Make sure only a single instance of a program is running

问题描述：

Is there a Pythonic way to have only one instance of a program running?

The only reasonable solution I've come up with is trying to run it as a server on some port, then second program trying to bind to same port - fails. But it's not really a great idea, maybe there's something more lightweight than this?

(Take into consideration that program is expected to fail sometimes, i.e. segfault - so things like "lock file" won't work)

解决方案 1：

The following code should do the job, it is cross-platform and runs on Python 2.4-3.2. I tested it on Windows, OS X and Linux.

from tendo import singleton
me = singleton.SingleInstance() # will sys.exit(-1) if other instance is running

The latest code version is available singleton.py. Please file bugs here.

You can install tend using one of the following methods:

• easy_install tendo

• pip install tendo

• manually by getting it from http://pypi.python.org/pypi/tendo

解决方案 2：

Simple, cross-platform solution, found in another question by zgoda:

import fcntl
import os
import sys

def instance_already_running(label="default"):
    """
    Detect if an an instance with the label is already running, globally
    at the operating system level.

    Using `os.open` ensures that the file pointer won't be closed
    by Python's garbage collector after the function's scope is exited.

    The lock will be released when the program exits, or could be
    released if the file pointer were closed.
    """

    lock_file_pointer = os.open(f"/tmp/instance_{label}.lock", os.O_WRONLY)

    try:
        fcntl.lockf(lock_file_pointer, fcntl.LOCK_EX | fcntl.LOCK_NB)
        already_running = False
    except IOError:
        already_running = True

    return already_running

A lot like S.Lott's suggestion, but with the code.

解决方案 3：

This code is Linux specific. It uses 'abstract' UNIX domain sockets, but it is simple and won't leave stale lock files around. I prefer it to the solution above because it doesn't require a specially reserved TCP port.

try:
    import socket
    s = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)
    ## Create an abstract socket, by prefixing it with null. 
    s.bind( 'postconnect_gateway_notify_lock') 
except socket.error as e:
    error_code = e.args[0]
    error_string = e.args[1]
    print "Process already running (%d:%s ). Exiting" % ( error_code, error_string) 
    sys.exit (0) 

The unique string postconnect_gateway_notify_lock can be changed to allow multiple programs that need a single instance enforced.

解决方案 4：

I don't know if it's pythonic enough, but in the Java world listening on a defined port is a pretty widely used solution, as it works on all major platforms and doesn't have any problems with crashing programs.

Another advantage of listening to a port is that you could send a command to the running instance. For example when the users starts the program a second time, you could send the running instance a command to tell it to open another window (that's what Firefox does, for example. I don't know if they use TCP ports or named pipes or something like that, 'though).

解决方案 5：

Never written python before, but this is what I've just implemented in mycheckpoint, to prevent it being started twice or more by crond:

import os
import sys
import fcntl
fh=0
def run_once():
    global fh
    fh=open(os.path.realpath(__file__),'r')
    try:
        fcntl.flock(fh,fcntl.LOCK_EX|fcntl.LOCK_NB)
    except:
        os._exit(0)

run_once()

Found Slava-N's suggestion after posting this in another issue (http://stackoverflow.com/questions/2959474). This one is called as a function, locks the executing scripts file (not a pid file) and maintains the lock until the script ends (normal or error).

解决方案 6：

Use a pid file. You have some known location, "/path/to/pidfile" and at startup you do something like this (partially pseudocode because I'm pre-coffee and don't want to work all that hard):

import os, os.path
pidfilePath = """/path/to/pidfile"""
if os.path.exists(pidfilePath):
   pidfile = open(pidfilePath,"r")
   pidString = pidfile.read()
   if <pidString is equal to os.getpid()>:
      # something is real weird
      Sys.exit(BADCODE)
   else:
      <use ps or pidof to see if the process with pid pidString is still running>
      if  <process with pid == 'pidString' is still running>:
          Sys.exit(ALREADAYRUNNING)
      else:
          # the previous server must have crashed
          <log server had crashed>
          <reopen pidfilePath for writing>
          pidfile.write(os.getpid())
else:
    <open pidfilePath for writing>
    pidfile.write(os.getpid())

So, in other words, you're checking if a pidfile exists; if not, write your pid to that file. If the pidfile does exist, then check to see if the pid is the pid of a running process; if so, then you've got another live process running, so just shut down. If not, then the previous process crashed, so log it, and then write your own pid to the file in place of the old one. Then continue.

解决方案 7：

The best solution for this on windows is to use mutexes as suggested by @zgoda.

import win32event
import win32api
from winerror import ERROR_ALREADY_EXISTS

mutex = win32event.CreateMutex(None, False, 'name')
last_error = win32api.GetLastError()

if last_error == ERROR_ALREADY_EXISTS:
   print("App instance already running")

Some answers use fctnl (included also in @sorin tendo package) which is not available on windows and should you try to freeze your python app using a package like pyinstaller which does static imports, it throws an error.

Also, using the lock file method, creates a read-only problem with database files( experienced this with sqlite3).

解决方案 8：

Here is my eventual Windows-only solution. Put the following into a module, perhaps called 'onlyone.py', or whatever. Include that module directly into your main python script file.

import win32event, win32api, winerror, time, sys, os
main_path = os.path.abspath(sys.modules['__main__'].__file__).replace("\\\", "/")

first = True
while True:
        mutex = win32event.CreateMutex(None, False, main_path + "_{<paste YOUR GUID HERE>}")
        if win32api.GetLastError() == 0:
            break
        win32api.CloseHandle(mutex)
        if first:
            print "Another instance of %s running, please wait for completion" % main_path
            first = False
        time.sleep(1)

Explanation

The code attempts to create a mutex with name derived from the full path to the script. We use forward-slashes to avoid potential confusion with the real file system.

Advantages

• No configuration or 'magic' identifiers needed, use it in as many different scripts as needed.

• No stale files left around, the mutex dies with you.

• Prints a helpful message when waiting

解决方案 9：

For anybody using wxPython for their application, you can use the function wx.SingleInstanceChecker documented here.

I personally use a subclass of wx.App which makes use of wx.SingleInstanceChecker and returns False from OnInit() if there is an existing instance of the app already executing like so:

import wx

class SingleApp(wx.App):
    """
    class that extends wx.App and only permits a single running instance.
    """

    def OnInit(self):
        """
        wx.App init function that returns False if the app is already running.
        """
        self.name = "SingleApp-%s".format(wx.GetUserId())
        self.instance = wx.SingleInstanceChecker(self.name)
        if self.instance.IsAnotherRunning():
            wx.MessageBox(
                "An instance of the application is already running", 
                "Error", 
                 wx.OK | wx.ICON_WARNING
            )
            return False
        return True

This is a simple drop-in replacement for wx.App that prohibits multiple instances. To use it simply replace wx.App with SingleApp in your code like so:

app = SingleApp(redirect=False)
frame = wx.Frame(None, wx.ID_ANY, "Hello World")
frame.Show(True)
app.MainLoop()