将变量传递给从命令行运行的 PHP 脚本

时间:2023-05-05
本文介绍了将变量传递给从命令行运行的 PHP 脚本的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

限时送ChatGPT账号..

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)

这篇关于将变量传递给从命令行运行的 PHP 脚本的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!

上一篇:OpenCart:下拉标题购物车信息的 div 在哪里? 下一篇:如何在 Windows 上使用命令行访问 PHP?

相关文章