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
首先,仔细研究邮件别名。邮件别名将在邮件系统内完成此任务,而无需您在守护进程或服务或任何类似的东西上瞎忙活。
您可以编写一个简单的脚本,每次将邮件消息发送到特定邮箱时,sendmail都会执行该脚本。
参见http://www.feep.net/sendmail/tutorial/intro/aliases.html
如果您真的想编写一个不必要的复杂服务器,您可以这样做。
nohup python myscript.py &
这就够了。您的脚本只是循环和休眠。
import time
def do_the_work():
# one round of polling -- checking email, whatever.
while True:
time.sleep( 600 ) # 10 min.
try:
do_the_work()
except:
pass
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。
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().
"""
要创建一些像service一样运行的东西,你可以使用这个东西:
你必须做的第一件事是安装水泥框架: 水泥框架是一个CLI框架,您可以在其上部署应用程序。
app命令行界面:
interface.py
from cement.core.foundation import CementApp
from cement.core.controller import CementBaseController, expose
from YourApp import yourApp
class Meta:
label = 'base'
description = "your application description"
arguments = [
(['-r' , '--run'],
dict(action='store_true', help='Run your application')),
(['-v', '--version'],
dict(action='version', version="Your app version")),
]
(['-s', '--stop'],
dict(action='store_true', help="Stop your application")),
]
@expose(hide=True)
def default(self):
if self.app.pargs.run:
#Start to running the your app from there !
YourApp.yourApp()
if self.app.pargs.stop:
#Stop your application
YourApp.yourApp.stop()
class App(CementApp):
class Meta:
label = 'Uptime'
base_controller = 'base'
handlers = [MyBaseController]
with App() as app:
app.run()
YourApp.py类:
import threading
class yourApp:
def __init__:
self.loger = log_exception.exception_loger()
thread = threading.Thread(target=self.start, args=())
thread.daemon = True
thread.start()
def start(self):
#Do every thing you want
pass
def stop(self):
#Do some things to stop your application
请记住,你的应用程序必须运行在一个线程作为守护进程
要运行应用程序,只需在命令行中这样做
Python interface.py——help
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。
你可以使用fork()将你的脚本从tty中分离出来,并让它继续运行,如下所示:
import os, sys
fpid = os.fork()
if fpid!=0:
# Running as daemon now. PID is fpid
sys.exit(0)
当然你还需要实现一个无限循环,比如
while 1:
do_your_check()
sleep(5)
希望这能让你开始。
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