美麗的湯提取谷歌搜索A HREF

Question

谷歌搜索使我對HTML以下第一個結果：

<h3 class="r"><a href="http://rads.stackoverflow.com/amzn/click/0470284889" class="l vst" onmousedown="return rwt(this,'','','','1','AFQjCNEv1W9YC2jcSKYdEo2kNqBMJ-Utmg','k89K9hF4cVNpxQYHtEKiUQ','0CCoQFjAA',null,event)"><em>Quantitative Trading</em>: <em>How to Build Your Own Algorithmic</em> <b>...</b> - Amazon</a></h3> 

我想從這個提取路段http://www.amazon.com/Quantitative-Trading-Build-Algorithmic-Business/dp/0470284889，但是當我用美麗的湯提取信息，我得到

soup.find("h3").find("a").get("href") 

我得到下面的字符串，而不是：

/url?q=http://www.amazon.com/Quantitative-Trading-Build-Algorithmic-Busines S/DP/0470284889 & SA = U & EI = P2ycT6OoNuasiAL2ncV5 & VED = 0CBIQFjAA & USG = AFQjCNEo_ujANAKnjheWDRlBKnJ1BGeA7A

我知道該鏈接是在那裏，我可以通過刪除/ URL解析呢？Q =和之後的一切&符號，但我想知道是否有更清潔的解決方案。

謝謝！

Answer 1

您可以使用urlparse.urlparse和urlparse.parse_qs組合，e.g

>>> import urlparse 
>>> url = '/url?q=http://www.amazon.com/Quantitative-Trading-Build-Algorithmic-Business/dp/0470284889&sa=U&ei=P2ycT6OoNuasiAL2ncV5&ved=0CBIQFjAA&usg=AFQjCNEo_ujANAKnjheWDRlBKnJ1BGe' 
>>> data = urlparse.parse_qs(
...  urlparse.urlparse(url).query 
...) 
>>> data 
{'ei': ['P2ycT6OoNuasiAL2ncV5'], 
'q': ['http://www.amazon.com/Quantitative-Trading-Build-Algorithmic-Business/dp/0470284889'], 
'sa': ['U'], 
'usg': ['AFQjCNEo_ujANAKnjheWDRlBKnJ1BGe'], 
'ved': ['0CBIQFjAA']} 
>>> data['q'][0] 
'http://www.amazon.com/Quantitative-Trading-Build-Algorithmic-Business/dp/0470284889'