Restarting a self-updating python script

I have written a script that will keep itself up to date by downloading the latest version from a website and overwriting the running script.

I am not sure what the best way to restart 开发者_StackOverflow中文版the script after it has been updated.

Any ideas?

I don't really want to have a separate update script. oh and it has to work on both linux/windows too.

---

In Linux, or any other form of unix, os.execl and friends are a good choice for this -- you just need to re-exec sys.executable with the same parameters it was executed with last time (sys.argv, more or less) or any variant thereof if you need to inform your next incarnation that it's actually a restart. On Windows, os.spawnl (and friends) is about the best you can do (though it will transiently take more time and memory than os.execl and friends would during the transition).

---

The CherryPy project has code that restarts itself. Here's how they do it:

    args = sys.argv[:]
    self.log('Re-spawning %s' % ' '.join(args))

    args.insert(0, sys.executable)
    if sys.platform == 'win32':
        args = ['"%s"' % arg for arg in args]

    os.chdir(_startup_cwd)
    os.execv(sys.executable, args)

I've used this technique in my own code, and it works great. (I didn't bother to do the argument-quoting step on windows above, but it's probably necessary if arguments could contain spaces or other special characters.)

---

I think the best solution whould be something like this:

Your normal program:

...

# ... part that downloaded newest files and put it into the "newest" folder

from subprocess import Popen

Popen("/home/code/reloader.py", shell=True) # start reloader

exit("exit for updating all files")

The update script: (e.g.: home/code/reloader.py)

from shutil import copy2, rmtree
from sys import exit

# maybie you could do this automatic:
copy2("/home/code/newest/file1.py", "/home/code/") # copy file
copy2("/home/code/newest/file2.py", "/home/code/")
copy2("/home/code/newest/file3.py", "/home/code/")
...

rmtree('/home/code/newest') # will delete the folder itself

Popen("/home/code/program.py", shell=True) # go back to your program

exit("exit to restart the true program")

I hope this will help you.

---

The cleanest solution is a separate update script!

Run your program inside it, report back (when exiting) that a new version is available. This allows your program to save all of its data, the updater to apply the update, and run the new version, which then loads the saved data and continues. To the user this can be completely transparent, as they just run the updater-shell which runs the real program.

---

To additionally support script calls with Python's "-m" parameter the following can be used (based on the Alex's answer; Windows version):

os.spawnl(os.P_WAIT, sys.executable, *([sys.executable] +
    (sys.argv if __package__ is None else ["-m", __loader__.name] + sys.argv[1:])))
sys.exit()

---

Main File:

if __name__ == '__main__':

if os.path.isfile('__config.py'):
    print 'Development'
    push.update_server()
else:
    e = update.check()
    if not e: sys.exit()

Update File:

def check():
    e = 1.....perform checks, if something needs updating, e=0;
    if not e:
        os.system("python main.pyw")
    return e

Here's the logic:

Main program calls the update function

1) If the update function needs to update, than it updates and calls a new instances of "main"

Then the original instance of "main" exits.

2) If the update function does not need to update, then "main" continues to run

---

Wouldn't it just be easier to do something like Very simple, no extra imports needed, and compatible with any OS depending on what you put in the os.system field

def restart_program():
  print("Restarting Now...")
  os.system('your program here')

---

You can use reload(module) to reload a module.