Question

I am running the following proc_open function. When the page is loaded, I get the error:

Use of undefined constant STDOUT - assumed 'STDOUT'`

How should I set STDOUT and STSDERR correctly?

PHP Snippet

$cmd = 'psql -p 5432 -d nominatim';

$descriptorspec = array(
   0 => array("pipe", "r"),  // stdin is a pipe that the child will read from
   1 => STDOUT,  // stdout is a pipe that the child will write to
   2 => STDERR // stderr is a file to write to
);

$pipes = null;

$process = proc_open($cmd, $descriptorspec, $pipes);

---

Update

<?php

    $cmd = 'psql -p 5432 -d nominatim';

    $descriptorspec = array(
        0 => array('pipe', 'r'), // stdin
        1 => array('pipe', 'w'), // stdout
        2 => array('pipe', 'a') // stderr
    );

    $pipes = null;

    $process = proc_open($cmd, $descriptorspec, $pipes);

?>

When I chmod 755 test.php and run ./test.php in the command line (CentOS), I get the error output:

: No such file or directory
: command not found
./test.php: line 3: =: command not found
: command not found
: command not found
./test.php: line 5: syntax error near unexpected token `('
'/test.php: line 5: `   $descriptorspec = array(

This is puzzling, = is not a command?

---

Update 2

#!/usr/bin/php <?php

    $cmd = 'psql -p 5432 -d nominatim';

    $descriptorspec = array(
        0 => array('pipe', 'r'), // stdin
        1 => array('pipe', 'w'), // stdout
        2 => array('pipe', 'a') // stderr
    );

    $pipes = null;

    $process = proc_open($cmd, $descriptorspec, $pipes);

?>

I get the output:

Status: 404 Not Found
X-Powered-By: PHP/5.3.16
Content-type: text/html

No input file specified.

Answer 1

You may use :

$descriptorspec = array(
    0 => array('pipe', 'r'), // stdin
    1 => array('pipe', 'w'), // stdout
    2 => array('pipe', 'a') // stderr
);

instead

Have a look to the manual