问题描述
我有一个需要包含此文件的 cron 作业:
I have a cron job that needs to include this file:
require '../includes/common.php';
但是，当它通过 cron 作业(而不是我的本地测试)运行时，相对路径不起作用.cron 作业运行以下文件(在实时服务器上):
however, when it is run via the cron job (and not my local testing), the relative path does not work. the cron job runs the following file (on the live server):
/home/username123/public_html/cron/mycronjob.php
这是错误:
Fatal error: require(): Failed opening required '../includes/common.php'
(include_path='.:/usr/lib/php:/usr/local/lib/php') in
/home/username123/public_html/cron/mycronjob.php on line 2
使用与 cron 作业相同的绝对格式，common.php 将位于
using the same absolute format as the cron job, common.php would be located at
/home/username123/public_html/includes/common.php
这是否意味着我必须将第 2 行替换为:
does that mean i have to replace my line 2 with:
require '/home/username123/public_html/includes/common.php';
?
谢谢！
推荐答案
从技术上看php脚本是在cron所在的地方运行的；前任.如果 cron 在/bin/cron 中，则该语句将在/bin/includes/common.php 中查找 common.php.
Technically seen the php script is run where cron is located; ex. If cron was in /bin/cron, then this statement would look for common.php in /bin/includes/common.php.
是的，您可能必须使用完整路径或使用 set_include_path
So yeah, you'll probably have to use fullpaths or use set_include_path
set_include_path('/home/username123/public_html/includes/');
require 'common.php';
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