How do I kill SimpleHTTPServer from within a Python script?

Question

I am trying to use http.server to test all the links in a Python project. I can get my script to work if I start the server before running my script, and then the server stops when I close the terminal window. But I'd really like the script itself to start and stop the server.

I made a test script to simply start the server, get a page and prove the server is running, and then stop the server. I can't seem to get the pid of the server. When I try to kill the pid that this script reports after the script runs, I get a message that there is no such process; but the server is still running.

How do I get the correct pid for the server, or more generally how do I stop the server from the script?

import os
import requests
from time import sleep

# Start server, in background.
print("Starting server...")
os.system('python -m http.server &')
# Make sure server has a chance to start before making request.
sleep(1)

print "Server pid: "
os.system('echo $$')

url = 'http://localhost:8000/index.html'
print("Testing request: ", url)
r = requests.get(url)
print("Status code: ", r.status_code)

You could use subprocess.Popen instead of os.system, it offers a lot of additional functionality including termination of the spawned subprocess. Or you could just import SimpleHTTPServer and use it directly in your script... — l4mpi
– l4mpi, Commented Oct 24, 2013 at 15:56
That is helpful, but do you mind saying a little more? The only references I seem to find about using SimpleHTTPServer talk about using it from the command line; I'm not quite sure how to stop and start the server from within a python script. I have also never used processes before, so I'm not quite sure how to start the server subprocess, and then stop it. Actively googling... — japhyr
– japhyr, Commented Oct 24, 2013 at 16:24
The subprocess documentation has all the info you need, see the sections about the Popen constructor and using Popen objects - you could do something like server = Popen(["python", "-m", "SimpleHTTPServer"]) and then use server.terminate() or server.kill() to end the process. As for starting a SimpleHTTPServer in a script, just look at the if __name__ == "__main__": block in SimpleHTTPServer.py - just keep a reference to the BaseServer instance and close it (it's a TCPServer subclass, so the TCPServer docs should help). — l4mpi
– l4mpi, Commented Oct 24, 2013 at 16:43

User · Accepted Answer · 2016-12-04 22:23:54Z

10

Here is what I am doing:

import threading

try: 
    from http.server import HTTPServer, SimpleHTTPRequestHandler # Python 3
except ImportError: 
    from SimpleHTTPServer import BaseHTTPServer
    HTTPServer = BaseHTTPServer.HTTPServer
    from SimpleHTTPServer import SimpleHTTPRequestHandler # Python 2

server = HTTPServer(('localhost', 0), SimpleHTTPRequestHandler)
thread = threading.Thread(target = server.serve_forever)
thread.daemon = True
thread.start()

def fin():
    server.shutdown()

print('server running on port {}'.format(server.server_port))

# here is your program

If you call fin in your program, then the server shuts down.

edited Dec 4, 2016 at 22:23

answered Oct 24, 2013 at 18:50

User

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6 Comments

japhyr Over a year ago

I am using Python 2 at the moment, but I'd love this to run on Python 3 as well. I got an error 'no module named HTTPServer', so I changed the Python 2 import to from BaseHTTPServer import HTTPServer. The import worked, but then I got an error that "__init__() got an unexpected keyword argument 'daemon'". I get this if I use my own version of this script, or if I just copy your script and fix the import. Any suggestions?

User Over a year ago

I fixed the deamon argument with an additional line. This should work under Python 3 and 2 now - hopefully.

Ciro Santilli OurBigBook.com Over a year ago

Are you able to kill the server with Ctrl-C?

User Over a year ago

@CiroSantilli巴拿馬文件六四事件法轮功 Since deamon is set to true, the program should finish once the main thread ends. Usually, the main thread can be terminated with Control+C.

Ciro Santilli OurBigBook.com Over a year ago

@User I didn't reproduce on Ubuntu 16.04, Python 2.7, gnome terminal. Nothing happens on Ctrl+C, the server just keeps running, and I have to close the terminal for it to stop.

|

Andrew Stewart · Accepted Answer · 2015-02-11 19:47:40Z

3

A slight modification to User's code above:

import threading
try: 
  from http.server import HTTPServer, BaseHTTPRequestHandler # Python 3
except ImportError: 
  import SimpleHTTPServer
  from BaseHTTPServer import HTTPServer # Python 2
  from SimpleHTTPServer import SimpleHTTPRequestHandler as BaseHTTPRequestHandler
server = HTTPServer(('localhost', 0), BaseHTTPRequestHandler)
thread = threading.Thread(target = server.serve_forever)
thread.deamon = True
def up():
  thread.start()
  print('starting server on port {}'.format(server.server_port))
def down():
  server.shutdown()
  print('stopping server on port {}'.format(server.server_port))

answered Feb 11, 2015 at 19:47

Andrew Stewart

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Comments

Levon · Accepted Answer · 2016-08-14 14:09:55Z

This is a closure solution to the problem. Works on python 3.

import os
import threading
import webbrowser
from http.server import HTTPServer, SimpleHTTPRequestHandler


def simple_http_server(host='localhost', port=4001, path='.'):

    server = HTTPServer((host, port), SimpleHTTPRequestHandler)
    thread = threading.Thread(target=server.serve_forever)
    thread.deamon = True

    cwd = os.getcwd()

    def start():
        os.chdir(path)
        thread.start()
        webbrowser.open_new_tab('http://{}:{}'.format(host, port))
        print('starting server on port {}'.format(server.server_port))

    def stop():
        os.chdir(cwd)
        server.shutdown()
        server.socket.close()
        print('stopping server on port {}'.format(server.server_port))

    return start, stop

simple_http_server which will return start and stop functions

>>> start, stop = simple_http_server(port=4005, path='/path/to/folder')

which you can use as

>>> start()
starting server on port 4005

127.0.0.1 - - [14/Aug/2016 17:49:31] "GET / HTTP/1.1" 200 -

>>> stop()
stopping server on port 4005

Tom Pohl · Accepted Answer · 2021-05-06 06:41:53Z

Here's a context-flavored version which I prefer because it cleans up automatically and you can specify the directory to serve:

from contextlib import contextmanager
from functools import partial
from http.server import SimpleHTTPRequestHandler, ThreadingHTTPServer
from threading import Thread


@contextmanager
def http_server(host: str, port: int, directory: str):
    server = ThreadingHTTPServer(
        (host, port), partial(SimpleHTTPRequestHandler, directory=directory)
    )
    server_thread = Thread(target=server.serve_forever, name="http_server")
    server_thread.start()

    try:
        yield
    finally:
        server.shutdown()
        server_thread.join()


def usage_example():
    import time

    with http_server("127.0.0.1", 8087, "."):
        # now you can use the web server
        time.sleep(100)

Community · Accepted Answer · 2017-05-23 12:16:47Z

1

I got this to run, but I'm curious to hear how this compares to User's answer above. I came up with this after looking at the accepted answer here.

import subprocess
import requests
import os
import signal
from time import sleep

print "Starting server..."
cmd = 'python -m SimpleHTTPServer'
pro = subprocess.Popen(cmd, shell=True, preexec_fn=os.setsid)

# Make sure server has a chance to start before making request.
sleep(1)

url = 'http://localhost:8000/index.html'
print "Testing request: ", url
r = requests.get(url)
print "Status code: ", r.status_code

os.killpg(pro.pid, signal.SIGTERM)

edited May 23, 2017 at 12:16

CommunityBot

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answered Oct 24, 2013 at 23:34

japhyr

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Comments

heltonitba · Accepted Answer · 2015-09-03 16:57:41Z

0

My solution with browser opening:

File: http.py

import SimpleHTTPServer
import SocketServer
import threading
import webbrowser
import platform
from socket import SOL_SOCKET,SO_REUSEADDR

class HTTPServer():

    def __init__(self,port=8000,url='http://localhost'):
        self.port = port
        self.thread = None
        self.httpd = None                
        self.run = False
        self.url = url

        os = platform.system()
        if os=='Linux':
            self.browser_path = "/usr/bin/google-chrome %s"            
        elif os == 'Windows':
            self.browser_path = "C:/Program Files (x86)/Google/Chrome/Application/chrome.exe %s"   
        else:
            print("Chrome not found!")


    def start(self):        
        self.run = True     
        self.httpd = SocketServer.TCPServer(("", self.port), SimpleHTTPServer.SimpleHTTPRequestHandler)
        self.httpd.socket.setsockopt(SOL_SOCKET, SO_REUSEADDR, 1)
        self.thread = threading.Thread(target = self._serve)
        self.thread.start()
        webbrowser.get(str(self.browser_path)).open(self.url+":"+str(self.port)+"/")


    def _serve(self):        
        while self.run:
            self.httpd.handle_request()

    def stop(self):
        self.run = False
        self.httpd.server_close()

After, just run:

from http import HTTPServer

server = HTTPServer()
server.start()

raw_input("Enter to close")

server.stop()

answered Sep 3, 2015 at 16:57

heltonitba

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1 Comment

Josep Valls Over a year ago

Instead of opening Chrome you can use start %s on Windows and open %s on OSX (not sure for Linux) to open the default browser.

Dun · Accepted Answer · 2019-11-24 14:57:26Z

0

In my opinion, what I did is:

go to activity monitor
then go to network
search for word 'python'

(only if you ran the cmd/terminal command in python, because I think the way to stop it in cmd/terminal is ctrl+c or just quit the cmd/terminal)

note: the pid is also there

quit the process 'python' (if it doesn't, use force quit process)
if this works in cmd/terminal, then you are done! enjoii~

Hope this helps :D!!

answered Nov 24, 2019 at 14:57

Dun

578 bronze badges

4 Comments

ZF007 Over a year ago

And how is this an "internal" approach?

Dun Over a year ago

well I have the same problem as yours, so I searched what is PID in google and shows the result in google like this: google.com/… then I have an idea of just quitting it then it does work lol

ZF007 Over a year ago

You're approach is in essence giving the result via an external "forced" process. Teh destinction here is that OP is asking for a solution from within the serverscript. This can be done via threading routines or latest asyncio methods. I don't have that problem and merely reviewed your answer as it was flagged for "low quality answer".. and judging whether it was needed to be removed or not. I hope my comment helps you to carefully look at the question again and revaluate whether your answer provides the correct solution OP was asking for.

Dun Over a year ago

Ok, I understand :)

Aman Kejriwal · Accepted Answer · 2020-07-30 15:21:10Z

0

Using Python 3.8

import http.server
import socketserver
import threading
PORT = 8000
Handler = http.server.SimpleHTTPRequestHandler
server = socketserver.TCPServer(("", PORT), Handler)
thread = threading.Thread(target = server.serve_forever)
thread.daemon = True
thread.start()

answered Jul 30, 2020 at 15:21

Aman Kejriwal

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