使用未定义的常量 STDOUT - 假定为“STDOUT”

Question

我正在运行以下proc_open 函数。加载页面时，我收到错误：

Use of undefined constant STDOUT - assumed 'STDOUT'`

如何正确设置STDOUT和STSDERR？

PHP 代码段

$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);

---

更新

<?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);

?>

当我 chmod 755 test.php 并在命令行 (CentOS) 中运行 ./test.php 时，我得到错误输出：

: 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(

这令人费解，=不是命令？

---

更新 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);

?>

我得到了输出：

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

No input file specified.

Answer 1

你可以使用：

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

改为

看看manual