Simple and light HTTP server in Python

server.py

from BaseHTTPServer import BaseHTTPRequestHandler, HTTPServer
from httplib import HTTPResponse
from os import curdir,sep

#Create a index.html aside the code
#Run: python server.py
#After run, try http://localhost:8080/

class RequestHandler(BaseHTTPRequestHandler):
    def do_GET(self):
        if self.path == '/':
            self.path  = '/index.html'
        try:
            sendReply = False
            if self.path.endswith(".html"):
                mimeType = 'text/html'
                sendReply = True
            if sendReply == True:
                f = open(curdir + sep + self.path)
                self.send_response(200)
                self.send_header('Content-type', mimeType)
                self.end_headers()
                self.wfile.write(f.read())
                f.close()
            return
        except IOError:
            self.send_error(404,'File not found!')


def run():
    print('http server is starting...')
    #by default http server port is 80
    server_address = ('127.0.0.1', 8080)
    httpd = HTTPServer(server_address, RequestHandler)
    try:
        print 'http server is running...'
        httpd.serve_forever()
    except KeyboardInterrupt:
        httpd.socket.close()

if __name__ == '__main__':
    run()

Hi Maurostorch, How to use a wildcard at ports or set a range of ports to listen on.

Here "server_address = ('127.0.0.1', 8080)"

each run() is a single server in a single port. You could call it multiple times with some small changes in this code and using threads.

Is this Python 2 or Python 3 ?

Python 3

Not currently working out of the box. For sure the print statement needs parenthesis by the way but even after fixing it, is not working. Tested on 3.9.9

@VAkris you say python3 but BaseHTTPServer isn't in python 3 and the print statements are python 2

I don't have the index.html. I tot it is supppose to come up with a directory of where the file exist?