PHP proc_open返回值始终等于127
我在PHP方面面临着一个非常大的问题我正在创建PHP应用程序，它将“使用”另一个应用程序。我想重定向输入和输出。我正在使用PHP proc_open返回值始终等于127,php,Php,我在PHP方面面临着一个非常大的问题我正在创建PHP应用程序，它将“使用”另一个应用程序。我想重定向输入和输出。我正在使用proc\u open这样做文件gsqasm在/var/www/http内，工作目录相同，因此/gsqasm应该可以实现这一点。太糟糕了，它没有返回代码始终等于127。在谷歌搜索了一段时间后，我发现127意味着找不到该文件。然后，为了确保，我检查了/tmp/error output.txt文件。它的内容是： sh: 1: gsqasm: not found sh: 1: /v
proc\u opengsqasm/var/www/http/gsqasm/tmp/error output.txtsh: 1: gsqasm: not found
sh: 1: /var/www/http/gsqasm: not found
sh: 1: /var/www/http/gsqasm: not found
sh: 1: /var/www/http/gsqasm: not found
sh: 1: ./gsqasm: not found
sh: 1: ./gsqasm: not found
sh: 1: .gsqasm: not found
sh: 1: /var/www/http/gsqasm: not found
sh: 1: /var/www/http/gsqasm: not found
<?php
$descriptorspec = array(
0 => array(
"pipe",
"r"
), // stdin is a pipe that the child will read from
1 => array(
"pipe",
"w"
), // stdout is a pipe that the child will write to
2 => array(
"file",
"/tmp/error-output.txt",
"a"
) // stderr is a file to write to
);
$cwd = '/tmp';
$env = array(
'some_option' => 'aeiou'
);
passthru('echo $PWD');
echo '<br>';
$process = proc_open('./gsqasm', $descriptorspec, $pipes, $cwd, $env);
if (!isset($_POST['input'])) {
echo "Input is empty.";
}
if (is_resource($process)) {
// $pipes now looks like this:
// 0 => writeable handle connected to child stdin
// 1 => readable handle connected to child stdout
// Any error output will be appended to /tmp/erroroutput.txt
fwrite($pipes[0], $_POST['input']);
fclose($pipes[0]);
echo stream_get_contents($pipes[1]);
fclose($pipes[1]);
// It is important that you close any pipes before calling
// proc_close in order to avoid a deadlock
$return_value = proc_close($process);
echo "command returned $return_value\n";
}
?>
在重定向流时，如何更改此代码以正确执行应用程序？这不是为在生产环境中运行而设计的代码，它是一些快速而肮脏的东西，在一段时间后会变得稳定。退出代码127实际上意味着“找不到命令”。该文件可能存在于
/var/www/http/gsqasm/tmp/gsqasm$cwd/tmp/gsqasm要解决此问题，您可以将
$cwdnull/var/www/http//var/www/http/gsqasm/tmp/gsqasm$cwd/tmp/gsqasm要解决此问题，您可以将
$cwdnull/var/www/http/ls-lathF | grep gsqasmpwdls-lathF | grep gsqasmpwd