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（PHP）当脚本作为cron作业执行时，如何获取文件名（或文件路径）？

[英](PHP) How to obtain the filename (or file path) when a script is executed as a cron job?

原文 2011-09-07 19:47:55 1 2 php/ cron

I have a PHP script that executes perfectly when called via the browser. 我有一个PHP脚本，当通过浏览器调用时可以完美执行。 It fails, however, when called as a cron job. 但是，当它被称为cron作业时，它将失败。 The failure is because the script needs to access its own filename, which I do via $thefilepath = $_SERVER['REQUEST_URI'] . 失败的原因是脚本需要访问自己的文件名，我可以通过$thefilepath = $_SERVER['REQUEST_URI'] 。 Unfortunately, the $_SERVER['REQUEST_URI'] variable isn't populated when the file is called as a cron job. 不幸的是，将文件作为cron作业调用时，不会填充$_SERVER['REQUEST_URI']变量。 (I've similarly failed with trying $_SERVER['PHP_SELF'] ). （同样，尝试$_SERVER['PHP_SELF']失败）。

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. 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 ），该整数用于计算数据库表中的行子集。 More specifically, myfile4.php takes rows with IDs from, say, 40-49, whereas myfile5.php takes rows with IDs from 50-59, etc. 更具体地说， myfile4.php接受ID为40-49的行，而myfile5.php接受ID为50-59的行， myfile5.php 。

So, my question is, given that $_SERVER['REQUEST_URI'] and $_SERVER['PHP_SELF'] fail to provide the filename (or filepath) for use in a cron job, can anyone please suggest an alternative method of accessing a script's own filename within this cron execution context? 因此，我的问题是，鉴于$_SERVER['REQUEST_URI']和$_SERVER['PHP_SELF']无法提供用于cron作业的文件名（或文件路径），任何人都可以建议使用另一种方法来访问脚本的在这个cron执行上下文中拥有自己的文件名？

Thank you! 谢谢！