将变量传递给从命令行运行的 PHP 脚本-php问题

Pass a variable to a PHP script running from the command line(将变量传递给从命令行运行的 PHP 脚本)

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问题描述

我有一个需要从命令行运行的 PHP 文件(通过 crontab).我需要将 type=daily 传递给文件，但我不知道如何.我试过了:

I have a PHP file that is needed to be run from the command line (via crontab). I need to pass type=daily to the file, but I don't know how. I tried:

php myfile.php?type=daily

但是返回了这个错误:

无法打开输入文件:myfile.php?type=daily

Could not open input file: myfile.php?type=daily

我能做什么?

推荐答案

?type=daily 参数(以 $_GET 数组结尾)仅对网页访问页面.

The ?type=daily argument (ending up in the $_GET array) is only valid for web-accessed pages.

您需要像 php myfile.php daily 一样调用它，并从 $argv 数组(这将是 $argv[1]，因为 $argv[0] 将是 myfile.php).

You'll need to call it like php myfile.php daily and retrieve that argument from the $argv array (which would be $argv[1], since $argv[0] would be myfile.php).

如果页面也用作网页，您可以考虑两种选择.使用 shell 脚本和 Wget 访问它，然后从 cron:

If the page is used as a webpage as well, there are two options you could consider. Either accessing it with a shell script and Wget, and call that from cron:

#!/bin/sh
wget http://location.to/myfile.php?type=daily

或者检查PHP文件是否从命令行调用:

Or check in the PHP file whether it's called from the command line or not:

if (defined('STDIN')) {
  $type = $argv[1];
} else {
  $type = $_GET['type'];
}

(注意:您可能需要/想要检查 $argv 是否确实包含足够的变量等)

(Note: You'll probably need/want to check if $argv actually contains enough variables and such)

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