Capturing stderr, but not stdout [duplicate]

Question

In my bash script, I read in a set of lines that look like this

arg $PROG arg arg

I want to be able to run the line, capture the STDERR as a variable, and prevent either the STDOUT or STDERR from printing to the screen. This is my solution so far but the error variable is always empty.

$PROG=/c/Program1
{ error=$($(eval $line) 2>&1 1>&$out); } {out}>&1
echo $error

Please explain solutions thoroughly. I am new to bash shell scripting and trying to learn.

Answer 1

For a command that prints to both stdout and stderr like this one:

mycommand () { echo "stdout"; echo "stderr" >&2; }

If I issue that normally, I see all the output:

$ mycommand
stdout
stderr

I can redirect stdout or stderr to /dev/null so it doesn't get printed:

$ mycommand >/dev/null    # Equivalent to 1>/dev/null
stderr
$ mycommand 2>/dev/null
stdout

If I want to capture only stderr, I first have to redirect stderr to where stdout out is pointing to at the moment, then redirect stdout to /dev/null:

error="$(2>&1 1>/dev/null mycommand)"

Now error only contains output from stderr:

$ echo "$error"
stderr

The position on the line doesn't really matter, but the order of redirections does, so this has the same effect:

error="$(mycommand 2>&1 1>/dev/null)"

but this doesn't:

error="$(1>/dev/null 2>&1 mycommand)"

The last command would redirect both stdout and stderr to /dev/null, and error would be empty.

A great resource for redirection is on the Bash Hackers Wiki.

---

A few pointers for what you tried:

• Assignment requires to not use the $ in the parameter name, so it should be PROG=/c/Program1 and not $PROG=/c/Program1.
• Uppercase variable names are discouraged because they are more likely to clash with environment variables.
• In 99% of the cases, there is a solution that avoids eval, see BashFAQ/048 why that is a good idea.