解决crond脚本执行并发冲突问题
问题
在crond配置脚本执行后,经常会出现重复执行的情况。如下面的脚本:
<?phpsleep(70);?>
如果,在crond中的配置项如下:
*/1 * * * * root php ./test.php
就出出现,之前执行的test.php还未结束,新的test.php又被执行。如下:
[hailong@vhost ~]$ ps aux | grep tt.php56667 5280 0.0 0.0 103388 4432 pts/2 T 08:06 0:00 vim ./test.phproot 5455 0.0 0.1 225288 8812 ? Ssl 08:08 0:00 /usr/bin/php /home/hailong/test.phproot 5665 5.0 0.1 225288 8748 ? Ssl 08:09 0:00 /usr/bin/php /home/hailong/test.php56667 5675 0.0 0.0 69460 852 pts/2 S+ 08:09 0:00 grep tt.php
进程一直堆积的话,可能会把系统资源给耗尽,导致系统宕机。
解决方案
解决方案有两种。各有利弊。
第一种,代码中控制并发
这种方法,就是对代码进行改造。增加是否有进程执行的判断。如下面的代码:
<?php $lockfile = '/tmp/mytest.lock'; if(file_exists($lockfile)){ exit(); }file_put_contents($lockfile, date("Y-m-d H:i:s")); sleep(70);unlink($lockfile); ?>
代码逻辑很简单,这里就不再解释。这种判断文件是否不存在的方式,会有一个问题。那就是有可能程序未执行到最后,也就是没有删除之前创建的mytest.lock文件。这会导致,之后程序将不能正常执行。
为了解决这个问题,我们对代码进行下改造。不再使用文件是否存在的判断,而是判断进程是否存在。代码修改后如下:
$fp = popen("ps aux | grep 'test.php' | wc -l", "r");$proc_num = fgets($fp);if ($proc_num > 3) { exit;}sleep(70);
这种方式有一个弊端,就是ps命令要写的精确。避免把不是执行test.php脚本的进程也统计到。如:
我们通过vim打开test.php文件。就会导致上面命令误统计。
另外,$proc_num 为什么要大于3而不是大于1,大家可以想想。如果,想不明白,可以加我微信 1798159444。
第二种,使用linux的flock命令
flock命令提供了文件锁的功能。命令参数如下:
[hailong@vhost ~]$ flock -hflock (util-linux 2.13-pre7)Usage: flock [-sxun][-w #] fd# flock [-sxon][-w #] file [-c] command... -s --shared Get a shared lock -x --exclusive Get an exclusive lock -u --unlock Remove a lock -n --nonblock Fail rather than wait -w --timeout Wait for a limited amount of time -o --close Close file descriptor before running command -c --command Run a single command string through the shell -h --help Display this text -V --version Display version
使用flock控制并发冲突,我们的crond配置如下:
*/1 * * * * root flock -xn /tmp/mytest.lock -c 'php ./test.php'
其实,使用flock也是有坑的。坑的详细信息,请查看 crond中使用flock命令的坑
原文链接: 解决crond脚本执行并发冲突问题,转载请注明来源!

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