我写了一个Python脚本,检查一个特定的电子邮件地址,并将新的电子邮件传递给一个外部程序。如何让这个脚本全天候执行,比如在Linux中将其转换为守护进程或服务。我是否还需要一个在程序中永不结束的循环,或者可以通过多次重新执行代码来完成?
当前回答
这是一个很好的类,从这里采取:
#!/usr/bin/env python
import sys, os, time, atexit
from signal import SIGTERM
class Daemon:
"""
A generic daemon class.
Usage: subclass the Daemon class and override the run() method
"""
def __init__(self, pidfile, stdin='/dev/null', stdout='/dev/null', stderr='/dev/null'):
self.stdin = stdin
self.stdout = stdout
self.stderr = stderr
self.pidfile = pidfile
def daemonize(self):
"""
do the UNIX double-fork magic, see Stevens' "Advanced
Programming in the UNIX Environment" for details (ISBN 0201563177)
http://www.erlenstar.demon.co.uk/unix/faq_2.html#SEC16
"""
try:
pid = os.fork()
if pid > 0:
# exit first parent
sys.exit(0)
except OSError, e:
sys.stderr.write("fork #1 failed: %d (%s)\n" % (e.errno, e.strerror))
sys.exit(1)
# decouple from parent environment
os.chdir("/")
os.setsid()
os.umask(0)
# do second fork
try:
pid = os.fork()
if pid > 0:
# exit from second parent
sys.exit(0)
except OSError, e:
sys.stderr.write("fork #2 failed: %d (%s)\n" % (e.errno, e.strerror))
sys.exit(1)
# redirect standard file descriptors
sys.stdout.flush()
sys.stderr.flush()
si = file(self.stdin, 'r')
so = file(self.stdout, 'a+')
se = file(self.stderr, 'a+', 0)
os.dup2(si.fileno(), sys.stdin.fileno())
os.dup2(so.fileno(), sys.stdout.fileno())
os.dup2(se.fileno(), sys.stderr.fileno())
# write pidfile
atexit.register(self.delpid)
pid = str(os.getpid())
file(self.pidfile,'w+').write("%s\n" % pid)
def delpid(self):
os.remove(self.pidfile)
def start(self):
"""
Start the daemon
"""
# Check for a pidfile to see if the daemon already runs
try:
pf = file(self.pidfile,'r')
pid = int(pf.read().strip())
pf.close()
except IOError:
pid = None
if pid:
message = "pidfile %s already exist. Daemon already running?\n"
sys.stderr.write(message % self.pidfile)
sys.exit(1)
# Start the daemon
self.daemonize()
self.run()
def stop(self):
"""
Stop the daemon
"""
# Get the pid from the pidfile
try:
pf = file(self.pidfile,'r')
pid = int(pf.read().strip())
pf.close()
except IOError:
pid = None
if not pid:
message = "pidfile %s does not exist. Daemon not running?\n"
sys.stderr.write(message % self.pidfile)
return # not an error in a restart
# Try killing the daemon process
try:
while 1:
os.kill(pid, SIGTERM)
time.sleep(0.1)
except OSError, err:
err = str(err)
if err.find("No such process") > 0:
if os.path.exists(self.pidfile):
os.remove(self.pidfile)
else:
print str(err)
sys.exit(1)
def restart(self):
"""
Restart the daemon
"""
self.stop()
self.start()
def run(self):
"""
You should override this method when you subclass Daemon. It will be called after the process has been
daemonized by start() or restart().
"""
其他回答
在linux下使用$nohup命令如何?
我用它在Bluehost服务器上运行命令。
如果我错了,请指教。
使用你的系统提供的任何服务管理器——例如在Ubuntu下使用upstart。这将为您处理所有细节,如启动时启动,崩溃时重新启动等。
cron is clearly a great choice for many purposes. However it doesn't create a service or daemon as you requested in the OP. cron just runs jobs periodically (meaning the job starts and stops), and no more often than once / minute. There are issues with cron -- for example, if a prior instance of your script is still running the next time the cron schedule comes around and launches a new instance, is that OK? cron doesn't handle dependencies; it just tries to start a job when the schedule says to.
如果您发现确实需要一个守护进程(一个永不停止运行的进程),请查看一下监控器。它提供了一种简单的方法来包装普通的、非守护进程化的脚本或程序,并使其像守护进程一样运行。这比创建本地Python守护进程好得多。
我推荐这个解决方案。您需要继承和重写方法运行。
import sys
import os
from signal import SIGTERM
from abc import ABCMeta, abstractmethod
class Daemon(object):
__metaclass__ = ABCMeta
def __init__(self, pidfile):
self._pidfile = pidfile
@abstractmethod
def run(self):
pass
def _daemonize(self):
# decouple threads
pid = os.fork()
# stop first thread
if pid > 0:
sys.exit(0)
# write pid into a pidfile
with open(self._pidfile, 'w') as f:
print >> f, os.getpid()
def start(self):
# if daemon is started throw an error
if os.path.exists(self._pidfile):
raise Exception("Daemon is already started")
# create and switch to daemon thread
self._daemonize()
# run the body of the daemon
self.run()
def stop(self):
# check the pidfile existing
if os.path.exists(self._pidfile):
# read pid from the file
with open(self._pidfile, 'r') as f:
pid = int(f.read().strip())
# remove the pidfile
os.remove(self._pidfile)
# kill daemon
os.kill(pid, SIGTERM)
else:
raise Exception("Daemon is not started")
def restart(self):
self.stop()
self.start()
你有两个选择。
Make a proper cron job that calls your script. Cron is a common name for a GNU/Linux daemon that periodically launches scripts according to a schedule you set. You add your script into a crontab or place a symlink to it into a special directory and the daemon handles the job of launching it in the background. You can read more at Wikipedia. There is a variety of different cron daemons, but your GNU/Linux system should have it already installed. Use some kind of python approach (a library, for example) for your script to be able to daemonize itself. Yes, it will require a simple event loop (where your events are timer triggering, possibly, provided by sleep function).
我不建议你选择2。,因为您实际上是在重复cron功能。Linux系统范例是让多个简单工具交互并解决您的问题。除非有其他原因需要创建守护进程(除了定期触发之外),否则选择其他方法。
此外,如果你使用daemonize进行循环并且发生了崩溃,那么没有人会在那之后检查邮件(正如Ivan Nevostruev在回答的评论中指出的那样)。而如果脚本作为cron作业添加,它将再次触发。
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