python - 如何使用 beautifulsoup 和 python 只获取 mp3 链接

Question

这是我的代码:

from bs4 import BeautifulSoup
import urllib.request
import re

url = urllib.request.urlopen("http://www.djmaza.info/Abhi-Toh-Party-Khubsoorat-Full-Song-MP3-2014-Singles.html")
content = url.read()
soup = BeautifulSoup(content)
for a in soup.findAll('a',href=True):
    if re.findall('http',a['href']):
        print ("URL:", a['href'])

这段代码的输出:

URL: http://twitter.com/mp3khan
URL: http://www.facebook.com/pages/MP3KhanCom-Music-Updates/233163530138863
URL: https://plus.google.com/114136514767143493258/posts
URL: http://www.djhungama.com
URL: http://www.djhungama.com
URL: http://songs.djmazadownload.com/music/Singles/Abhi Toh Party (Khoobsurat) -190Kbps [DJMaza.Info].mp3
URL: http://songs.djmazadownload.com/music/Singles/Abhi Toh Party (Khoobsurat) -190Kbps [DJMaza.Info].mp3
URL: http://songs.djmazadownload.com/music/Singles/Abhi Toh Party (Khoobsurat) -320Kbps [DJMaza.Info].mp3
URL: http://songs.djmazadownload.com/music/Singles/Abhi Toh Party (Khoobsurat) -320Kbps [DJMaza.Info].mp3
URL: http://www.htmlcommentbox.com
URL: http://www.djmaza.com
URL: http://www.djhungama.com

我只需要 .mp3 链接。

那么，我该如何重写代码呢？

谢谢

Answer 1

更改您的 findAll 以使用正则表达式进行匹配，例如:

for a in soup.findAll('a',href=re.compile('http.*\.mp3')):
    print ("URL:", a['href'])

与评论相关的更新:

I need to store those links on an array for downloading . how can i do that ?

您可以改为使用列表理解来构建列表:

links = [a['href'] for a in soup.find_all('a',href=re.compile('http.*\.mp3'))]