运行PHP文件时出现Cron作业错误
我试图通过cron作业运行php文件,手动运行时该文件工作正常,但在cron作业中运行时,出现以下错误:运行PHP文件时出现Cron作业错误,php,cron,crontab,Php,Cron,Crontab,我试图通过cron作业运行php文件,手动运行时该文件工作正常,但在cron作业中运行时,出现以下错误: Warning: include(classes/EmailAddressValidator.php): failed to open stream: No such file or directory in /var/www/onecent_dev/classes/MiscFunctions.php on line 3 Warning: include(): Failed opening
Warning: include(classes/EmailAddressValidator.php): failed to open stream: No such file or directory in /var/www/onecent_dev/classes/MiscFunctions.php on line 3
Warning: include(): Failed opening 'classes/EmailAddressValidator.php' for inclusion (include_path='.:/usr/share/php:/usr/share/pear') in /var/www/onecent_dev/classes/MiscFunctions.php on line 3
miscsfunctions.php和EmailAddressValidator.php都是现有的文件,并且都位于正确的位置,它们提供了什么
谢谢请看这个问题:
您的include\u路径不包含您正在执行的脚本的路径。请参阅以下问题:
您的include\u路径不包含正在执行的脚本的路径。看起来您的include\u路径正在解析。将当前目录设置为的任何cron,而不是脚本所在的目录。首先尝试将crontab编辑为cd:
0 * * * * cd /path/to/script && php script.php
或者显式提供include_路径:
0 * * * * php -d include_path=/path/to/script script.php
看起来您的include\u路径正在解析。将当前目录设置为的任何cron,而不是脚本所在的目录。首先尝试将crontab编辑为cd:
0 * * * * cd /path/to/script && php script.php
或者显式提供include_路径:
0 * * * * php -d include_path=/path/to/script script.php