问题描述

我有一个运行 PHP 文件的 cronjob，该文件运行用 PHP 编写的 DAEMON，但我只想运行 DAEMON，如果没有其他实例正在运行，我如何获取正在运行的 PHP 进程列表以便查找我的 DAEMON 是否正在运行.我想过某种 exec 会生成一个我可以存储在数组中的列表.有任何想法吗?谢谢

I have a cronjob that runs a PHP file that runs a DAEMON written in PHP, but I only want to run the DAEMON if no other instances of it are running, how can I get a list of PHP processes running in order to find if my DAEMON is running. I thought about some kind of exec that will generate a list that I can store in an array. Any ideas? thanks

推荐答案

要获取 PHP 进程列表，请参阅此问题:

To get the list of PHP processes see this question:

如何获取正在运行的php列表使用 PHP exec() 的脚本?

另一种选择是您可以获取文件的锁，然后在运行前检查它:例如:

Another option is that you can acquire a lock of the file and then check it before running: for example:

$thisfilepath = $_SERVER['SCRIPT_FILENAME'];
$thisfilepath = fopen($thisfilepath,'r');
if (!flock($thisfilepath,LOCK_EX | LOCK_NB))
{
  customlogfunctionandemail("File is Locked");
  exit();
}
elseif(flock($thisfilepath,LOCK_EX | LOCK_NB)) // Acquire Lock
{
  // Write your code


 // Unlock before finish
 flock($thisfilepath,LOCK_UN);  // Unlock the file and Exit
 customlogfunctionandemail("Process completed. File is unlocked");
 exit();
}

基本上在上面的例子中，你首先检查文件是否被锁定，如果它没有被锁定(意味着过程完成)，你可以获取锁定并开始你的代码.

Basically in the above example you are first checking if the file is locked or not and if it is not locked (means process is completed) you can acquire lock and begin your code.

谢谢

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