Python3 创建守护进程

Question

我试图在Python中创建一个守护进程，以便在Ubuntu服务器上运行。下面的代码是我有问题的代码。

import sys
import time
import threading
import logging
import logging.handlers
from daemon import runner
class Main(object):
    def run(self):
        my_logger = logging.getLogger('NameGeneratorDeamon')
        my_logger.setLevel(logging.DEBUG)
        handler = logging.handlers.SysLogHandler(address=('192.168.0.69', 514),facility=LOG_DAEMON)
        my_logger.addHandler(handler)
        try:
            my_logger.info('Started')
            while True:
                pass
        except Exception as inst:
            #Send error to syslog server
            my_logger.critical(inst)
class App():
    def __init__(self):
        self.stdin_path = '/dev/null'
        self.stdout_path = '/dev/null'
        self.stderr_path = '/dev/null'
        self.pidfile_path =  '/tmp/foo.pid'
        self.pidfile_timeout = 5
    def run(self):
        service = Main()
        service.run()
app = App()
daemon_runner = runner.DaemonRunner(app)
daemon_runner.do_action()

运行代码时收到的错误消息如下：

   File "Main.py", line 35, in <module>
     daemon_runner = runner.DaemonRunner(app)
   File "/usr/local/lib/python3.4/dist-packages/daemon/runner.py", line 111, in __init__
     self.daemon_context.stdout = open(app.stdout_path, 'w+t')
io.UnsupportedOperation: File or stream is not seekable.

有谁知道如何解决这个问题，或者你有更好的方法来在 Python 中创建守护进程吗？

Answer 1

就我而言，这没有错误：

# vi /usr/local/lib/python3.5/dist-packages/daemon/runner.py 
# 118 -120
        self.daemon_context = DaemonContext()
        self.daemon_context.stdin = open(app.stdin_path, 'wb+',buffering=0)
        self.daemon_context.stdout = open(app.stdout_path,  'wb+',buffering=0)
        self.daemon_context.stderr = open(
                app.stderr_path, 'wb+', buffering=0)