How to write regular a tag url (python or js)

Question

<a target="blank"href="http://a.b.c.d/abc.php?viewkey=11111111111d5c2a51d1e2&amp;page=1&amp;viewtype=basic&amp;category=rf"></a>
<a target="blank"href="http://a.b.c.d/abc.php?viewkey=6d7a7f6a6e9c2a5191e2&amp;page=1&amp;viewtype=basic&amp;category=rf"></a>

<a target="blank"href="http://a.b.c.d/abc.php?viewkey=6d7a7f6a6e9c2a5191e2&amp;page=1&amp;viewtype=basic&amp;category=rf"></a>


<a target="blank"href="http://a.b.c.d/abc.php?viewkey=6d7a7f6a6e9c2a5191e2&amp"></a>

<a target="blank"href="http://a.b.c.d/abc"></a>


<a target="blank"href="http://a.b.c.d/123"></a>

I want to get the link in href

The first three of the 6 links meet the conditions. How to write the regular form (that is, the link must have the parameter viewkey page viewtype category)

The second and third links are the same. How to repeat them (under python)

Answer 1

# python 2.7

import re

a = '''<a target="blank"href="http://a.b.c.d/abc.php?viewkey=11111111111d5c2a51d1e2&amp;page=1&amp;viewtype=basic&amp;category=rf"></a>
<a target="blank"href="http://a.b.c.d/abc.php?viewkey=6d7a7f6a6e9c2a5191e2&amp;page=1&amp;viewtype=basic&amp;category=rf"></a>

<a target="blank"href="http://a.b.c.d/abc.php?viewkey=6d7a7f6a6e9c2a5191e2&amp;page=1&amp;viewtype=basic&amp;category=rf"></a>


<a target="blank"href="http://a.b.c.d/abc.php?viewkey=6d7a7f6a6e9c2a5191e2&amp"></a>

<a target="blank"href="http://a.b.c.d/abc"></a>


<a target="blank"href="http://a.b.c.d/123"></a>'''

print set(re.findall('''(?=.*(?:viewkey))(?=.*(?:page))(?=.*(?:viewtype))(?=.*(?:category))href=["']([^'"]+)''', a))

Answer 2

Extract the first three links:

links= re.findall(r'href=\"(.*?=rf)\"',l_string,re.S)

Remove duplicates:

new_links=set(links)