views:

134

answers:

1

Hello,

I have this very simple python code to read xml for the wikipedia api:

import urllib
from xml.dom import minidom

usock = urllib.urlopen("http://en.wikipedia.org/w/api.php?action=query&titles=Fractal&prop=links&pllimit=500")
xmldoc=minidom.parse(usock)
usock.close()
print xmldoc.toxml() 

But this code returns with these errors:

Traceback (most recent call last):
  File "/home/user/workspace/wikipediafoundations/src/list.py", line 5, in <module><br>
    xmldoc=minidom.parse(usock)<br>
  File "/usr/lib/python2.6/xml/dom/minidom.py", line 1918, in parse<br>
    return expatbuilder.parse(file)<br>
  File "/usr/lib/python2.6/xml/dom/expatbuilder.py", line 928, in parse<br>
    result = builder.parseFile(file)<br>
  File "/usr/lib/python2.6/xml/dom/expatbuilder.py", line 207, in parseFile<br>
    parser.Parse(buffer, 0)<br>
xml.parsers.expat.ExpatError: syntax error: line 1, column 62<br>

I have no clue as I just learning python. Is there a way to get an error with more detail? Does anyone know the solution? Also, please recommend a better language to do this in.

Thank You,
Venkat Rao

+7  A: 

The URL you're requesting is an HTML representation of the XML that would be returned:

http://en.wikipedia.org/w/api.php?action=query&amp;titles=Fractal&amp;prop=links&amp;pllimit=500

So the XML parser fails. You can see this by pasting the above in a browser. Try adding a format=xml at the end:

http://en.wikipedia.org/w/api.php?action=query&amp;titles=Fractal&amp;prop=links&amp;pllimit=500&amp;format=xml

as documented on the linked page:

ars
Thank YouThat was easy.
Venkat S. Rao
@user, since @ars's answer solved you problem, **accept it** -- that is, clic on the checkmark-shaped icon on the left of his answer's text. This is fundamental SO etiquette!
Alex Martelli