Capture both stdout and stderr in Bash

Question

I know this syntax

var=`myscript.sh`

or

var=$(myscript.sh)

Will capture the result (stdout) of myscript.sh into var. I could redirect stderr into stdout if I wanted to capture both. How to save each of them to separate variables?

My use case here is if the return code is nonzero I want to echo stderr and suppress otherwise. There may be other ways to do this but this approach seems it will work, if it's actually possible.

Answer 1

There's a really ugly way to capture stderr and stdout in two separate variables without temporary files (if you like plumbing), using process substitution, source, and declare appropriately. I'll call your command banana. You can mimic such a command with a function:

banana() {
    echo "banana to stdout"
    echo >&2 "banana to stderr"
}

I'll assume you want standard output of banana in variable bout and standard error of banana in variable berr. Here's the magic that'll achieve that (Bash≥4 only):

. <({ berr=$({ bout=$(banana); } 2>&1; declare -p bout >&2); declare -p berr; } 2>&1)

---

So, what's happening here?

Let's start from the innermost term:

bout=$(banana)

This is just the standard way to assign to bout the standard output of banana, the standard error being displayed on your terminal.

Then:

{ bout=$(banana); } 2>&1

will still assign to bout the stdout of banana, but the stderr of banana is displayed on terminal via stdout (thanks to the redirection 2>&1.

Then:

{ bout=$(banana); } 2>&1; declare -p bout >&2

will do as above, but will also display on the terminal (via stderr) the content of bout with the declare builtin: this will be reused soon.

Then:

berr=$({ bout=$(banana); } 2>&1; declare -p bout >&2); declare -p berr

will assign to berr the stderr of banana and display the content of berr with declare.

At this point, you'll have on your terminal screen:

declare -- bout="banana to stdout"
declare -- berr="banana to stderr"

with the line

declare -- bout="banana to stdout"

being displayed via stderr.

A final redirection:

{ berr=$({ bout=$(banana); } 2>&1; declare -p bout >&2); declare -p berr; } 2>&1

will have the previous displayed via stdout.

Finally, we use a process substitution to source the content of these lines.

---

You mentioned the return code of the command too. Change banana to:

banana() {
    echo "banana to stdout"
    echo >&2 "banana to stderr"
    return 42
}

We'll also have the return code of banana in the variable bret like so:

. <({ berr=$({ bout=$(banana); bret=$?; } 2>&1; declare -p bout bret >&2); declare -p berr; } 2>&1)

---

You can do without sourcing and a process substitution by using eval too (and it works with Bash<4 too):

eval "$({ berr=$({ bout=$(banana); bret=$?; } 2>&1; declare -p bout bret >&2); declare -p berr; } 2>&1)"

And all this is safe, because the only stuff we're sourceing or evaling are obtained from declare -p and will always be properly escaped.

---

Of course, if you want the output in an array (e.g., with mapfile, if you're using Bash≥4—otherwise replace mapfile with a while–read loop), the adaptation is straightforward.

For example:

banana() {
    printf 'banana to stdout %d\n' {1..10}
    echo >&2 'banana to stderr'
    return 42
}

. <({ berr=$({ mapfile -t bout < <(banana); } 2>&1; declare -p bout >&2); declare -p berr; } 2>&1)

and with return code:

. <({ berr=$({ mapfile -t bout< <(banana; bret=$?; declare -p bret >&3); } 3>&2 2>&1; declare -p bout >&2); declare -p berr; } 2>&1)