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Home/ Questions/Q 96497
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Editorial Team
Asked: May 10, 20262026-05-10T23:49:27+00:00 2026-05-10T23:49:27+00:00

I’ve written this to try and log onto a forum (phpBB3). import urllib2, re

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I’ve written this to try and log onto a forum (phpBB3).

import urllib2, re import urllib, re logindata = urllib.urlencode({'username': 'x', 'password': 'y'}) page = urllib.urlopen('http://www.woarl.com/board/ucp.php?mode=login'[logindata]) output = page.read() 

However when I run it it comes up with;

Traceback (most recent call last):   File 'C:/Users/Mike/Documents/python/test urllib2', line 4, in <module>     page = urllib.urlopen('http://www.woarl.com/board/ucp.php?mode=login'[logindata]) TypeError: string indices must be integers 

any ideas as to how to solve this?

edit

adding a comma between the string and the data gives this error instead

Traceback (most recent call last):   File 'C:/Users/Mike/Documents/python/test urllib2', line 4, in <module>     page = urllib.urlopen('http://www.woarl.com/board/ucp.php?mode=login',[logindata])   File 'C:\Python25\lib\urllib.py', line 84, in urlopen     return opener.open(url, data)   File 'C:\Python25\lib\urllib.py', line 192, in open     return getattr(self, name)(url, data)   File 'C:\Python25\lib\urllib.py', line 327, in open_http     h.send(data)   File 'C:\Python25\lib\httplib.py', line 711, in send     self.sock.sendall(str)   File '<string>', line 1, in sendall TypeError: sendall() argument 1 must be string or read-only buffer, not list 

edit2

I’ve changed the code from what it was to;

import urllib2, re import urllib, re logindata = urllib.urlencode({'username': 'x', 'password': 'y'}) page = urllib2.urlopen('http://www.woarl.com/board/ucp.php?mode=login', logindata) output = page.read() 

This doesn’t throw any error messages, it just gives 3 blank lines. Is this because I’m trying to read from the log in page which disappears after logging in. If so how do I get it to display the index which is what should appear after hitting log in.

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  1. 2026-05-10T23:49:28+00:00Added an answer on May 10, 2026 at 11:49 pm

    Your line

    page = urllib.urlopen('http://www.woarl.com/board/ucp.php?mode=login'[logindata]) 

    is semantically invalid Python. Presumably you meant

    page = urllib.urlopen('http://www.woarl.com/board/ucp.php?mode=login', [logindata]) 

    which has a comma separating the arguments. However, what you ACTUALLY want is simply

    page = urllib2.urlopen('http://www.woarl.com/board/ucp.php?mode=login', logindata) 

    without trying to enclose logindata into a list and using the more up-to-date version of urlopen is the urllib2 library.

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