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I have a PHP script that I want to call on some form action.
This command works fine:
/usr/local/bin/php -q script.php > /dev/null 2>&1 &But it all go wrong when I add arguments in the command:
/usr/local/bin/php -q script.php --var=value > /dev/null 2>&1 &it returns "Ambiguous output redirect".
I have read this http://www.xaprb.com/blog/2006/06/06/what-does-devnull-21-mean/ article too, and have tried almost everything possible as far as I can understand, and still have no idea how to fix it.
Please help me.
815
If your script's redirect contains a variable, and the script body defines that variable in a section enclosed by parenthesis, you will get the "ambiguous redirect" error.
&1 is used to reference the value of the file descriptor 1 (stdout). Now to the point 2>&1 means “Redirect the stderr to the same place we are redirecting the stdout” Now you can do this.
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