Python Beautifulsoup Extract Specific Urls

Is it possible to get only specific URLs? Like: next ... , href=re.compile('http://www\.iwashere\.com/'))

which matches (for your example):

[<a href="http://www.iwashere.com/washere.html">next</a>, <a href="http://www.iwashere.com/wasnot.html">next</a>]

so any <a> tag with a href attribute that has a value that starts with the string http://www.iwashere.com/.

You can loop over the results and pick out just the href attribute:

>>> for elem in soup.find_all('a', href=re.compile('http://www\.iwashere\.com/')):
...     print elem['href']
... 
http://www.iwashere.com/washere.html
http://www.iwashere.com/wasnot.html

To match all relative paths instead, use a negative look-ahead assertion that tests if the value does not start with a schem (e.g. http: or mailto:), or a double slash (//hostname/path); any such value must be a relative path instead:

soup.find_all('a', href=re.compile(r'^(?!(?:[a-zA-Z][a-zA-Z0-9+.-]*:|//))'))

Solution 2:

If you're using BeautifulSoup 4.0.0 or greater:

soup.select('a[href^="http://www.iwashere.com/"]')

Solution 3:

You could solve this with partial matching in gazpacho:

Input:

html = """\
<a href="http://www.iwashere.com/washere.html">next</a>
<span class="class">...</span>
<a href="http://www.heelo.com/hello.html">next</a>
<span class="class">...</span>
<a href="http://www.iwashere.com/wasnot.html">next</a>
<span class="class">...</span>
"""

Code:

from gazpacho import Soup

soup = Soup(html)
links = soup.find('a', {'href': "http://www.iwashere.com/"}, partial=True)
[link.attrs['href'] for link in links]

Which will output:

# ['http://www.iwashere.com/washere.html', 'http://www.iwashere.com/wasnot.html']