I'm trying to create multithreaded web server in python, but it only responds to one request at a time and I can't figure out why. Can you help me, please?
#!/usr/bin/env python2
# -*- coding: utf-8 -*-
from SocketServer import ThreadingMixIn
from BaseHTTPServer import HTTPServer
from SimpleHTTPServer import SimpleHTTPRequestHandler
from time import sleep
class ThreadingServer(ThreadingMixIn, HTTPServer):
pass
class RequestHandler(SimpleHTTPRequestHandler):
def do_GET(self):
self.send_response(200)
self.send_header('Content-type', 'text/plain')
sleep(5)
response = 'Slept for 5 seconds..'
self.send_header('Content-length', len(response))
self.end_headers()
self.wfile.write(response)
ThreadingServer(('', 8000), RequestHandler).serve_forever()
Check this post from Doug Hellmann's blog.
from BaseHTTPServer import HTTPServer, BaseHTTPRequestHandler
from SocketServer import ThreadingMixIn
import threading
class Handler(BaseHTTPRequestHandler):
def do_GET(self):
self.send_response(200)
self.end_headers()
message = threading.currentThread().getName()
self.wfile.write(message)
self.wfile.write('\n')
return
class ThreadedHTTPServer(ThreadingMixIn, HTTPServer):
"""Handle requests in a separate thread."""
if __name__ == '__main__':
server = ThreadedHTTPServer(('localhost', 8080), Handler)
print 'Starting server, use <Ctrl-C> to stop'
server.serve_forever()
I have developed a PIP Utility called ComplexHTTPServer that is a multi-threaded version of SimpleHTTPServer.
To install it, all you need to do is:
pip install ComplexHTTPServer
Using it is as simple as:
python -m ComplexHTTPServer [PORT]
(By default, the port is 8000.)
In python3, you can use the code below (https or http):
from http.server import HTTPServer, BaseHTTPRequestHandler
from socketserver import ThreadingMixIn
import threading
USE_HTTPS = True
class Handler(BaseHTTPRequestHandler):
def do_GET(self):
self.send_response(200)
self.end_headers()
self.wfile.write(b'Hello world\t' + threading.currentThread().getName().encode() + b'\t' + str(threading.active_count()).encode() + b'\n')
class ThreadingSimpleServer(ThreadingMixIn, HTTPServer):
pass
def run():
server = ThreadingSimpleServer(('0.0.0.0', 4444), Handler)
if USE_HTTPS:
import ssl
server.socket = ssl.wrap_socket(server.socket, keyfile='./key.pem', certfile='./cert.pem', server_side=True)
server.serve_forever()
if __name__ == '__main__':
run()
You will figure out this code will create a new thread to deal with every request.
Command below to generate self-sign certificate:
openssl req -x509 -newkey rsa:4096 -nodes -out cert.pem -keyout key.pem -days 365
If you are using Flask, this blog is great.
It's amazing how many votes these solutions that break streaming are getting. If streaming might be needed down the road, then ThreadingMixIn and gunicorn are no good because they just collect up the response and write it as a unit at the end (which actually does nothing if your stream is infinite).
Your basic approach of combining BaseHTTPServer with threads is fine. But the default BaseHTTPServer settings re-bind a new socket on every listener, which won't work in Linux if all the listeners are on the same port. Change those settings before the serve_forever() call. (Just like you have to set self.daemon = True on a thread to stop ctrl-C from being disabled.)
The following example launches 100 handler threads on the same port, with each handler started through BaseHTTPServer.
import time, threading, socket, SocketServer, BaseHTTPServer
class Handler(BaseHTTPServer.BaseHTTPRequestHandler):
def do_GET(self):
if self.path != '/':
self.send_error(404, "Object not found")
return
self.send_response(200)
self.send_header('Content-type', 'text/html; charset=utf-8')
self.end_headers()
# serve up an infinite stream
i = 0
while True:
self.wfile.write("%i " % i)
time.sleep(0.1)
i += 1
# Create ONE socket.
addr = ('', 8000)
sock = socket.socket (socket.AF_INET, socket.SOCK_STREAM)
sock.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
sock.bind(addr)
sock.listen(5)
# Launch 100 listener threads.
class Thread(threading.Thread):
def __init__(self, i):
threading.Thread.__init__(self)
self.i = i
self.daemon = True
self.start()
def run(self):
httpd = BaseHTTPServer.HTTPServer(addr, Handler, False)
# Prevent the HTTP server from re-binding every handler.
# https://stackoverflow.com/questions/46210672/
httpd.socket = sock
httpd.server_bind = self.server_close = lambda self: None
httpd.serve_forever()
[Thread(i) for i in range(100)]
time.sleep(9e9)
A multithreaded https server in python3.7
from http.server import BaseHTTPRequestHandler, HTTPServer
from socketserver import ThreadingMixIn
import threading
import ssl
hostName = "localhost"
serverPort = 8080
class MyServer(BaseHTTPRequestHandler):
def do_GET(self):
self.send_response(200)
self.send_header("Content-type", "text/html")
self.end_headers()
self.wfile.write(bytes("<html><head><title>https://pythonbasics.org</title></head>", "utf-8"))
self.wfile.write(bytes("<p>Request: %s</p>" % self.path, "utf-8"))
self.wfile.write(bytes("<p>Thread: %s</p>" % threading.currentThread().getName(), "utf-8"))
self.wfile.write(bytes("<p>Thread Count: %s</p>" % threading.active_count(), "utf-8"))
self.wfile.write(bytes("<body>", "utf-8"))
self.wfile.write(bytes("<p>This is an example web server.</p>", "utf-8"))
self.wfile.write(bytes("</body></html>", "utf-8"))
class ThreadingSimpleServer(ThreadingMixIn,HTTPServer):
pass
if __name__ == "__main__":
webServer = ThreadingSimpleServer((hostName, serverPort), MyServer)
webServer.socket = ssl.wrap_socket(webServer.socket, keyfile='./privkey.pem',certfile='./certificate.pem', server_side=True)
print("Server started http://%s:%s" % (hostName, serverPort))
try:
webServer.serve_forever()
except KeyboardInterrupt:
pass
webServer.server_close()
print("Server stopped.")
you can test it in a browser: https://localhost:8080
the running result is:
enter image description here
enter image description here
remind that you can generate your own keyfile and certificate use
$openssl req -newkey rsa:2048 -keyout privkey.pem -x509 -days 36500 -out certificate.pem
To learn details about creating self-signed certificate with openssl:https://www.devdungeon.com/content/creating-self-signed-ssl-certificates-openssl
Related
I'm trying out some PHP on my pc and made a little python server to host the files, one problem:
It can't do POST, I always get the error 501. I've heard that you can implement POST in these servers, but I didn't find how to do this, can someone help?
Here's my current server:
import http.server
import socketserver
PORT = 8080
Handler = http.server.SimpleHTTPRequestHandler
with socketserver.TCPServer(("", PORT), Handler) as httpd:
print("serving at port", PORT)
httpd.serve_forever()
This is the script I personally use for when I need this kind of functionality:
#!/usr/env python3
import http.server
import os
import logging
try:
import http.server as server
except ImportError:
# Handle Python 2.x
import SimpleHTTPServer as server
class HTTPRequestHandler(server.SimpleHTTPRequestHandler):
"""
SimpleHTTPServer with added bonus of:
- handle PUT requests
- log headers in GET request
"""
def do_GET(self):
server.SimpleHTTPRequestHandler.do_GET(self)
logging.warning(self.headers)
def do_PUT(self):
"""Save a file following a HTTP PUT request"""
filename = os.path.basename(self.path)
# Don't overwrite files
if os.path.exists(filename):
self.send_response(409, 'Conflict')
self.end_headers()
reply_body = '"%s" already exists\n' % filename
self.wfile.write(reply_body.encode('utf-8'))
return
file_length = int(self.headers['Content-Length'])
with open(filename, 'wb') as output_file:
output_file.write(self.rfile.read(file_length))
self.send_response(201, 'Created')
self.end_headers()
reply_body = 'Saved "%s"\n' % filename
self.wfile.write(reply_body.encode('utf-8'))
if __name__ == '__main__':
server.test(HandlerClass=HTTPRequestHandler)
But perhaps a more fitting, and simpler script would be the following, as found on Flavio Copes' blog:
from http.server import BaseHTTPRequestHandler, HTTPServer
class handler(BaseHTTPRequestHandler):
def do_GET(self):
self.send_response(200)
self.send_header('Content-type','text/html')
self.end_headers()
message = "Hello, World! Here is a GET response"
self.wfile.write(bytes(message, "utf8"))
def do_POST(self):
self.send_response(200)
self.send_header('Content-type','text/html')
self.end_headers()
message = "Hello, World! Here is a POST response"
self.wfile.write(bytes(message, "utf8"))
with HTTPServer(('', 8000), handler) as server:
server.serve_forever()
I'm trying to figure out why curl http://localhost:2113/healthz fails with curl: (56) Recv failure: Connection reset by peer
main.py starts a HealthServer:
from healthServer import HealthServer
if __name__ == '__main__':
HealthServer().start()
healthServer.py defines HealthServer as a thread which serves HTTP with a handler that derives from BaseHTTPRequestHandler:
from configuration import config
from http.server import HTTPStatus, HTTPServer, BaseHTTPRequestHandler
import logging
from threading import Thread
class HealthHTTPRequestHandler(BaseHTTPRequestHandler):
def do_GET(self):
'''Respond to a GET request.'''
if self.path == '/healthz':
logging.info('GET /Healthz')
self.send_response(HTTPStatus.OK)
self.end_headers()
self.wfile.write(b'ok')
else:
logging.error('health endpoint received a GET for path {}'.format(self.path))
self.send_response(HTTPStatus.NOT_FOUND)
self.end_headers()
class HealthServer(Thread):
def __init__(self):
super().__init__()
def run(self):
port=config['server']['ports']['healthz'].get()
logging.info('Starting HTTP server on port {} to serve health on /healthz'.format(port))
httpd = HTTPServer(('localhost', port), HealthHTTPRequestHandler)
httpd.serve_forever()
logging.warning('health server exited')
Starting the application logs
Starting HTTP server on port 2113 to serve health on /healthz.
I have a scenario where I need to first respond with HTTP 200 to a server request (due to a time limit) and then continue processing with the actual work.
I also can not use threads, processes, tasks, queues or any other method that would allow me to do this by starting a parallel "process".
My approach is to use the build in "Simple HTTP" server and I am looking for a way to force the server to respond with HTTP 200 and then be able to continue processing.
The current code will receive a POST request and print its content after a 3 seconds. I put a placeholder where I would like to send the response.
from http.server import BaseHTTPRequestHandler, HTTPServer
import time
class MyWebServer(BaseHTTPRequestHandler):
def do_POST(self):
content_length = int(self.headers['Content-Length'])
post_data = self.rfile.read(content_length)
self.send_response_only(200)
self.end_headers()
# force server to send request ???
time.sleep(3)
print(post_data)
def run(server_class=HTTPServer, handler_class=MyWebServer, port=8000):
server_address = ('', port)
httpd = server_class(server_address, handler_class)
print('Starting httpd...')
httpd.serve_forever()
if __name__ == "__main__":
run()
I figured out a workaround solution. You can force the server to send a 200 OK and continue processing after with these two commands:
self.finish()
self.connection.close()
This solution is from this SO question: SimpleHTTPRequestHandler close connection before returning from do_POST method
However, this will apparently close the internal IO buffer that the server uses and it won't be able to server any additional requests after that.
To avoid running into an exception it works to terminate the program (which works for me). However this is just a workaround and I would still be looking for a solution that allows the server to keep processing new requests.
from http.server import BaseHTTPRequestHandler, HTTPServer
import time
class MyHandler(BaseHTTPRequestHandler):
def do_POST(self):
content_length = int(self.headers['Content-Length'])
post_data = self.rfile.read(content_length)
self.send_response_only(200)
self.end_headers()
self.finish()
self.connection.close()
time.sleep(3)
print(post_data)
quit()
def run(server_class=HTTPServer, handler_class=MyHandler, port=8000):
server_address = ('', port)
httpd = server_class(server_address, handler_class)
print('Starting httpd...')
httpd.serve_forever()
if __name__ == "__main__":
run()
After a lot of trouble to make a ros and a python based http server work, I have another problem to combine two different codes.
Here is what I have written so far to write a code that publish ros topics yet it respond to http server at the same time. The problem is that server is OK unless I replace
httpd.serve_forever()
with
httpd.service_actions
to make it non-blocking. Then the server does not respond. Any way to resolve this problem yet keeping the code non-blcoking?
#!/usr/bin/env python3
import rospy
from std_msgs.msg import String
from http.server import BaseHTTPRequestHandler, HTTPServer
class S(BaseHTTPRequestHandler):
def _set_headers(self):
self.send_response(200)
self.send_header('Content-type', 'text/html')
self.end_headers()
def do_GET(self):
self._set_headers()
self.wfile.write(b"<html><body><h1>hi!</h1></body></html>")
def do_HEAD(self):
self._set_headers()
def do_POST(self):
# Doesn't do anything with posted data
self._set_headers()
self.wfile.write(b"<html><body><h1>POST!</h1></body></html>")
def http_server_init(port,server_class=HTTPServer, handler_class=S):
server_address = ('', port)
httpd = server_class(server_address, handler_class)
print('Starting httpd...')
return (httpd)
def http_server_loop(httpd):
#httpd.serve_forever()
httpd.service_actions()
def talker_init():
pub = rospy.Publisher('chatter', String, queue_size=10)
rospy.init_node('talker', anonymous=True)
rate = rospy.Rate(10) # 10hz
return (pub,rospy,rate)
def talker_loop(pub,rospy,rate):
hello_str = "hello world %s" % rospy.get_time()
rospy.loginfo(hello_str)
pub.publish(hello_str)
rate.sleep()
if __name__ == '__main__':
from sys import argv
if len(argv) == 2:
http_port=int(argv[1])
else:
http_port=8080
httpd=http_server_init(http_port)
try:
pub,rospy,rate=talker_init()
while not rospy.is_shutdown():
http_server_loop(httpd)
talker_loop(pub,rospy,rate)
except rospy.ROSInterruptException:
pass
By the way, in ROS, this code should be called via
rosrun <packagename> <script>.py
Calling from bash leads to an error. Unless you remove the ROS-related codes.
I've written a simple HTTP server and made it into a Windows service using pywin32. The server successfully processes requests when run in the debugger, inside actual service it gets the request but hangs on send_response operation. What might be the reason?
from http.server import BaseHTTPRequestHandler, HTTPServer
import win32serviceutil
import win32service
import sys
PORT_NUMBER = 6363
class myHandler(BaseHTTPRequestHandler):
def do_GET(self):
self.send_response(200)
self.send_header('Content-type', 'text/html')
self.end_headers()
self.wfile.write(bytes("Hello World !", "utf-8"))
return
class TestSvc(win32serviceutil.ServiceFramework):
_svc_name_ = "TestSvc"
_svc_display_name_ = "Test Service"
_svc_description_ = "Tests server inside service."
def __init__(self, args):
win32serviceutil.ServiceFramework.__init__(self, args)
self.server = HTTPServer(('', PORT_NUMBER), myHandler)
def SvcDoRun(self):
self.ReportServiceStatus(win32service.SERVICE_RUNNING)
print('Started httpserver on port ', PORT_NUMBER)
self.server.serve_forever()
def SvcStop(self):
self.ReportServiceStatus(win32service.SERVICE_STOP_PENDING)
self.server.shutdown()
#sys.frozen = 'windows_exe'
if __name__ == '__main__':
win32serviceutil.HandleCommandLine(TestSvc, argv=sys.argv)
Actually it was hanging when executing sys.stderr.write (which is the default logging output for BaseHTTPRequestHandler). So I've overridden log_message function in my request handler class and it works fine now.