[Python] crawl Baidu mp3 music song

python crawl Baidu mp3 music song, the current success rate is not 100%, because I only caught every song again, not to judge the success of the situation and grab the link speed,
as well as the name of the song I get way too little out of place, songs longer name of the song is to search for the source of sometimes hunt to, using the exact name of the song after not have this problem.

1. [Python crawl Baidu mp3 music song code] [Python] Code

#-*- coding: UTF-8 -*- 
'''
Created on 2012-3-8
 
@author: tiantian
www.iplaypy.com python
'''
import urllib
import re
 
top500 = 'http://list.mp3.baidu.com/top/top500.html'
 
songs = []
 
def main():
     
    divr = '<div class="rank-top-wrap clearfix".*?<ul.*?</ul>.*?<ul.*?</ul>.*?</div>'   
    mf = urllib.urlopen(top500)
    content = mf.read()
    content = content.decode('gbk')
 
    content = re.sub('\n+',' ',content)
    alldiv = re.findall(divr,content)
    i =0
    for div in alldiv:
        ulr = '<ul.*?</ul>'
        allul = re.findall(ulr,div)
         
        for ul in allul:
            lir = '<li.*?</li>'
            allli = re.findall(lir,ul)
             
            for li in allli:
                if i<245:
                    i = i+1
                    continue
                i = i+1
                songName = '<div class="music-name">.*?<a.*?>(.*?)</a>.*?</div>'
                name = re.findall(songName,li)
                songAuthor = '<div class="singer">.*?<a.*?>(.*?)</a>.*?</div>'
                author = re.findall(songAuthor,li)
                 
                songs.append([name[0],author[0]])
                 
                songUrl = getSongUrl(name[0],author[0])
                try:
                    urllib.urlretrieve(songUrl,'songs/'+name[0]+'-'+author[0]+'.mp3')
                    # 异常检查并不能判断是否下载成功，需要进行其他判断
                    print i,name[0],author[0],'下载成功'
                     
                except Exception :
                    print i,name[0],author[0],'没下载成功'
                     
 
def getSongUrl(songName,authorName):
    '''这里由于歌曲名称和作者名称的不完整，可能导致无法得到url，'''
    songUrl = 'http://box.zhangmen.baidu.com/x?op=12&count=1&mtype=1&title=%s$$%s$$$$&url=&listenreelect=0&.r=0.1696378872729838' % (urllib.quote(songName.encode('gbk')),urllib.quote(authorName.encode('gbk')))
    f = urllib.urlopen(songUrl)
    c = f.read()
    url1 = re.findall('<encode>.*?CDATA\[(.*?)\]].*?</encode>',c)
    url2 = re.findall('<decode>.*?CDATA\[(.*?)\]].*?</decode>',c)
    if len(url1) <1:
        return 'http://box.zhangmen.baidu.com/unknow.mp3'
     
    try:
        return url1[0][:url1[0].rindex('/')+1] + url2[0]
    except Exception:
        return url1[0]
     
if __name__ == '__main__':
    main()

Recommended reading:

Basics zero Python's most detailed source of tutorials

2019 Python Reptile Learning Roadmap full version

Why Python can be firmly secured the first card AI Artificial Intelligence Language

Python rise, TIOBE list of programming languages ​​a new high!