html in do_GET() method of a simple Python webserver
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Published on 2011-06-30T06:50:23Z
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2011/06/30
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I am relatively new to Python but have been doing a lot of different things with it recently and I am liking it a lot. However, I ran into trouble/block with the following code.
import http.server
import socketserver
import glob
import random
class Server(http.server.SimpleHTTPRequestHandler):
def do_GET(self):
self.send_response(200, 'OK')
self.send_header('Content-type', 'html')
self.end_headers()
self.wfile.write(bytes("<html> <head><title> Hello World </title> </head> <body>", 'UTF-8'))
images = glob.glob('*.jpg')
rand = random.randint(0,len(images)-1)
imagestring = "<img src = \"" + images[rand] + "\" height = 1028 width = 786 align = \"right\"/> </body> </html>"
self.wfile.write(bytes(imagestring, 'UTF-8'))
def serve_forever(port):
socketserver.TCPServer(('', port), Server).serve_forever()
if __name__ == "__main__":
Server.serve_forever(8000)
What I am trying to do here is grab a random image from multiple images in the directory and add it into the response to a web request. The code works fine but when I access the server via browser, the image is not displayed. The html of the page is as intended though. The permissions on the files are 755. Also I tried to create an index.html file in the do_GET method. That didn't work either. I mean the index.html was generated fine, but the response in the browser this time did not show anything (not even the hello world in the title).
Am I missing anything very simple here? I was thinking should I overload the handle_request of the underlying SocketServer.BaseServer as the documentation says you should never override BaseHTTPServer's handle() method and should rather override the corresponding do_* method?
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