PHP：如何手动/编程启动cron脚本 [英] PHP: how to manually/programatically kickoff cron scripts

问题描述

您好，我正在做一个cron状态页面，显示cron运行的脚本上的所有状态信息。我知道如何建立显示状态信息，但我需要帮助和建议以下...

Hi I am making a cron status page that shows all the status info on scripts run by cron. I know how to build the "showing status info" but I need help and suggestion on the following...

我需要允许用户看到状态页有手动启动cron脚本的权力。像一个按钮，它会调用cron来运行一个特定的脚本，如现在。

I need to allow a user seeing the status page to have the power to kickoff a cron script manually. Like a button that would call the cron to run a specific script like "now".

这是可能还是有一个解决方法我可以做吗？

Is that possible or is there a workaround I can do?

请帮助并提前感谢。

推荐答案

您可以手动执行cron脚本：

You can manually execute the cron script:

exec("cron script here");

如果需要，可以获取可用的cron脚本（未测试）：

If you want, you can get the available cron scripts (untested):

$crontab = file_get_contents("path/to/cron");
$cron_jobs = array_filter(explode(PHP_EOL, $crontab), function($cron_job) {
    return $cron_job[0] !== "#"; // Remove comments.
});
$cron_jobs = array_map(function($cron_job) {
    $fields = explode(" ", $cron_job);
    $fields = array_splice($fields, 0, 5); // Get rid of timing information.
    return implode(" ", $fields);
}, $cron_jobs);

$cron_script = $_GET["cron_script"];
if ($cron_script < sizeof($cron_jobs)) {
    exec($cron_jobs[$cron_script]);
}

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