Ubuntu有一个非常简单的方法来管理服务。 对于python来说，不同之处在于所有依赖项(包)都必须在运行主文件的同一目录中。

我只是设法创建了这样一个服务，为我的客户提供天气信息。 步骤:

Create your python application project as you normally do. Install all dependencies locally like: sudo pip3 install package_name -t . Create your command line variables and handle them in code (if you need any) Create the service file. Something (minimalist) like: [Unit] Description=1Droid Weather meddleware provider [Service] Restart=always User=root WorkingDirectory=/home/ubuntu/weather ExecStart=/usr/bin/python3 /home/ubuntu/weather/main.py httpport=9570 provider=OWMap [Install] WantedBy=multi-user.target Save the file as myweather.service (for example) Make sure that your app runs if started in the current directory python3 main.py httpport=9570 provider=OWMap The service file produced above and named myweather.service (important to have the extension .service) will be treated by the system as the name of your service. That is the name that you will use to interact with your service. Copy the service file: sudo cp myweather.service /lib/systemd/system/myweather.service Refresh demon registry: sudo systemctl daemon-reload Stop the service (if it was running) sudo service myweather stop Start the service: sudo service myweather start Check the status (log file with where your print statements go): tail -f /var/log/syslog Or check the status with: sudo service myweather status Back to the start with another iteration if needed

此服务现在正在运行，即使您退出也不会受到影响。 如果主机关闭并重新启动，则此服务将重新启动…

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

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().
                """

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。

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

Ubuntu有一个非常简单的方法来管理服务。 对于python来说，不同之处在于所有依赖项(包)都必须在运行主文件的同一目录中。

我只是设法创建了这样一个服务，为我的客户提供天气信息。 步骤:

Create your python application project as you normally do. Install all dependencies locally like: sudo pip3 install package_name -t . Create your command line variables and handle them in code (if you need any) Create the service file. Something (minimalist) like: [Unit] Description=1Droid Weather meddleware provider [Service] Restart=always User=root WorkingDirectory=/home/ubuntu/weather ExecStart=/usr/bin/python3 /home/ubuntu/weather/main.py httpport=9570 provider=OWMap [Install] WantedBy=multi-user.target Save the file as myweather.service (for example) Make sure that your app runs if started in the current directory python3 main.py httpport=9570 provider=OWMap The service file produced above and named myweather.service (important to have the extension .service) will be treated by the system as the name of your service. That is the name that you will use to interact with your service. Copy the service file: sudo cp myweather.service /lib/systemd/system/myweather.service Refresh demon registry: sudo systemctl daemon-reload Stop the service (if it was running) sudo service myweather stop Start the service: sudo service myweather start Check the status (log file with where your print statements go): tail -f /var/log/syslog Or check the status with: sudo service myweather status Back to the start with another iteration if needed

此服务现在正在运行，即使您退出也不会受到影响。 如果主机关闭并重新启动，则此服务将重新启动…