I want to lint a file, and print the stderr (error message), but do not print the stdout (saying the file is OK). <pre class="prettyprint"><code>php -l "foo/bar.php" </code></pre> <ul> <li>If there are no errors, it prints a "No errors" message to stdout.</li> <li>If there are errors, it prints a detailed message to stderr. I want only this one.</li> </ul> I figure it will be some magic with <code>>&</code>, but I never understood how those work. What I want is to consume all stdout, keep stderr. (Sorry if this is dulicate, but I haven't found any question exactly about this)

Just send the stdout to null: <pre class="prettyprint"><code>cmd > /dev/null </code></pre> This retains stderr, but suppresses stdout.

Bash print stderr only, not stdout [duplicate]

I want to lint a file, and print the stderr (error message), but do not print the stdout (saying the file is OK).

php -l "foo/bar.php"

If there are no errors, it prints a "No errors" message to stdout.
If there are errors, it prints a detailed message to stderr. I want only this one.

I figure it will be some magic with >&, but I never understood how those work.

What I want is to consume all stdout, keep stderr.

(Sorry if this is dulicate, but I haven't found any question exactly about this)

Just send the stdout to null:

cmd > /dev/null

This retains stderr, but suppresses stdout.