在php文件上调用exec并传递参数? [英] calling exec on a php file and passing parameters?

问题描述

我想使用 exec 调用php文件.

I am wanting to call a php file using exec.

当我调用它时，我希望能够通过(一个id)传递一个变量.

When I call it I want to be able to pass a variable through (an id).

我可以调用 echo exec("php/var/www/unity/src/emailer.php"); 很好，但是当我添加 echo exec("php/var/www/unity/src/emailer.php?id=123); exec调用失败.

I can call echo exec("php /var/www/unity/src/emailer.php"); fine, but the moment I add anything like echo exec("php /var/www/unity/src/emailer.php?id=123"); the exec call fails.

我该怎么做?

推荐答案

您的调用失败，因为您正在通过命令行调用使用Web样式的语法(?parameter = value ).我了解您在想什么，但这根本行不通.

Your call is failing because you're using a web-style syntax (?parameter=value) with a command-line invokation. I understand what you're thinking, but it simply doesn't work.

您将改为使用 $ argv .参见 PHP手册.

要查看实际效果，请将此单行代码写入文件:

To see this in action, write this one-liner to a file:

<?php print_r($argv); ?>

然后从命令行使用参数调用它:

Then invoke it from the command-line with arguments:

php -f /path/to/the/file.php firstparam secondparam

您会看到 $ argv 包含脚本本身的名称，作为元素零，其后是您传入的其他任何参数.

You'll see that $argv contains the name of the script itself as element zero, followed by whatever other parameters you passed in.

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