How to store standard error in a variable

BashShellRedirectVariablesStderr

Bash Problem Overview


Let's say I have a script like the following:

useless.sh

echo "This Is Error" 1>&2
echo "This Is Output" 

And I have another shell script:

alsoUseless.sh

./useless.sh | sed 's/Output/Useless/'

I want to capture "This Is Error", or any other stderr from useless.sh, into a variable. Let's call it ERROR.

Notice that I am using stdout for something. I want to continue using stdout, so redirecting stderr into stdout is not helpful, in this case.

So, basically, I want to do

./useless.sh 2> $ERROR | ...

but that obviously doesn't work.

I also know that I could do

./useless.sh 2> /tmp/Error
ERROR=`cat /tmp/Error`

but that's ugly and unnecessary.

Unfortunately, if no answers turn up here that's what I'm going to have to do.

I'm hoping there's another way.

Anyone have any better ideas?

Bash Solutions


Solution 1 - Bash

It would be neater to capture the error file thus:

ERROR=$(</tmp/Error)

The shell recognizes this and doesn't have to run 'cat' to get the data.

The bigger question is hard. I don't think there's an easy way to do it. You'd have to build the entire pipeline into the sub-shell, eventually sending its final standard output to a file, so that you can redirect the errors to standard output.

ERROR=$( { ./useless.sh | sed s/Output/Useless/ > outfile; } 2>&1 )

Note that the semi-colon is needed (in classic shells - Bourne, Korn - for sure; probably in Bash too). The '{}' does I/O redirection over the enclosed commands. As written, it would capture errors from sed too.

>WARNING: Formally untested code - use at own risk.

Solution 2 - Bash

Redirected stderr to stdout, stdout to /dev/null, and then use the backticks or $() to capture the redirected stderr:

ERROR=$(./useless.sh 2>&1 >/dev/null)

Solution 3 - Bash

alsoUseless.sh

This will allow you to pipe the output of your useless.sh script through a command such as sed and save the stderr in a variable named error. The result of the pipe is sent to stdout for display or to be piped into another command.

It sets up a couple of extra file descriptors to manage the redirections needed in order to do this.

#!/bin/bash

exec 3>&1 4>&2 #set up extra file descriptors

error=$( { ./useless.sh | sed 's/Output/Useless/' 2>&4 1>&3; } 2>&1 )

echo "The message is \"${error}.\""

exec 3>&- 4>&- # release the extra file descriptors

Solution 4 - Bash

There are a lot of duplicates for this question, many of which have a slightly simpler usage scenario where you don't want to capture stderr and stdout and the exit code all at the same time.

if result=$(useless.sh 2>&1); then
    stdout=$result
else
    rc=$?
    stderr=$result
fi

works for the common scenario where you expect either proper output in the case of success, or a diagnostic message on stderr in the case of failure.

Note that the shell's control statements already examine $? under the hood; so anything which looks like

cmd
if [ $? -eq 0 ], then ...

is just a clumsy, unidiomatic way of saying

if cmd; then ...

Solution 5 - Bash

For the benefit of the reader, this recipe here

  • can be re-used as oneliner to catch stderr into a variable
  • still gives access to the return code of the command
  • Sacrifices a temporary file descriptor 3 (which can be changed by you of course)
  • And does not expose this temporary file descriptors to the inner command

If you want to catch stderr of some command into var you can do

{ var="$( { command; } 2>&1 1>&3 3>&- )"; } 3>&1;

Afterwards you have it all:

echo "command gives $? and stderr '$var'";

If command is simple (not something like a | b) you can leave the inner {} away:

{ var="$(command 2>&1 1>&3 3>&-)"; } 3>&1;

Wrapped into an easy reusable bash-function (probably needs version 3 and above for local -n):

: catch-stderr var cmd [args..]
catch-stderr() { local -n v="$1"; shift && { v="$("$@" 2>&1 1>&3 3>&-)"; } 3>&1; }

Explained:

  • local -n aliases "$1" (which is the variable for catch-stderr)
  • 3>&1 uses file descriptor 3 to save there stdout points
  • { command; } (or "$@") then executes the command within the output capturing $(..)
  • Please note that the exact order is important here (doing it the wrong way shuffles the file descriptors wrongly):
    • 2>&1 redirects stderr to the output capturing $(..)
    • 1>&3 redirects stdout away from the output capturing $(..) back to the "outer" stdout which was saved in file descriptor 3. Note that stderr still refers to where FD 1 pointed before: To the output capturing $(..)
    • 3>&- then closes the file descriptor 3 as it is no more needed, such that command does not suddenly has some unknown open file descriptor showing up. Note that the outer shell still has FD 3 open, but command will not see it.
    • The latter is important, because some programs like lvm complain about unexpected file descriptors. And lvm complains to stderr - just what we are going to capture!

You can catch any other file descriptor with this recipe, if you adapt accordingly. Except file descriptor 1 of course (here the redirection logic would be wrong, but for file descriptor 1 you can just use var=$(command) as usual).

Note that this sacrifices file descriptor 3. If you happen to need that file descriptor, feel free to change the number. But be aware, that some shells (from the 1980s) might understand 99>&1 as argument 9 followed by 9>&1 (this is no problem for bash).

Also note that it is not particluar easy to make this FD 3 configurable through a variable. This makes things very unreadable:

: catch-var-from-fd-by-fd variable fd-to-catch fd-to-sacrifice command [args..]
catch-var-from-fd-by-fd()
{
local -n v="$1";
local fd1="$2" fd2="$3";
shift 3 || return;

eval exec "$fd2>&1";
v="$(eval '"$@"' "$fd1>&1" "1>&$fd2" "$fd2>&-")";
eval exec "$fd2>&-";
}

> Security note: The first 3 arguments to catch-var-from-fd-by-fd must not be taken from a 3rd party. Always give them explicitly in a "static" fashion. > > So no-no-no catch-var-from-fd-by-fd $var $fda $fdb $command, never do this! > > If you happen to pass in a variable variable name, at least do it as follows: > local -n var="$var"; catch-var-from-fd-by-fd var 3 5 $command > > This still will not protect you against every exploit, but at least helps to detect and avoid common scripting errors.

Notes:

  • catch-var-from-fd-by-fd var 2 3 cmd.. is the same as catch-stderr var cmd..
  • shift || return is just some way to prevent ugly errors in case you forget to give the correct number of arguments. Perhaps terminating the shell would be another way (but this makes it hard to test from commandline).
  • The routine was written such, that it is more easy to understand. One can rewrite the function such that it does not need exec, but then it gets really ugly.
  • This routine can be rewritten for non-bash as well such that there is no need for local -n. However then you cannot use local variables and it gets extremely ugly!
  • Also note that the evals are used in a safe fashion. Usually eval is considerered dangerous. However in this case it is no more evil than using "$@" (to execute arbitrary commands). However please be sure to use the exact and correct quoting as shown here (else it becomes very very dangerous).

Solution 6 - Bash

# command receives its input from stdin.
# command sends its output to stdout.
exec 3>&1
stderr="$(command </dev/stdin 2>&1 1>&3)"
exitcode="${?}"
echo "STDERR: $stderr"
exit ${exitcode}

Solution 7 - Bash

POSIX

STDERR can be captured with some redirection magic:

$ { error=$( { { ls -ld /XXXX /bin | tr o Z ; } 1>&3 ; } 2>&1); } 3>&1
lrwxrwxrwx 1 rZZt rZZt 7 Aug 22 15:44 /bin -> usr/bin/

$ echo $error
ls: cannot access '/XXXX': No such file or directory

Note that piping of STDOUT of the command (here ls) is done inside the innermost { }. If you're executing a simple command (eg, not a pipe), you could remove these inner braces.

You can't pipe outside the command as piping makes a subshell in bash and zsh, and the assignment to the variable in the subshell wouldn't be available to the current shell.

bash

In bash, it would be better not to assume that file descriptor 3 is unused:

{ error=$( { { ls -ld /XXXX /bin | tr o Z ; } 1>&$tmp ; } 2>&1); } {tmp}>&1; 
exec {tmp}>&-  # With this syntax the FD stays open

Note that this doesn't work in zsh.


Thanks to this answer for the general idea.

Solution 8 - Bash

Iterating a bit on Tom Hale's answer, I've found it possible to wrap the redirection yoga into a function for easier reuse. For example:

#!/bin/sh

capture () {
    { captured=$( { { "$@" ; } 1>&3 ; } 2>&1); } 3>&1
}

# Example usage; capturing dialog's output without resorting to temp files
# was what motivated me to search for this particular SO question
capture dialog --menu "Pick one!" 0 0 0 \
        "FOO" "Foo" \
        "BAR" "Bar" \
        "BAZ" "Baz"
choice=$captured

clear; echo $choice

It's almost certainly possible to simplify this further. Haven't tested especially-thoroughly, but it does appear to work with both bash and ksh.


EDIT: an alternative version of the capture function which stores the captured STDERR output into a user-specified variable (instead of relying on a global $captured), taking inspiration from Léa Gris's answer while preserving the ksh (and zsh) compatibility of the above implementation:

capture () {
    if [ "$#" -lt 2 ]; then
        echo "Usage: capture varname command [arg ...]"
        return 1
    fi
    typeset var captured; captured="$1"; shift
    { read $captured <<<$( { { "$@" ; } 1>&3 ; } 2>&1); } 3>&1
}

And usage:

capture choice dialog --menu "Pick one!" 0 0 0 \
        "FOO" "Foo" \
        "BAR" "Bar" \
        "BAZ" "Baz"

clear; echo $choice

Solution 9 - Bash

Here's how I did it :

#
# $1 - name of the (global) variable where the contents of stderr will be stored
# $2 - command to be executed
#
captureStderr()
{
    local tmpFile=$(mktemp)

    $2 2> $tmpFile
    
    eval "$1=$(< $tmpFile)"
    
    rm $tmpFile
}

Usage example :

captureStderr err "./useless.sh"

echo -$err-

It does use a temporary file. But at least the ugly stuff is wrapped in a function.

Solution 10 - Bash

A simple solution

{ ERROR=$(./useless.sh 2>&1 1>&$out); } {out}>&1
echo "-"
echo $ERROR

Will produce:

This Is Output
-
This Is Error

Solution 11 - Bash

This is an interesting problem to which I hoped there was an elegant solution. Sadly, I end up with a solution similar to Mr. Leffler, but I'll add that you can call useless from inside a Bash function for improved readability:

#!/bin/bash

function useless { /tmp/useless.sh | sed 's/Output/Useless/' }

ERROR=$(useless) echo $ERROR

All other kind of output redirection must be backed by a temporary file.

Solution 12 - Bash

This post helped me come up with a similar solution for my own purposes:

MESSAGE=`{ echo $ERROR_MESSAGE | format_logs.py --level=ERROR; } 2>&1`

Then as long as our MESSAGE is not an empty string, we pass it on to other stuff. This will let us know if our format_logs.py failed with some kind of python exception.

Solution 13 - Bash

In zsh:

{ . ./useless.sh > /dev/tty } 2>&1 | read ERROR
$ echo $ERROR
( your message )

Solution 14 - Bash

Capture AND Print stderr

ERROR=$( ./useless.sh 3>&1 1>&2 2>&3 | tee /dev/fd/2 )

Breakdown

You can use $() to capture stdout, but you want to capture stderr instead. So you swap stdout and stderr. Using fd 3 as the temporary storage in the standard swap algorithm.

If you want to capture AND print use tee to make a duplicate. In this case the output of tee will be captured by $() rather than go to the console, but stderr(of tee) will still go to the console so we use that as the second output for tee via the special file /dev/fd/2 since tee expects a file path rather than a fd number.

NOTE: That is an awful lot of redirections in a single line and the order matters. $() is grabbing the stdout of tee at the end of the pipeline and the pipeline itself routes stdout of ./useless.sh to the stdin of tee AFTER we swapped stdin and stdout for ./useless.sh.

Using stdout of ./useless.sh

The OP said he still wanted to use (not just print) stdout, like ./useless.sh | sed 's/Output/Useless/'.

No problem just do it BEFORE swapping stdout and stderr. I recommend moving it into a function or file (also-useless.sh) and calling that in place of ./useless.sh in the line above.

However, if you want to CAPTURE stdout AND stderr, then I think you have to fall back on temporary files because $() will only do one at a time and it makes a subshell from which you cannot return variables.

Solution 15 - Bash

Improving on YellowApple's answer:

This is a Bash function to capture stderr into any variable

stderr_capture_example.sh:

#!/usr/bin/env bash

# Capture stderr from a command to a variable while maintaining stdout
# @Args:
# $1: The variable name to store the stderr output
# $2: Vararg command and arguments
# @Return:
# The Command's Returnn-Code or 2 if missing arguments
function capture_stderr {
  [ $# -lt 2 ] && return 2
  local stderr="$1"
  shift
  {
    printf -v "$stderr" '%s' "$({ "$@" 1>&3; } 2>&1)"
  } 3>&1
}

# Testing with a call to erroring ls
LANG=C capture_stderr my_stderr ls "$0" ''

printf '\nmy_stderr contains:\n%s' "$my_stderr"

Testing:

bash stderr_capture_example.sh

Output:

 stderr_capture_example.sh

my_stderr contains:
ls: cannot access '': No such file or directory

This function can be used to capture the returned choice of a dialog command.

Solution 16 - Bash

I think you want to capture stderr, stdout and exitcode if that is your intention you can use this code:

## Capture error when 'some_command() is executed
some_command_with_err() {
    echo 'this is the stdout'
    echo 'this is the stderr' >&2
    exit 1
}

run_command() {
    {
        IFS=$'\n' read -r -d '' stderr;
        IFS=$'\n' read -r -d '' stdout;
        IFS=$'\n' read -r -d '' stdexit;
    } < <((printf '\0%s\0%d\0' "$(some_command_with_err)" "${?}" 1>&2) 2>&1)
    stdexit=${stdexit:-0};
}

echo 'Run command:'
if ! run_command; then
    ## Show the values
    typeset -p stdout stderr stdexit
else
    typeset -p stdout stderr stdexit
fi

This scripts capture the stderr, stdout as well as the exitcode.

But Teo how it works?

First, we capture the stdout as well as the exitcode using printf '\0%s\0%d\0'. They are separated by the \0 aka 'null byte'.

After that, we redirect the printf to stderr by doing: 1>&2 and then we redirect all back to stdout using 2>&1. Therefore, the stdout will look like:

"<stderr>\0<stdout>\0<exitcode>\0"

Enclosing the printf command in <( ... ) performs process substitution. Process substitution allows a process’s input or output to be referred to using a filename. This means <( ... ) will pipe the stdout of (printf '\0%s\0%d\0' "$(some_command_with_err)" "${?}" 1>&2) 2>&1into the stdin of the command group using the first <.

Then, we can capture the piped stdout from the stdin of the command group with read. This command reads a line from the file descriptor stdin and split it into fields. Only the characters found in $IFS are recognized as word delimiters. $IFS or Internal Field Separator is a variable that determines how Bash recognizes fields, or word boundaries, when it interprets character strings. $IFS defaults to whitespace (space, tab, and newline), but may be changed, for example, to parse a comma-separated data file. Note that $* uses the first character held in $IFS.

## Shows whitespace as a single space, ^I(horizontal tab), and newline, and display "$" at end-of-line.
echo "$IFS" | cat -vte
# Output:
# ^I$
# $

## Reads commands from string and assign any arguments to pos params
bash -c 'set w x y z; IFS=":-;"; echo "$*"'
# Output:
# w:x:y:z

for l in $(printf %b 'a b\nc'); do echo "$l"; done
# Output: 
# a
# b
# c

IFS=$'\n'; for l in $(printf %b 'a b\nc'); do echo "$l"; done
# Output: 
# a b
# c

That is why we defined IFS=$'\n' (newline) as delimiter. Our script uses read -r -d '', where read -r does not allow backslashes to escape any characters, and -d '' continues until the first character '' is read, rather than newline.

Finally, replace some_command_with_err with your script file and you can capture and handle the stderr, stdout as well as the exitcode as your will.

Solution 17 - Bash

If you want to bypass the use of a temporary file you may be able to use process substitution. I haven't quite gotten it to work yet. This was my first attempt:

$ .useless.sh 2> >( ERROR=$(<) )
-bash: command substitution: line 42: syntax error near unexpected token `)'
-bash: command substitution: line 42: `<)'

Then I tried

$ ./useless.sh 2> >( ERROR=$( cat <() )  )
This Is Output
$ echo $ERROR   # $ERROR is empty

However

$ ./useless.sh 2> >( cat <() > asdf.txt )
This Is Output
$ cat asdf.txt
This Is Error

So the process substitution is doing generally the right thing... unfortunately, whenever I wrap STDIN inside >( ) with something in $() in an attempt to capture that to a variable, I lose the contents of $(). I think that this is because $() launches a sub process which no longer has access to the file descriptor in /dev/fd which is owned by the parent process.

Process substitution has bought me the ability to work with a data stream which is no longer in STDERR, unfortunately I don't seem to be able to manipulate it the way that I want.

Solution 18 - Bash

$ b=$( ( a=$( (echo stdout;echo stderr >&2) ) ) 2>&1 )
$ echo "a=>$a b=>$b"
a=>stdout b=>stderr

Solution 19 - Bash

For error proofing your commands:

execute [INVOKING-FUNCTION] [COMMAND]

execute () {
    function="${1}"
    command="${2}"
    error=$(eval "${command}" 2>&1 >"/dev/null")

    if [ ${?} -ne 0 ]; then
        echo "${function}: ${error}"
        exit 1
    fi
}

Inspired in Lean manufacturing:

Solution 20 - Bash

I'll use find command

find / -maxdepth 2 -iname 'tmp' -type d

as non superuser for the demo. It should complain 'Permission denied' when acessing / dir.

#!/bin/bash

echo "terminal:"
{ err="$(find / -maxdepth 2 -iname 'tmp' -type d 2>&1 1>&3 3>&- | tee /dev/stderr)"; } 3>&1 | tee /dev/fd/4 2>&1; out=$(cat /dev/fd/4)
echo "stdout:" && echo "$out"
echo "stderr:" && echo "$err"

that gives output:

terminal:
find: ‘/root’: Permission denied
/tmp
/var/tmp
find: ‘/lost+found’: Permission denied
stdout:
/tmp
/var/tmp
stderr:
find: ‘/root’: Permission denied
find: ‘/lost+found’: Permission denied

The terminal output has also /dev/stderr content the same way as if you were running that find command without any script. $out has /dev/stdout and $err has /dev/stderr content.

use:

#!/bin/bash

echo "terminal:"
{ err="$(find / -maxdepth 2 -iname 'tmp' -type d 2>&1 1>&3 3>&-)"; } 3>&1 | tee /dev/fd/4; out=$(cat /dev/fd/4)
echo "stdout:" && echo "$out"
echo "stderr:" && echo "$err"

if you don't want to see /dev/stderr in the terminal output.

terminal:
/tmp
/var/tmp
stdout:
/tmp
/var/tmp
stderr:
find: ‘/root’: Permission denied
find: ‘/lost+found’: Permission denied

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Solution 1 - BashJonathan LefflerView Answer on Stackoverflow
Solution 2 - BashChas. OwensView Answer on Stackoverflow
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