问题描述
我有一个cronjob,运行PHP文件,运行用PHP编写的DAEMON,但我只想运行DAEMON,如果没有其他实例运行,我如何得到一个PHP进程运行列表为了发现我的DAEMON是否正在运行。我想到某种类型的exec,将生成一个列表,我可以存储在一个数组。有任何想法吗?感谢
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,请参阅以下问题:
To get the list of PHP processes see this question:
另一个选择是你可以获取文件的锁,然后在运行之前检查它:
:
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
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.
感谢
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