Python之python 守护程序锁定文件在 lock.acquire() 上超时

我正在使用 python-daemon 模块来管理我的 Python 脚本的守护进程。 但是，在运行我根本无法弄清楚的脚本时，我感到很头疼。我也不知道如何开始调试它。

我有代码:

def run_application(): 
    #Do something here... 
 
class App(): 
    def __init__(self): 
        self.stdin_path = '/dev/null' 
        self.stdout_path = 'stdout.txt' 
        self.stderr_path = 'stdlog.log' 
        self.pidfile_path = 'filelock.pid' 
        self.pidfile_timeout = 5 
    def run(self): 
        run_application() 
 
 
app = App() 
daemon_runner = runner.DaemonRunner(app) 
daemon_runner.do_action() 

运行时，它总是将以下内容写入stdlog.log:

Traceback (most recent call last): 
  File "MyApp.py", line 335, in <module> 
    daemon_runner.do_action() 
  File "/anaconda/lib/python2.7/site-packages/daemon/runner.py", line 189, in do_action 
    func(self) 
  File "/anaconda/lib/python2.7/site-packages/daemon/runner.py", line 124, in _start 
    self.daemon_context.open() 
  File "/anaconda/lib/python2.7/site-packages/daemon/daemon.py", line 346, in open 
    self.pidfile.__enter__() 
  File "/anaconda/lib/python2.7/site-packages/lockfile/__init__.py", line 229, in __enter__ 
    self.acquire() 
  File "/anaconda/lib/python2.7/site-packages/daemon/pidfile.py", line 42, in acquire 
    super(TimeoutPIDLockFile, self).acquire(timeout, *args, **kwargs) 
  File "/anaconda/lib/python2.7/site-packages/lockfile/pidlockfile.py", line 88, in acquire 
    self.path) 
lockfile.LockTimeout: Timeout waiting to acquire lock for /MyApp/filelock.pid 

因此在尝试锁定 filelock.pid 时似乎超时。我不知道这是为什么。我已经删除了 filelock.pid，我已经更改了权限；每次都出现同样的错误。

我怎样才能开始调试这个？？？我很茫然。

我正在使用 python-daemon version 1.6(如果重要的话)。

更新:

遵循建议here ，我现在看到已经有一个进程在运行。现在我怎样才能弄清楚如何确定正在运行的守护进程的 PID。

请您参考如下方法:

我同意@ExploWare 就他如何证明您可以捕获那些LockTimeout 异常而言。

至于调试和查看哪个进程持有此锁的方法，这里是您可以运行的外部代码位...

import daemon.pidfile 
import os 
import lockfile 
 
# We know the lockfile name. 
pidfile = daemon.pidfile.PIDLockFile( 
                            os.path.join("/MyApp/","filelock.pid")) 
 
# This current process id... 
os.getpid() 
# 46337 

那么是什么进程获得了这个锁呢？

pidfile.is_locked() 
# True 
 
pidfile.read_pid() 
# 96856 

当我们的 PIDLockFile 实例尝试“获取”时，

pidfile.__dict__ 
# {'unique_name': '/MyApp/filelock.pid', 'lock_file': '/MyApp/filelock.pid.lock', 'hostname':  
# 'MyMachine.local', 'pid': 46337, 'timeout': None, 'tname': '', 'path': '/MyApp/filelock.pid'} 
 
pidfile.acquire() 
# 
# (Had to Control-C quit because I didnt set a timeout on PIDLockFile ) 
# 
# ^CTraceback (most recent call last): 
#   File "<stdin>", line 1, in <module> 
#   File "/Users/michal/venf/lib/python2.7/site-packages/lockfile/pidlockfile.py", line 92, in acquire 
#     time.sleep(timeout is not None and timeout/10 or 0.1) 
# KeyboardInterrupt 

因此，改为使用@ExploWare 的异常捕获。

# Wait only 5 seconds. 
pidfile.timeout = 5 
 
try: 
    pidfile.acquire() 
except lockfile.LockTimeout: 
    print 'locked . need to wait or move on.' 
#  
# locked . need to wait or move on.