Cara menulis url tag biasa (python atau 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>

Saya ingin mendapatkan pautan di href

Tiga pertama daripada 6 pautan memenuhi syarat Cara menulis ungkapan biasa (iaitu, pautan mesti mempunyai kategori jenis paparan halaman kunci pandang)

Pautan kedua dan ketiga adalah sama

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

Ekstrak tiga pautan pertama:

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

Alih keluar pendua:

new_links=set(links)