; echo $?
> }
$ g
1

Notice that only function f, which captures x to a local, can express the behavior. Particularly, function g which does the same thing, but without the 'local' keyword, works.

One can therefore not use a local variable, and perhaps 'unset' it after use.

EDIT NVRAM points out that the local declaration can be made beforehand to avoid the issue:

$ function h {
>   local x
>   x=$(false 2>&1) ; echo $?
> }
$ h
1
user_5804120
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