(PHP) 当脚本作为 cron 作业执行时如何获取文件名（或文件路径）？
我有一个 PHP 脚本，通过浏览器调用时可以完美执行。然而，当作为 cron 作业调用时，它会失败。失败是因为脚本需要访问自己的文件名，我是通过 $thefilepath = $_SERVER['REQUEST_URI'] 来访问的。不幸的是，当文件作为 cron 作业调用时，$_SERVER['REQUEST_URI'] 变量不会被填充。 （我尝试 $_SERVER['PHP_SELF'] 也同样失败了）。
如果有人感兴趣，脚本需要访问自己的文件名，因为文件名包含一个整数（例如，filename = myfile4.php），该整数用于计算 a 中的行子集。数据库表。更具体地说，myfile4.php 获取 ID 为 40-49 的行，而 myfile5.php 获取 ID 为 50-59 等的行
。问题是，鉴于 $_SERVER['REQUEST_URI'] 和 $_SERVER['PHP_SELF'] 无法提供用于以下用途的文件名（或文件路径）一个 cron 作业，任何人都可以建议一种在此 cron 执行上下文中访问脚本自己的文件名的替代方法吗？
谢谢你！
I have a PHP script that executes perfectly when called via the browser. It fails, however, when called as a cron job. The failure is because the script needs to access its own filename, which I do via $thefilepath = $_SERVER['REQUEST_URI']. Unfortunately, the $_SERVER['REQUEST_URI'] variable isn't populated when the file is called as a cron job. (I've similarly failed with trying $_SERVER['PHP_SELF']).
In case anyone is interested, the script needs to access its own filename because the filename contains an integer (for example, filename = myfile4.php) that is used in the calculation of a subset of rows in a database table. More specifically, myfile4.php takes rows with IDs from, say, 40-49, whereas myfile5.php takes rows with IDs from 50-59, etc.
So, my question is, given that $_SERVER['REQUEST_URI'] and $_SERVER['PHP_SELF'] fail to provide the filename (or filepath) for use in a cron job, can anyone please suggest an alternative method of accessing a script's own filename within this cron execution context?
Thank you!
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使用
__FILE__或__DIR__或dirname(__FILE__)http://www.php.net/manual/en/language.constants .预定义.php
Use
__FILE__or__DIR__ordirname(__FILE__)http://www.php.net/manual/en/language.constants.predefined.php
$argv[0]将始终包含脚本的名称：http://www.php.net/manual/en/reserved.variables.argv.php$argv[0]will always contain the name of the script: http://www.php.net/manual/en/reserved.variables.argv.php