You may want to use urllib.parse: >>> from urllib.parse import urlparse, parse_qs >>> url=”http://example.com/?foo=bar&one=1″ >>> parse_qs(urlparse(url).query) {‘foo’: [‘bar’], ‘one’: [‘1’]} For Python 2, the module is named urlparse instead of url.parse.
Your program in python 3.xx does work right out of the box – except for one minor problem. The issue is not in your code but the place where you are writing these lines: self.wfile.write(“Hello World !”) You are trying to write “string” in there, but bytes should go there. So you need to convert … Read more
You can access POST body in do_POST method like this: for python 2 content_len = int(self.headers.getheader(‘content-length’, 0)) for python 3 content_len = int(self.headers.get(‘Content-Length’)) and then read the data post_body = self.rfile.read(content_len)
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) … Read more
This will probably do it: from BaseHTTPServer import HTTPServer, BaseHTTPRequestHandler class MyHandler(BaseHTTPRequestHandler): def do_GET(self): self.send_response(200) self.send_header(‘Content-type’, ‘text/html’) self.end_headers() self.wfile.write(‘<html><body><p>OK</p></body></html>’) def log_message(self, format, *args): return httpd = HTTPServer((”, 8001), MyHandler) httpd.serve_forever()