Andrew Berg wrote:
ElementTree doesn't seem to have been updated in a long time, so I'll
assume it won't work with Python 3.
I don't know how to use it, but you'll find ElementTree as xml.etree in
Python 3.
~Ethan~
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On 05/19/2011 11:35 PM, Andrew Berg wrote:
On 2011.05.16 02:26 AM, Karim wrote:
Use regular expression for bad HTLM or beautifulSoup (google it), below
a exemple to extract all html links:
Actually, using regex wasn't so bad:
import re
import urllib.request
url = 'http://x264.nl/x264/?dir=./6
On 2011.05.16 02:26 AM, Karim wrote:
> Use regular expression for bad HTLM or beautifulSoup (google it), below
> a exemple to extract all html links:
Actually, using regex wasn't so bad:
> import re
> import urllib.request
>
> url = 'http://x264.nl/x264/?dir=./64bit/8bit_depth'
> page = str(urllib
Andrew Berg, 19.05.2011 02:39:
On 2011.05.18 03:30 AM, Stefan Behnel wrote:
Well, it pretty clearly states that on the PyPI page, but I also added it
to the project home page now. lxml 2.3 works with any CPython version from
2.3 to 3.2.
Thank you. I never would've looked at PyPI for info on a p
On 2011.05.18 03:30 AM, Stefan Behnel wrote:
> Well, it pretty clearly states that on the PyPI page, but I also added it
> to the project home page now. lxml 2.3 works with any CPython version from
> 2.3 to 3.2.
Thank you. I never would've looked at PyPI for info on a project that
has its own sit
Andrew Berg, 17.05.2011 03:05:
lxml looks promising, but it doesn't say anywhere whether it'll work on
Python 3 or not
Well, it pretty clearly states that on the PyPI page, but I also added it
to the project home page now. lxml 2.3 works with any CPython version from
2.3 to 3.2.
Stefan
--
On 05/17/2011 03:05 AM, Andrew Berg wrote:
On 2011.05.16 02:26 AM, Karim wrote:
Use regular expression for bad HTLM or beautifulSoup (google it), below
a exemple to extract all html links:
linksList = re.findall('.*?',htmlSource)
for link in linksList:
print link
I was afraid I might hav
On 2011.05.16 02:26 AM, Karim wrote:
> Use regular expression for bad HTLM or beautifulSoup (google it), below
> a exemple to extract all html links:
>
> linksList = re.findall('.*?',htmlSource)
> for link in linksList:
> print link
I was afraid I might have to use regexes (mostly because I c
On 05/16/2011 03:06 AM, David Robinow wrote:
On Sun, May 15, 2011 at 4:45 PM, Andrew Berg wrote:
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):
def h
On Sun, May 15, 2011 at 4:45 PM, Andrew Berg wrote:
> 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):
> def handle_starttag(self, tag, attrs):
>
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):
def handle_starttag(self, tag, attrs):
if tag == 'a' and attrs:
print(tag,'-',attrs
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