自动创建pid文件，并加锁

最新推荐文章于 2024-08-13 04:27:18 发布

转载最新推荐文章于 2024-08-13 04:27:18 发布 · 742 阅读

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本文介绍了如何使用Python实现进程锁定文件，确保同一进程只运行一次实例，避免资源冲突。

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#!/usr/bin/env python
#coding=utf-8
import sys
import os
import re
import traceback
import fcntl
class PidFile():
def __init__(self, file = None):
if file == None:
argv0_list = sys.argv[0].split("\\")
scname = argv0_list[len(argv0_list) - 1]
if re.search(".py$", scname):
scname = scname[0:-3]
self.file = "/tmp/%s.pid" % (scname)
else:
self.file = file
self.pidfile = None
def _create(self, file):
f = open(file, "w")
pid = os.getpid()
f.write("%s" % (pid))
f.close()
def __enter__(self):
if not os.path.isfile(self.file):
self._create(self.file)
self.pidfile = open(self.file, 'a+')
try:
fcntl.flock(self.pidfile.fileno(), fcntl.LOCK_EX|fcntl.LOCK_NB)
return True
except:
traceback.print_exc()
fpid = open(self.file, "r").read()
raise SystemExit("Process id %s Already running , pid to %s " % (fpid, self.file))
return False
def __exit__(self, *args):
try:
self.pidfile.close()
except :
pass
os.remove(self.file)
if __name__ == "__main__":
# use
with PidFile() as t:
print "lock file %s" % (file)
time.sleep(10)
exit(0)