php : Capturing the command output

Question

I am using exec function to execute the specific executable files in php .

exec ( $file , $output , $return_value  ) ;

When the given file executed successfully I am able to get the output in the second argument by checking the return values , So, It is working fine. But My requirement is when the command execution getting failed due to some reason. I need to get that error message of that executed program . What I need to do to get the error . through second argument we can get the successful output only . Not error message.

Thanks.

Answer 1

The second argument $output only captures STDOUT from your executable. Error messages are usually sent to STDERR so that they easily can be written to an error log or similar, but this means that you won't see them when you call exec.

If this is a linux system, you could append 2>&1 to your command, in order to redirect STDERR to STDOUT. I haven't tried this, but it should forward the error messages to your $output variable.

Edit:

I've read up on it on www.php.net/exec, and it seems this would work.

exec($file.' 2>&1', $outputAndErrors, $return_value);

It is also possible to redirect the errors to a temporary file and read them separately.

exec($file.' 2> '.$tmpFile, $outputOnly, $return_value);

Edit 2

It seems windows also uses this Bourne style output redirecting syntax, so the examples should work for windows too.

More on input and output streams

Answer 2

If you need to keep stderr and stdout separate, try proc_open: http://php.net/manual/en/function.proc-open.php