Getting a "TypeError: an integer is required" in my script
Asked Answered
A

3

1

I am trying to make a botnet using Python, for educational reasons, and I keep getting the following error:

TypeError: an integer is required

Here is my script:

import os
import socket
import random
import string

# string.letters = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
a = random.choice(string.letters) 
b = random.choice(string.letters)
c = random.choice(string.letters)
d = random.choice(string.letters)   
e = random.choice(string.letters)
name = 'bot' + a + b + c + d + e

network = 'chat.freenode.net'
port = 6667
irc = socket.socket ( socket.AF_INET, socket.SOCK_STREAM )
irc.connect ( ( network, port ) )
print irc.recv ( 4096 )

irc.send ('NICK', name + '\r\n')
irc.send ( 'USER', name, name, name, ':Python IRC\r\n' )
irc.send ( 'JOIN #occult_hand\r\n' )
irc.send ( 'PRIVMSG #occult_hand :Hello World.\r\n' )

while True:
    data = irc.recv ( 4096 )
    if data.find ( 'PING' ) != -1:
        irc.send ( 'PONG ' + data.split() [ 1 ] + '\r\n' )
    if data.find ( '!shutdown' ) != -1:
        irc.send ( "PRIVMSG #occult_hand :Fine, if you don't want me\r\n" )
        irc.send ( 'QUIT\r\n' )
    if data.find ( '!list' ) != -1:
        irc.send ( 'PRIVMSG #occult_hand :' + name, 'ONLINE\r\n' )
    if data.find ( '!ddos' ) != -1:
        irc.send ( 'PRIVMSG #occult_hand :Enter a target\r\n' )
    if data.find ( 'KICK' ) != -1:
        irc.send ( 'JOIN #occult_hand\r\n' )
    if data.find ('cheese') != -1:
        irc.send ( 'PRIVMSG #occult_hand :WHERE!!!!!!\r\n' )
    print data
Alluvion answered 10/5, 2014 at 22:19 Comment(2)
Please format your code and post the full tracebackLenora
Which line throws the error?Bornholm
V
0

I looked up for this issue, seems like you should put a integer ( UNIX socket constant ) there to let it work, but from your later code, I assume you just want to give messages but this optional parameter, so you could change

irc.send ('NICK', name + '\r\n')
irc.send ( 'USER', name, name, name, ':Python IRC\r\n' )

to

irc.send ('NICK' + name + '\r\n')
irc.send ( 'USER' + name + name + name + ':Python IRC\r\n' )

I guess you're mixing python's string concatenation operator + with , from some other languages :) I tested the changed version in my box, it won't cause the error anymore, though I still don't know what this code does, could you please explain a bit :)

Valles answered 10/5, 2014 at 22:42 Comment(0)
H
1

The script errors on line 21, irc.send('NICK', name + '\r\n'):

Traceback (most recent call last):
  File "botnet.py", line 21, in 
    irc.send('NICK', name + '\r\n')
TypeError: an integer is required

It's because the socket.send method has the following signature, as per the docs:

socket.send(string[, flags])

The string argument is the data to be sent. The flags argument is the optional flags, that are the same as described in Unix man 2 recv. See http://man7.org/linux/man-pages/man2/recv.2.html for details.

Basically, the flags argument is an integer value and defaults to 0. The flag values, as described by Unix man page, are available in the socket module as constants and you get the value by combining the required flag values using an OR logical operation, e.g.:

socket.send(data, flags=socket.MSG_OOB | socket.MSG_DONTROUTE)

To fix your script, you have to concatenate all the data you want to send into one string, and pass that as the first argument to the socket.send method everywhere:

irc.send('NICK' + name + '\r\n')
irc.send('USER' + name + name + name + ':Python IRC\r\n')
# ...
Hypercorrection answered 10/5, 2014 at 22:53 Comment(0)
P
0

Please post the actual error message. For now what I can say is that you should have another look at your send method. For example, with

irc.send ('NICK', name + '\r\n')

you are passing two strings to send. According to https://docs.python.org/2/library/socket.html and

man 2 recv

the flag argument(s) must be integer.

Perlie answered 10/5, 2014 at 22:30 Comment(0)
V
0

I looked up for this issue, seems like you should put a integer ( UNIX socket constant ) there to let it work, but from your later code, I assume you just want to give messages but this optional parameter, so you could change

irc.send ('NICK', name + '\r\n')
irc.send ( 'USER', name, name, name, ':Python IRC\r\n' )

to

irc.send ('NICK' + name + '\r\n')
irc.send ( 'USER' + name + name + name + ':Python IRC\r\n' )

I guess you're mixing python's string concatenation operator + with , from some other languages :) I tested the changed version in my box, it won't cause the error anymore, though I still don't know what this code does, could you please explain a bit :)

Valles answered 10/5, 2014 at 22:42 Comment(0)

© 2022 - 2024 — McMap. All rights reserved.