Running SimpleXMLRPCServer in separate thread and shutting down - python

I have a class that I wish to test via SimpleXMLRPCServer in python. The way I have my unit test set up is that I create a new thread, and start SimpleXMLRPCServer in that. Then I run all the test, and finally shut down.
This is my ServerThread:
class ServerThread(Thread):
running = True
def run(self):
self.server = #Creates and starts SimpleXMLRPCServer
while (self.running):
self.server.handle_request()
def stop(self):
self.running = False
self.server.server_close()
The problem is, that calling ServerThread.stop(), followed by Thread.stop() and Thread.join() will not cause the thread to stop properly if it's already waiting for a request in handle_request. And since there doesn't seem to be any interrupt or timeout mechanisms here that I can use, I am at a loss for how I can cleanly shut down the server thread.

I had the same problem and after hours of research i solved it by switching from using my own handle_request() loop to serve_forever() to start the server.
serve_forever() starts an internal loop like yours. This loop can be stopped by calling shutdown(). After stopping the loop it is possible to stop the server with server_close().
I don't know why this works and the handle_request() loop don't, but it does ;P
Here is my code:
from threading import Thread
from xmlrpc.server import SimpleXMLRPCServer
from pyWebService.server.service.WebServiceRequestHandler import WebServiceRquestHandler
class WebServiceServer(Thread):
def __init__(self, ip, port):
super(WebServiceServer, self).__init__()
self.running = True
self.server = SimpleXMLRPCServer((ip, port),requestHandler=WebServiceRquestHandler)
self.server.register_introspection_functions()
def register_function(self, function):
self.server.register_function(function)
def run(self):
self.server.serve_forever()
def stop_server(self):
self.server.shutdown()
self.server.server_close()
print("starting server")
webService = WebServiceServer("localhost", 8010)
webService.start()
print("stopping server")
webService.stop_server()
webService.join()
print("server stopped")

Two suggestions.
Suggestion One is to use a separate process instead of a separate thread.
Create a stand-alone XMLRPC server program.
Start it with subprocess.Popen().
Kill it when the test is done. In standard OS's (not Windows) the kill works nicely. In Windows, however, there's no trivial kill function, but there are recipes for this.
The other suggestion is to have a function in your XMLRPC server which causes server self-destruction. You define a function that calls sys.exit() or os.abort() or raises a similar exception that will stop the process.

This is my way. send SIGTERM to self. (Works for me)
Server code
import os
import signal
import xmlrpc.server
server = xmlrpc.server.SimpleXMLRPCServer(("0.0.0.0", 8000))
server.register_function(lambda: os.kill(os.getpid(), signal.SIGTERM), 'quit')
server.serve_forever()
Client code
import xmlrpc.client
c = xmlrpc.client.ServerProxy("http://localhost:8000")
try:
c.quit()
except ConnectionRefusedError:
pass

Related

Run Python HTTPServer in Background and Continue Script Execution

I am trying to figure out how to run my overloaded customized BaseHTTPServer instance in the background after running the "".serve_forever() method.
Normally when you run the method execution will hang until you execute a keyboard interrupt, but I would like it to serve requests in the background while continuing script execution. Please help!
You can start the server in a different thread: https://docs.python.org/3/library/_thread.html#thread.start_new_thread
So something like:
def start_server():
# Setup stuff here...
server.serve_forever()
# start the server in a background thread
thread.start_new_thread(start_server)
print('The server is running but my script is still executing!')
I was trying to do some long-term animation using async and thought I'd have to rewrite server to use aiohttp (https://docs.aiohttp.org/en/v0.12.0/web.html), but Olivers technique of using seperate thread saved me all that pain. My code looks like this, where MyHTTPServer is simply my custom sublass of HTTPServer
import threading
import asyncio
from http.server import BaseHTTPRequestHandler, HTTPServer
import socketserver
import io
import threading
async def tick_async(server):
while True:
server.animate_something()
await asyncio.sleep(1.0)
def start_server():
httpd.serve_forever()
try:
print('Server listening on port 8082...')
httpd = MyHTTPServer(('', 8082), MyHttpHandler)
asyncio.ensure_future(tick_async(httpd))
loop = asyncio.get_event_loop()
t = threading.Thread(target=start_server)
t.start()
loop.run_forever()

Shutdown an SimpleXMLRPCServer server in python

Currently I am writing an application using the SimpleXMLRPCServer module in Python.
The basic aim of this application is to keep running on a server and keep checking a Queue for any task. If it encounters any new request in the Queue, serve the request.
Snapshot of what I am trying to do :
class MyClass():
"""
This class will have methods which will be exposed to the clients
"""
def __init__(self):
taskQ = Queue.Queue()
def do_some_task(self):
while True:
logging.info("Checking the Queue for any Tasks..")
task = taskQ.get()
# Do some processing based on the availability of some task
Main
if name == "main":
server = SimpleXMLRPCServer.SimpleXMLRPCServer((socket.gethostname(), Port)
classObj = MyClass()
rpcserver.register_function(classObj.do_some_task)
rpcserver.serve_forever()
Once the server is started it remains in the loop forever inside do_some_task method to keep checking the Queue for any task. This is what i wanted to achieve. But now i want to gracefully shutdown the server. In this case i am unable to shutdown the server.
Till now I have Tried using a global flag STOP_SERVER for 'True' and checking its status in the do_some_task while loop to get out of it and stop the server. But no help.
Tried using SHUTDOWN() method of the SimpleXMLRPCServer but it seems it is getting into a infinite loop of somekind.
Could you suggest some proper way to gracefully shutdown the server.
Thanks in advance
You should use handle_request() instead of serve_forever() if you want to close it manualy. Because SimpleXMLRPCServer is implemented as a single thread and the serve_forever() will make the server instance run into an infinite loop.
You can refer to this article. This is an example cited from there:
from SimpleXMLRPCServer import *
class MyServer(SimpleXMLRPCServer):
def serve_forever(self):
self.quit = 0
while not self.quit:
self.handle_request()
def kill():
server.quit = 1
return 1
server = MyServer(('127.0.0.1', 8000))
server.register_function(kill)
server.serve_forever()
By using handle_request(), this code use a state variable self.quit to indicate whether to quit the infinite loop.
The serve_forever function is inherited from a base class in the socketserver module called BaseServer. If you look at this fucntion you'll see it has an attribute called __shutdown_request, and this can be used to break the serving while loop. Because of the double underscore you'll have to access the variable with its mangled name: _BaseServer__shutdown_request.
Putting that all together you can make a very simple quit function as follows:
from xmlrpc.server import SimpleXMLRPCServer
class MyXMLRPCServer(SimpleXMLRPCServer):
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
self.register_function(self.quit)
def quit(self):
self._BaseServer__shutdown_request = True
return 0

Why can't this Python multiprocessing process be terminated from Kivy?

I'm trying to run a django development server from within a Kivy application. This did work out quite well so far.
Now i want to allow the user to continue working with the program while the server is running. My idea was to create a multiprocessing.Process for the httpd.serve_forever() to avoid a complete lock of the main program. Did work well. This is the code in my internal_django module:
import multiprocessing
import os
import time
from wsgiref.simple_server import make_server
def django_wsgi_application():
PROJECT_ROOT = os.path.dirname(os.path.abspath(__file__))
settings_module = "djangosettings"#%s.djangosettings" % PROJECT_ROOT.split(os.sep)[-1]
os.environ.update({"DJANGO_SETTINGS_MODULE":settings_module})
from django.core.wsgi import get_wsgi_application
application = get_wsgi_application()
return application
class Singleton(type):
_instances = {}
def __call__(cls, *args, **kwargs):
if cls not in cls._instances:
cls._instances[cls] = super(Singleton, cls).__call__(*args, **kwargs)
return cls._instances[cls]
class DjangoServer():
__metaclass__ = Singleton
def start(self):
self.httpd = make_server('', 8000, django_wsgi_application())
self.server = multiprocessing.Process(target=self.httpd.serve_forever)
self.server.start()
print "Now serving on port 8000..."
print "Server Process PID = %s" %self.server.pid
def stop(self):
print("shutdown initiated")
print "Server Process PID = %s" %self.server.pid
while self.server.is_alive():
self.server.terminate()
print("Server should have shut down")
time.sleep(1)
print("Server is_alive: %s" %self.server.is_alive())
self.server.join()
print("server process joined")
if __name__ == "__main__":
server = DjangoServer()
server.start()
time.sleep(3)
server.stop()
When i run this code, everything works as expected. This is what is being put out in the console:
Now serving on port 8000...
Server Process PID = 1406
shutdown initiated
Server Process PID = 1406
Server should have shut down
Server is_alive: False
server process joined
Next step was to provide a way to stop the server from within the Kivy application. For that i just wanted to use my DjangoServer class as i did before:
from internal_django import DjangoServer
class StartScreen(Screen):
def start_server(self):
server = DjangoServer()
server.start()
class StopScreen(Screen):
def stop_server(self):
server = DjangoServer()
server.stop()
But when doing so, the process once started never quits. My first idea was that the Singleton did not work as expected, and that i try to quit the wrong process. but as you can see in the output, the PID's are identical. The server receives the terminate command, but just continues to work. This is what the console looks like:
Now serving on port 8000...
Server Process PID = 1406
shutdown initiated
Server Process PID = 1406
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
Server should have shut down
(and so on, until i manually kill the server process)
Am i using multiprocessing in a completely wrong way? Is Kivy somehow interfering with the process?
I think the problems here might be two:
A signal handler is intercepting the TERM request sent by Process.terminate() and ignores it. To verify that simply use the signal.getsignal(signal.SIGTERM) from within the new process and print the results. To circumvent such issue you can reset the default behavior with signal.signal(signal.SIGTERM, signal.SIG_DFL), nevertheless keep in mind that there might be a reason why SIGTERM is silenced by the frameworks (I'm not familiar neither with Django nor with Kivy).
If you're using Python 2 you must consider that the interpreter does not process signals if it's blocked on a synchronization primitive from threading library (Locks, Semaphores..) or on a native C call. The serve_forever() function might fall in these cases (use the force of the source!). Quick check could be trying to run the code on Python 3 and see whether it works or not.
A quick and dirty solution consists in waiting a small amount of time and send a SIGKILL if the process is still alive.
import os
import signal
process.terminate()
process.join(1)
if process.is_alive() and os.name != 'nt':
try:
os.kill(process.pid, signal.SIGKILL)
process.join()
except OSError:
return # process might have died while checking it
On windows you cannot kill a process in such simple way that's why I test the os.name.
It's a pretty raw approach so I'd rather recommend to find the cause of the issue.
What happens if you call terminate(), then join() and skip the while loop? Also, I shuffle the code a little and factor some code into _create_server(). Please let me know if this works out for you.
class DjangoServer():
__metaclass__ = Singleton
def _create_server(self):
httpd = make_server('', 8000, django_wsgi_application())
print "Now serving on port {}...".format(httpd.server_port)
httpd.serve_forever()
def start(self):
self.server = multiprocessing.Process(target=self._create_server)
self.server.start()
print "Server Process PID = %s" %self.server.pid
def stop(self):
print("shutdown initiated")
print "Server Process PID = %s" %self.server.pid
self.server.terminate()
self.server.join()
print("server process terminated")

How to really test signal handling in Python?

My code is simple:
def start():
signal(SIGINT, lambda signal, frame: raise SystemExit())
startTCPServer()
So I register my application with signal handling of SIGINT, then I start a start a TCP listener.
here are my questions:
How can I using python code to send a SIGINT signal?
How can I test whether if the application receives a signal of SIGINT, it will raise a SystemExit exception?
If I run start() in my test, it will block and how can I send a signal to it?
First of, testing the signal itself is a functional or integration test, not a unit test. See What's the difference between unit, functional, acceptance, and integration tests?
You can run your Python script as a subprocess with subprocess.Popen(), then use the Popen.send_signal() method to send signals to that process, then test that the process has exited with Popen.poll().
How can I using python code to send a SIGINT signal?
You can use os.kill, which slightly misleadingly, can used to send any signal to any process by its ID. The process ID of the application/test can be found by os.getpid(), so you would have...
pid = os.getpid()
# ... other code discussed later in the answer ...
os.kill(pid, SIGINT)
How can I test whether if the application receives a signal of SIGINT, it will raise a SystemExit exception?
The usual way in a test you can check that some code raises SystemExit, is with unittest.TestCase::assertRaises...
import start
class TestStart(unittest.TestCase):
def test_signal_handling(self):
# ... other code discussed later in the answer ...
with self.assertRaises(SystemExit):
start.start()
If I run start() in my test, it will block and how can I send a signal to it?
This is the trick: you can start another thread which then sends a signal back to the main thread which is blocking.
Putting it all together, assuming your production start function is in start.py:
from signal import (
SIGINT,
signal,
)
import socketserver
def startTCPServer():
# Taken from https://docs.python.org/3.4/library/socketserver.html#socketserver-tcpserver-example
class MyTCPHandler(socketserver.BaseRequestHandler):
def handle(self):
self.data = self.request.recv(1024).strip()
self.request.sendall(self.data.upper())
HOST, PORT = "localhost", 9999
server = socketserver.TCPServer((HOST, PORT), MyTCPHandler)
server.serve_forever()
def start():
def raiseSystemExit(_, __):
raise SystemExit
signal(SIGINT, raiseSystemExit)
startTCPServer()
Then your test code could be like the following, say in test.py
import os
from signal import (
SIGINT,
)
import threading
import time
import unittest
import start
class TestStart(unittest.TestCase):
def test_signal_handling(self):
pid = os.getpid()
def trigger_signal():
# You could do something more robust, e.g. wait until port is listening
time.sleep(1)
os.kill(pid, SIGINT)
thread = threading.Thread(target=trigger_signal)
thread.daemon = True
thread.start()
with self.assertRaises(SystemExit):
start.start()
if __name__ == '__main__':
unittest.main()
and run using
python test.py
The above is the same technique as in the answer at https://stackoverflow.com/a/49500820/1319998

How to stop BaseHTTPServer.serve_forever() in a BaseHTTPRequestHandler subclass?

I am running my HTTPServer in a separate thread (using the threading module which has no way to stop threads...) and want to stop serving requests when the main thread also shuts down.
The Python documentation states that BaseHTTPServer.HTTPServer is a subclass of SocketServer.TCPServer, which supports a shutdown method, but it is missing in HTTPServer.
The whole BaseHTTPServer module has very little documentation :(
Another way to do it, based on http://docs.python.org/2/library/basehttpserver.html#more-examples, is: instead of serve_forever(), keep serving as long as a condition is met, with the server checking the condition before and after each request. For example:
import CGIHTTPServer
import BaseHTTPServer
KEEP_RUNNING = True
def keep_running():
return KEEP_RUNNING
class Handler(CGIHTTPServer.CGIHTTPRequestHandler):
cgi_directories = ["/cgi-bin"]
httpd = BaseHTTPServer.HTTPServer(("", 8000), Handler)
while keep_running():
httpd.handle_request()
I should start by saying that "I probably wouldn't do this myself, but I have in the past". The serve_forever (from SocketServer.py) method looks like this:
def serve_forever(self):
"""Handle one request at a time until doomsday."""
while 1:
self.handle_request()
You could replace (in subclass) while 1 with while self.should_be_running, and modify that value from a different thread. Something like:
def stop_serving_forever(self):
"""Stop handling requests"""
self.should_be_running = 0
# Make a fake request to the server, to really force it to stop.
# Otherwise it will just stop on the next request.
# (Exercise for the reader.)
self.make_a_fake_request_to_myself()
Edit: I dug up the actual code I used at the time:
class StoppableRPCServer(SimpleXMLRPCServer.SimpleXMLRPCServer):
stopped = False
allow_reuse_address = True
def __init__(self, *args, **kw):
SimpleXMLRPCServer.SimpleXMLRPCServer.__init__(self, *args, **kw)
self.register_function(lambda: 'OK', 'ping')
def serve_forever(self):
while not self.stopped:
self.handle_request()
def force_stop(self):
self.server_close()
self.stopped = True
self.create_dummy_request()
def create_dummy_request(self):
server = xmlrpclib.Server('http://%s:%s' % self.server_address)
server.ping()
The event-loops ends on SIGTERM, Ctrl+C or when shutdown() is called.
server_close() must be called after server_forever() to close the listening socket.
import http.server
class StoppableHTTPServer(http.server.HTTPServer):
def run(self):
try:
self.serve_forever()
except KeyboardInterrupt:
pass
finally:
# Clean-up server (close socket, etc.)
self.server_close()
Simple server stoppable with user action (SIGTERM, Ctrl+C, ...):
server = StoppableHTTPServer(("127.0.0.1", 8080),
http.server.BaseHTTPRequestHandler)
server.run()
Server running in a thread:
import threading
server = StoppableHTTPServer(("127.0.0.1", 8080),
http.server.BaseHTTPRequestHandler)
# Start processing requests
thread = threading.Thread(None, server.run)
thread.start()
# ... do things ...
# Shutdown server
server.shutdown()
thread.join()
In my python 2.6 installation, I can call it on the underlying TCPServer - it still there inside your HTTPServer:
TCPServer.shutdown
>>> import BaseHTTPServer
>>> h=BaseHTTPServer.HTTPServer(('',5555), BaseHTTPServer.BaseHTTPRequestHandler)
>>> h.shutdown
<bound method HTTPServer.shutdown of <BaseHTTPServer.HTTPServer instance at 0x0100D800>>
>>>
I think you can use [serverName].socket.close()
In python 2.7, calling shutdown() works but only if you are serving via serve_forever, because it uses async select and a polling loop. Running your own loop with handle_request() ironically excludes this functionality because it implies a dumb blocking call.
From SocketServer.py's BaseServer:
def serve_forever(self, poll_interval=0.5):
"""Handle one request at a time until shutdown.
Polls for shutdown every poll_interval seconds. Ignores
self.timeout. If you need to do periodic tasks, do them in
another thread.
"""
self.__is_shut_down.clear()
try:
while not self.__shutdown_request:
# XXX: Consider using another file descriptor or
# connecting to the socket to wake this up instead of
# polling. Polling reduces our responsiveness to a
# shutdown request and wastes cpu at all other times.
r, w, e = select.select([self], [], [], poll_interval)
if self in r:
self._handle_request_noblock()
finally:
self.__shutdown_request = False
self.__is_shut_down.set()
Heres part of my code for doing a blocking shutdown from another thread, using an event to wait for completion:
class MockWebServerFixture(object):
def start_webserver(self):
"""
start the web server on a new thread
"""
self._webserver_died = threading.Event()
self._webserver_thread = threading.Thread(
target=self._run_webserver_thread)
self._webserver_thread.start()
def _run_webserver_thread(self):
self.webserver.serve_forever()
self._webserver_died.set()
def _kill_webserver(self):
if not self._webserver_thread:
return
self.webserver.shutdown()
# wait for thread to die for a bit, then give up raising an exception.
if not self._webserver_died.wait(5):
raise ValueError("couldn't kill webserver")
This is a simplified version of Helgi's answer for python 3.7:
import threading
import time
from http.server import ThreadingHTTPServer, SimpleHTTPRequestHandler
class MyServer(threading.Thread):
def run(self):
self.server = ThreadingHTTPServer(('localhost', 8000), SimpleHTTPRequestHandler)
self.server.serve_forever()
def stop(self):
self.server.shutdown()
if __name__ == '__main__':
s = MyServer()
s.start()
print('thread alive:', s.is_alive()) # True
time.sleep(2)
s.stop()
print('thread alive:', s.is_alive()) # False
This method I use successfully (Python 3) to stop the server from the web application itself (a web page):
import http.server
import os
import re
class PatientHTTPRequestHandler(http.server.SimpleHTTPRequestHandler):
stop_server = False
base_directory = "/static/"
# A file to use as an "server stopped user information" page.
stop_command = "/control/stop.html"
def send_head(self):
self.path = os.path.normpath(self.path)
if self.path == PatientHTTPRequestHandler.stop_command and self.address_string() == "127.0.0.1":
# I wanted that only the local machine could stop the server.
PatientHTTPRequestHandler.stop_server = True
# Allow the stop page to be displayed.
return http.server.SimpleHTTPRequestHandler.send_head(self)
if self.path.startswith(PatientHTTPRequestHandler.base_directory):
return http.server.SimpleHTTPRequestHandler.send_head(self)
else:
return self.send_error(404, "Not allowed", "The path you requested is forbidden.")
if __name__ == "__main__":
httpd = http.server.HTTPServer(("127.0.0.1", 8080), PatientHTTPRequestHandler)
# A timeout is needed for server to check periodically for KeyboardInterrupt
httpd.timeout = 1
while not PatientHTTPRequestHandler.stop_server:
httpd.handle_request()
This way, pages served via base address http://localhost:8080/static/ (example http://localhost:8080/static/styles/common.css) will be served by the default handler, an access to http://localhost:8080/control/stop.html from the server's computer will display stop.html then stop the server, any other option will be forbidden.
I tried all above possible solution and ended up with having a "sometime" issue - somehow it did not really do it - so I ended up making a dirty solution that worked all the time for me:
If all above fails, then brute force kill your thread using something like this:
import subprocess
cmdkill = "kill $(ps aux|grep '<name of your thread> true'|grep -v 'grep'|awk '{print $2}') 2> /dev/null"
subprocess.Popen(cmdkill, stdout=subprocess.PIPE, shell=True)
import http.server
import socketserver
import socket as sck
import os
import threading
class myserver:
def __init__(self, PORT, LOCATION):
self.thrd = threading.Thread(None, self.run)
self.Directory = LOCATION
self.Port = PORT
hostname = sck.gethostname()
ip_address = sck.gethostbyname(hostname)
self.url = 'http://' + ip_address + ':' + str(self.Port)
Handler = http.server.SimpleHTTPRequestHandler
self.httpd = socketserver.TCPServer(("", PORT), Handler)
print('Object created, use the start() method to launch the server')
def run(self):
print('listening on: ' + self.url )
os.chdir(self.Directory)
print('myserver object started')
print('Use the objects stop() method to stop the server')
self.httpd.serve_forever()
print('Quit handling')
print('Sever stopped')
print('Port ' + str(self.Port) + ' should be available again.')
def stop(self):
print('Stopping server')
self.httpd.shutdown()
self.httpd.server_close()
print('Need just one more request before shutting down'
def start(self):
self.thrd.start()
def help():
helpmsg = '''Create a new server-object by initialising
NewServer = webserver3.myserver(Port_number, Directory_String)
Then start it using NewServer.start() function
Stop it using NewServer.stop()'''
print(helpmsg)
Not a experience python programmer, just wanting to share my comprehensive solution. Mostly based on snippets here and there. I usually import this script in my console and it allows me to set up multiple servers for different locations using their specific ports, sharing my content with other devices on the network.
Here's a context-flavored version for Python 3.7+ 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)

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