python 的问题

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码如下：
PID = 'lKGXIQa'

GetPlayer = "http://213432.com/liveforapp.php?cid={cid}&appid={appid}" .format(cid=PID,appid="01")

from urllib.request import urlopen
from pprint import pprint
from bs4 import BeautifulSoup

url = urlopen(GetPlayer)
print(url.read())
这样我就能得出
b'<a href="http://213432.com/live.php?token=dH0xNDc2XTU3NDc2JmM9bEtHWElRYSZhPTAxJnY9ZWJhZGY|">http://213432com/live.php?token=dH0xNDc2XTU3NDc2JmM9bEtHWElRYSZhPTAxJnY9ZWJhZGY|</a>'
问题是我只想得出 url ，因为待会要用...html 标签什么的没用，网上查了查好多没用的资料 TT ，有什么办法吗？

import

getplayer

pprint

urlopen

2 replies • 2016-10-20 22:55:30 +08:00

182247236

Oct 20, 2016

我之前是这么写的
PID = 'lKGXIQa'

GetPlayer = "http://12345.cutv.com/liveforapp.php?cid=%s" %PID +'&appid=S01'

from urllib.request import urlopen
from pprint import pprint
from bs4 import BeautifulSoup

html = urlopen(GetPlayer)

bs_obj = BeautifulSoup(html,'html.parser')

for link in bs_obj.findAll('a'):
print(link.get_text())
这么可以得出我需要的 url ，可是我觉得太复杂不好，想改简单点

182247236

Oct 20, 2016

要沉吗 TT