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I am being unable to run my code as soon as I add these send line in client and recv line in the server. I don't know what's wrong. Without them, the program is running perfectly. I am pasting just the smallest possible code for the ease of debugging because without these lines, code is perfect.

Client.py

    print ('Opening file for editting.....')
with open('C:\\Users\\dell\\Desktop\\received_file.txt', 'ab') as e:
    e.write('Ya Allah tera shukar')
e.close()
print ('Editting Done!!!')
s.send("Trial Message")
s.close()
print ("Connection closed")

Server.py

while (l): #Keep sending it until EOF id found
    conn.send(l) #Keep sending the opened file
    print(repr(l)) #repr(1) gives the string representation of contents of the file. The repr() of a string adds string quotes and backslashes:
    l = f.read(1024) #Keep reading the opened file so the contents appear on the server too
f.close()
print('Done sending the file')
print conn.recv(1024) #This is not receiving the trial message sent by client
conn.send('Thank you for connecting')
conn.close()
user3601076
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  • How did you defined "s" in the client.py? – Yaron May 15 '16 at 13:33
  • I suspect you are trying to use string as an argument of byte-accepting method. – Basilevs May 15 '16 at 13:34
  • What is the problem? – Daniel May 15 '16 at 13:35
  • No by mistake I attached the wrong code. Check this edited one. In the client when I write s.send("any message"). How do I receive it on server side and display it. I am trying to send basically the link of a file from client to server. Receiving it via the line conn.recv(1024) is causing my program to get stuck :( – user3601076 May 15 '16 at 13:36
  • I'll be so grateful if you just guide me through this sending and receiving procedure between c&s. I want to send a full path for a file to server. – user3601076 May 15 '16 at 13:39