如何使用CRON运行单个PHP实例？ [英] How do you run a single PHP instance with CRON?

问题描述

我无法使用CRON运行PHP脚本的单个实例。也许有人可以帮助解释需要什么。当前，我有一个 startup 脚本，该脚本由 crontab 调用，该脚本检查以确保PHP脚本的实例是

I'm having trouble running a single instance of a PHP script using CRON. Perhaps someone can help explain what is needed. Currenty, I have a startup script that is called by crontab which checks to make sure an instance of a PHP script isn't already running before calling the PHP instance.

crontab -e 条目：

* * * * * /var/www/private/script.php >> /var/www/private/script.log 2>&1 &

./ startup

./startup

#!/bin/bash

if ps -ef | grep '[s]cript';
        then
                exit;
        else
                /usr/bin/php /var/www/private/script.php  >>/var/www/private/script.log 2>&1 &
                echo 'started'
fi

这似乎不起作用并且似乎无法记录任何错误以知道如何进行调试。

This doesn't seem to be working and I can't seem to get any errors logged to know how to proceed to debug this.

推荐答案

您可以将lockrun用于： http://www.unixwiz.net/tools/lockrun.html

You can use lockrun for that: http://www.unixwiz.net/tools/lockrun.html

* * * * * /usr/local/bin/lockrun --lockfile=/var/run/script.lockrun -- php /home/script.php

或使用Perl：

system('php /home/script.php') if ( ! `ps -aef|grep -v grep|grep script.php`);

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