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Date: Sun, 15 May 2011 21:06:49 -0400
Subject: Re: Trying to understand html.parser.HTMLParser
From: David Robinow <drobinow@gmail.com>
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Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:5454

On Sun, May 15, 2011 at 4:45 PM, Andrew Berg <bahamutzero8825@gmail.com> wr=
ote:
> I'm trying to understand why HMTLParser.feed() isn't returning the whole
> page. My test script is this:
>
> import urllib.request
> import html.parser
> class MyHTMLParser(html.parser.HTMLParser):
> =A0 =A0def handle_starttag(self, tag, attrs):
> =A0 =A0 =A0 =A0if tag =3D=3D 'a' and attrs:
> =A0 =A0 =A0 =A0 =A0 =A0print(tag,'-',attrs)
>
> url =3D 'http://x264.nl/x264/?dir=3D./64bit/8bit_depth'
> page =3D urllib.request.urlopen(url).read()
> parser =3D MyHTMLParser()
> parser.feed(str(page))
>
> I can do print(page) and get the entire HTML source, but
> parser.feed(str(page)) only spits out the information for the top links
> and none of the "revisionxxxx" links. Ultimately, I just want to find
> the name of the first "revisionxxxx" link (right now it's
> "revision1995", when a new build is uploaded it will be "revision2000"
> or whatever). I figure this is a relatively simple page; once I
> understand all of this, I can move on to more complicated pages.
You've got bad HTML. Look closely and you'll see the there's no space
between the "revisionxxxx" strings and the style tag following.
The parser doesn't like this. I don't know a solution other than
fixing the html.
(I created a local copy, edited it and it worked.)