Bash script - store stderr in a variable [duplicate]

Question

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• How to store standard error in a variable 15 answers

I'm writing a script to backup a database. I have the following line:

mysqldump --user=$dbuser --password=$dbpswd  \
   --host=$host $mysqldb | gzip > $filename

I want to assign the stderr to a variable, so that it will send an email to myself letting me know what happened if something goes wrong. I've found solutions to redirect stderr to stdout, but I can't do that as the stdout is already being sent (via gzip) to a file. How can I separately store stderr in a variable $result ?

Answer 1

Try redirecting stderr to stdout and using $() to capture that. In other words:

VAR=$((your-command-including-redirect) 2>&1)

Since your command redirects stdout somewhere, it shouldn't interfere with stderr. There might be a cleaner way to write it, but that should work.

Edit:

This really does work. I've tested it:

#!/bin/bash                                                                                                                                                                         
BLAH=$((
(
echo out >&1
echo err >&2
) 1>log
) 2>&1)

echo "BLAH=$BLAH"

will print BLAH=err and the file log contains out.

Answer 2

For any generic command in Bash, you can do something like this:

{ error=$(command 2>&1 1>&$out); } {out}>&1

Regular output appears normally, anything to stderr is captured in $error (quote it as "$error" when using it to preserve newlines). To capture stdout to a file, just add a redirection at the end, for example:

{ error=$(ls /etc/passwd /etc/bad 2>&1 1>&$out); } {out}>&1 >output

Breaking it down, reading from the outside in, it:

• creates a file description $out for the whole block, duplicating stdout
• captures the stdout of the whole command in $error (but see below)
• the command itself redirects stderr to stdout (which gets captured above) then stdout to the original stdout from outside the block, so only the stderr gets captured

Answer 3

You can save the stdout reference from before it is redirected in another file number (e.g. 3) and then redirect stderr to that:

result=$(mysqldump --user=$dbuser --password=$dbpswd  \
   --host=$host $mysqldb 3>&1 2>&3 | gzip > $filename)

So 3>&1 will redirect file number 3 to stdout (notice this is before stdout is redirected with the pipe). Then 2>&3 redirects stderr to file number 3, which now is the same as stdout. Finally stdout is redirected by being fed into a pipe, but this is not affecting file numbers 2 and 3 (notice that redirecting stdout from gzip is unrelated to the outputs from the mysqldump command).

Edit: Updated the command to redirect stderr from the mysqldump command and not gzip, I was too quick in my first answer.

Answer 4

dd writes both stdout and stderr:

$ dd if=/dev/zero count=50 > /dev/null 
50+0 records in
50+0 records out

the two streams are independent and separately redirectable:

$ dd if=/dev/zero count=50 2> countfile | wc -c
25600
$ cat countfile 
50+0 records in
50+0 records out
$ mail -s "countfile for you" thornate < countfile

if you really needed a variable:

$ variable=`cat countfile`