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)
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 tendopip install tendo- manually by getting it from http://pypi.python.org/pypi/tendo
Simple, cross-platform solution, found in another question by zgoda:
import fcntl, sys
pid_file = 'program.pid'
fp = open(pid_file, 'w')
try:
fcntl.lockf(fp, fcntl.LOCK_EX | fcntl.LOCK_NB)
except IOError:
# another instance is running
sys.exit(0)
A lot like S.Lott's suggestion, but with the code.
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( '\0postconnect_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.
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).
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).
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.
You already found reply to similar question in another thread, so for completeness sake see how to achieve the same on Windows uning named mutex.
This may work.
Attempt create a PID file to a known location. If you fail, someone has the file locked, you're done.
When you finish normally, close and remove the PID file, so someone else can overwrite it.
You can wrap your program in a shell script that removes the PID file even if your program crashes.
You can, also, use the PID file to kill the program if it hangs.
Using a lock-file is a quite common approach on unix. If it crashes, you have to clean up manually. You could stor the PID in the file, and on startup check if there is a process with this PID, overriding the lock-file if not. (However, you also need a lock around the read-file-check-pid-rewrite-file). You will find what you need for getting and checking pid in the os-package. The common way of checking if there exists a process with a given pid, is to send it a non-fatal signal.
Other alternatives could be combining this with flock or posix semaphores.
Opening a network socket, as saua proposed, would probably be the easiest and most portable.
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()
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
I'm posting this as an answer because I'm a new user and Stack Overflow won't let me vote yet.
Sorin Sbarnea's solution works for me under OS X, Linux and Windows, and I am grateful for it.
However, tempfile.gettempdir() behaves one way under OS X and Windows and another under other some/many/all(?) *nixes (ignoring the fact that OS X is also Unix!). The difference is important to this code.
OS X and Windows have user-specific temp directories, so a tempfile created by one user isn't visible to another user. By contrast, under many versions of *nix (I tested Ubuntu 9, RHEL 5, OpenSolaris 2008 and FreeBSD 8), the temp dir is /tmp for all users.
That means that when the lockfile is created on a multi-user machine, it's created in /tmp and only the user who creates the lockfile the first time will be able to run the application.
A possible solution is to embed the current username in the name of the lock file.
It's worth noting that the OP's solution of grabbing a port will also misbehave on a multi-user machine.
I use single_process on my gentoo;
pip install single_process
example:
from single_process import single_process
@single_process
def main():
print 1
if __name__ == "__main__":
main()
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).
I keep suspecting there ought to be a good POSIXy solution using process groups, without having to hit the file system, but I can't quite nail it down. Something like:
On startup, your process sends a 'kill -0' to all processes in a particular group. If any such processes exist, it exits. Then it joins the group. No other processes use that group.
However, this has a race condition - multiple processes could all do this at precisely the same time and all end up joining the group and running simultaneously. By the time you've added some sort of mutex to make it watertight, you no longer need the process groups.
This might be acceptable if your process only gets started by cron, once every minute or every hour, but it makes me a bit nervous that it would go wrong precisely on the day when you don't want it to.
I guess this isn't a very good solution after all, unless someone can improve on it?
I ran into this exact problem last week, and although I did find some good solutions, I decided to make a very simple and clean python package and uploaded it to PyPI. It differs from tendo in that it can lock any string resource name. Although you could certainly lock __file__ to achieve the same effect.
Install with: pip install quicklock
Using it is extremely simple:
[nate@Nates-MacBook-Pro-3 ~/live] python
Python 2.7.6 (default, Sep 9 2014, 15:04:36)
[GCC 4.2.1 Compatible Apple LLVM 6.0 (clang-600.0.39)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from quicklock import singleton
>>> # Let's create a lock so that only one instance of a script will run
...
>>> singleton('hello world')
>>>
>>> # Let's try to do that again, this should fail
...
>>> singleton('hello world')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/Users/nate/live/gallery/env/lib/python2.7/site-packages/quicklock/quicklock.py", line 47, in singleton
raise RuntimeError('Resource <{}> is currently locked by <Process {}: "{}">'.format(resource, other_process.pid, other_process.name()))
RuntimeError: Resource <hello world> is currently locked by <Process 24801: "python">
>>>
>>> # But if we quit this process, we release the lock automatically
...
>>> ^D
[nate@Nates-MacBook-Pro-3 ~/live] python
Python 2.7.6 (default, Sep 9 2014, 15:04:36)
[GCC 4.2.1 Compatible Apple LLVM 6.0 (clang-600.0.39)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> from quicklock import singleton
>>> singleton('hello world')
>>>
>>> # No exception was thrown, we own 'hello world'!
Take a look: https://pypi.python.org/pypi/quicklock
linux example
This method is based on the creation of a temporary file automatically deleted after you close the application. the program launch we verify the existence of the file; if the file exists ( there is a pending execution) , the program is closed ; otherwise it creates the file and continues the execution of the program.
from tempfile import *
import time
import os
import sys
f = NamedTemporaryFile( prefix='lock01_', delete=True) if not [f for f in os.listdir('/tmp') if f.find('lock01_')!=-1] else sys.exit()
YOUR CODE COMES HERE
On a Linux system one could also ask
pgrep -a for the number of instances, the script
is found in the process list (option -a reveals the
full command line string). E.g.
import os
import sys
import subprocess
procOut = subprocess.check_output( "/bin/pgrep -u $UID -a python", shell=True,
executable="/bin/bash", universal_newlines=True)
if procOut.count( os.path.basename(__file__)) > 1 :
sys.exit( ("found another instance of >{}<, quitting."
).format( os.path.basename(__file__)))
Remove -u $UID if the restriction should apply to all users.
Disclaimer: a) it is assumed that the script's (base)name is unique, b) there might be race conditions.
import sys,os
# start program
try: # (1)
os.unlink('lock') # (2)
fd=os.open("lock", os.O_CREAT|os.O_EXCL) # (3)
except:
try: fd=os.open("lock", os.O_CREAT|os.O_EXCL) # (4)
except:
print "Another Program running !.." # (5)
sys.exit()
# your program ...
# ...
# exit program
try: os.close(fd) # (6)
except: pass
try: os.unlink('lock')
except: pass
sys.exit()
来源：https://stackoverflow.com/questions/380870/make-sure-only-a-single-instance-of-a-program-is-running