Ch. 2 Violent Python syntax error: ValueError: invalid literal for int() with base 10: '21,'

Question

I am working my way through 'Violent Python'. I am in chapter 2, and I receive this syntax error when attempting to run the below code:

python2 portscanner.py -H perl6.org -p 21, 22, 80

['21,'] [+] Scan results for: www.p6c.org Scanning port 21, Traceback (most recent call last): File "/home/wolf/python/Violent-Python-Examples/Chapter-2/portscan.py", line 53, in main() File "/home/wolf/python/Violent-Python-Examples/Chapter-2/portscan.py", line 49, in main portScan(tgtHost, tgtPorts) File "/home/wolf/python/Violent-Python-Examples/Chapter-2/portscan.py", line 32, in portScan connScan(tgtHost, int(tgtPort)) ValueError: invalid literal for int() with base 10: '21,'

I copied it mostly directly from the code, but I did a few things to make it Python3 instead of Python2, such as adding the parenthesis for the print() function. However, I being new to Python, I don't understand what the problem is here. I assume it is something with the split() method?

import optparse
import socket
from socket import *


def connScan(tgtHost, tgtPort):
    try:
        connSkt = socket(AF_INET, SOCK_STREAM)
        connSkt.connect((tgtHost, tgtPort))
        connSkt.send('ViolentPython\r\n')
        results = connSkt.recv(100)
        print('[+]%d/tcp open' % tgtPort)
        connSkt.close()
    except:
        print('[-]%d/tcp closed' % tgtPort)


def portScan(tgtHost, tgtPorts):
    try:
        tgtIP = gethostbyname(tgtHost)
    except:
        print("[-] Cannot resolve '%s': Unknown host" % tgtHost)
        return
    try:
        tgtName = gethostbyaddr(tgtIP)
        print('\n[+] Scan results for: ' + tgtName[0])
    except:
        print('\n[+] Scan results for: ' + tgtIP)
    setdefaulttimeout(1)
    for tgtPort in tgtPorts:
        print('Scanning port ' + tgtPort)
        connScan(tgtHost, int(tgtPort))


def main():
    parser = optparse.OptionParser("usage%prog -H <target host> +\
                                    -p <target port>")
    parser.add_option('-H', dest='tgtHost', type='string',
                      help='specify target host')
    parser.add_option('-p', dest='tgtPort', type='string',
                      help='specify target port')
    (options, args) = parser.parse_args()
    tgtHost = options.tgtHost
    tgtPorts = str(options.tgtPort).split(', ')
    print(tgtPorts)
    if (tgtHost is None) | (tgtPorts[0] is None):
        print('[-] You must specify a target host and port[s].')
        exit(0)
    portScan(tgtHost, tgtPorts)


if __name__ == '__main__':
    main()

Actually, it works now that I have no spaces in between the ports e.g. -p 21,22,80, but the book has spaces in between them. — Jonathan Dewein, Jul 15 '18 at 03:44
Probably want to check your commas, looks like it's trying to do : `int('21,')` instead of `int('21')` — jeremycg, Jul 15 '18 at 03:44
The issue isn't your modifications. The unmodified code from the 2013 edition of the book (using the command in the book) gives the same error. — Zev, Jul 15 '18 at 04:38
In case it's not obvious, quoting the argument `-p "21, 22, 80"` would also work around the problem. — tripleee, Jul 15 '18 at 06:37

Zev · Accepted Answer · 2018-07-15T12:12:16.620

The options you are parsing is just looking for one port -p <target port> so it cuts off everything after "21,".

str(options.tgtPort).split(', ') is looking for a comma followed by a space. "21," just has the comma.

See Split by comma and strip whitespace in Python if you need to split by both but really, just send the command python2 portscanner.py -H perl6.org -p 21 and it should work. Or as Tripleee pointed out, just put quotes around the argument.

If you need to support multiple arguments without quotes, this question will help: Processing multiple values for one single option using getopt/optparse?

By the way, if it makes you feel any better, this is an issue with the original unmodified code from the book. The book's Python 2 code gives the same error for the command the book suggests.

Ch. 2 Violent Python syntax error: ValueError: invalid literal for int() with base 10: '21,'

1 Answers1