<p>Given an HTML link like</p>

<pre><code>&lt;a href="urltxt" class="someclass" close="true"&gt;texttxt&lt;/a&gt;
</code></pre>

<p>how can I isolate the url and the text? </p>

<p><strong>Updates</strong></p>

<p>I'm using Beautiful Soup, and am unable to figure out how to do that. </p>

<p>I did </p>

<pre><code>soup = BeautifulSoup.BeautifulSoup(urllib.urlopen(url))

links = soup.findAll('a')

for link in links:
    print "link content:", link.content," and attr:",link.attrs
</code></pre>

<p>i get </p>

<pre><code>*link content: None  and attr: [(u'href', u'_redirectGeneric.asp?genericURL=/root    /support.asp')]*  ...
...
</code></pre>

<p>Why am i missing the content? </p>

<p>edit: elaborated on 'stuck' as advised :)</p>

Use
Beautiful Soup
.  Doing it yourself is harder than it looks, you'll be better off using a tried and tested module.
EDIT:
I think you want:
```
```
soup = BeautifulSoup.BeautifulSoup(urllib.urlopen(url).read())
```
```
By the way, it's a bad idea to try opening the URL there, as if it goes wrong it could get ugly.
EDIT 2:
This should show you all the links in a page:
```
```
import urlparse, urllib
from BeautifulSoup import BeautifulSoup

url = "http://www.example.com/index.html"
source = urllib.urlopen(url).read()

soup = BeautifulSoup(source)

for item in soup.fetchall('a'):
    try:
        link =  urlparse.urlparse(item['href'].lower())
    except:
        # Not a valid link
        pass
    else:
        print link
```
```