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So, i am trying to create a simple server on python and trying to access a html file in the same directory through it, but as the output i keep on getting ready to serve... output

EDIT: Put an HTML file (e.g., HelloWorld.html) in the same directory that the server is in. Run the server program. Determine the IP address of the host that is running the server (e.g., 128.238.251.26). From another host, open a browser and provide the corresponding URL. For example: http://128.238.251.26:6789/HelloWorld.html ‘HelloWorld.html’ is the name of the file you placed in the server directory. Note also the use of the port number after the colon. You need to replace this port number with whatever port you have used in the server code. In the above example, we have used the port number 6789. The browser should then display the contents of HelloWorld.html. If you omit ":6789", the browser will assume port 80 and you will get the web page from the server only if your server is listening at port 80. Then try to get a file that is not present at the server. You should get a “404 Not Found” message. 

#import socket module
from socket import *
serverSocket = socket(AF_INET, SOCK_STREAM)
#Prepare a sever socket
serverSocket.bind(('', 12006))
serverSocket.listen(1)
while True:
    print 'Ready to serve...'
    #Establish the connection
    connectionSocket, addr = serverSocket.accept()
    try:
        message = connectionSocket.recv(1024)
        filename = message.split()[1]
        f = open(filename[1:])
        outputdata = f.read()
        f.close()
        #Send one HTTP header line into socket
        connectionSocket.send('HTTP/1.0 200 OK\r\n\r\n')
        #Send the content of the requested file to the client
        for i in range(0, len(outputdata)):
            connectionSocket.send(outputdata[i])
        connectionSocket.close()
    except IOError:
        #Send response message for file not found
        connectionSocket.send('404 Not Found')
        #Close client socket
        connectionSocket.close()
serverSocket.close()
Shane Gurusingha
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  • Your output is a __standart output__, that used through `print` function. you should to make a request to your server and you'll get the correct output – sakost Apr 28 '19 at 22:54
  • Oh thank you, but do i just use a random ip address? – Shane Gurusingha Apr 28 '19 at 22:57
  • no. If your server on your local machine, you should use `localhost` address; if not, you should use your server ip. Also you should to specify a port. 12006 in your case. `localhost:12006` as an example – sakost Apr 28 '19 at 22:59
  • I am getting an erroe now ``` TypeError: a bytes-like object is required, not 'str' ``` caused from this line ``` connectionSocket.send('HTTP/1.0 200 OK\r\n\r\n') ``` – Shane Gurusingha Apr 28 '19 at 23:04
  • Based on this last comment this is a duplicate of [TypeError: a bytes-like object is required, not 'str'](https://stackoverflow.com/questions/33003498/typeerror-a-bytes-like-object-is-required-not-str) and many similar questions (just google this error in connection with "socket"). – Steffen Ullrich Apr 29 '19 at 04:07
  • This is not valid Python 3, as `print` is used as a statement, without parentheses. Either the code or the tags are wrong. I guess the TypeError won't be seen in Python 2, either. – JohanL Apr 29 '19 at 04:55

1 Answers1

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Your output is a standart output, that used through print function. you should to make a request to your server and you'll get the correct output

If your server on your local machine, you should use localhost address; if not, you should use your server ip. Also you should to specify a port. 12006 in your case. localhost:12006 as an example

Also socket.send method requires a byte-like object. not string

If it's only a string literal, you should to add a b character before the first quotation mark

Example:

connectionSocket.send(b'HTTP/1.0 200 OK\r\n\r\n')

If it is a string object, you should to encode it:

connectionSocket.send(outputdata[i].encode())

Check out the documentation

sakost
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