Question

我用这个评论做了一个cron工作：＆＃34; /home/username/public_html/cron.php"

我的php（cron.php）

    <?php 

    $server = 'localhost';
    $user = 'name';
    $pass = 'password';
    $dbname = 'dbname';

    $conn = new mysqli($server, $user, $pass, $dbname);

    $sql = "Update table;";
    $conn->query($sql);

    $conn->close();
    ?>

这是错误：

/home/username/public_html/cron.php: line 1: ?php: No such file or directory
/home/username/public_html/cron.php: line 3: =: command not found
/home/username/public_html/cron.php: line 4: =: command not found
/home/username/public_html/cron.php: line 5: =: command not found
/home/username/public_html/cron.php: line 6: =: command not found
/home/username/public_html/cron.php: line 8: syntax error near unexpected token `('
/home/username/public_html/cron.php: line 8: `$conn = new mysqli($server, $user, $pass, $dbname);'

任何人都可以帮忙解决这个问题？谢谢。

Answer 1

您的脚本被解释为shell脚本。要在命令行运行PHP脚本：

php /home/username/public_html/cron.php

或者将其添加到文件的顶部并使其可执行：

#!/usr/bin/php

（或者PHP可执行文件的路径）

Cron作业PHP脚本错误

1 个答案: