使用python扫描本地音乐并下载歌词，python扫描, 先简单的说下吧，百度提

使用python扫描本地音乐并下载歌词，python扫描, 先简单的说下吧，百度提

这次这个真的是干货哦，昨晚弄了半晚上，从8点吃完饭就开始写，一直到了快12点才弄好，新手伤不起呀。
先简单的说下吧，百度提供了一个音乐搜索的api，你想百度请求类似于
http://box.zhangmen.baidu.com/x?op=12&count=1&title=最佳损友 XHTML

This XML file does not appear to have any style information associated with it. The document tree is shown below.

1





8
2829
1




]]>
</encode>
<decode>
<![CDATA[
7345405.mp3?xcode=e6b69cf593ea22ac78e1478e78479dc19e8e4650995cb99a&mid=0.31929107437537
]]>
</decode>
<type>8</type>
<lrcid>2829</lrcid>
<flag>1</flag>
</durl>
<p2p>
<hash>f98b6772aa97966550ec80617879becee0233bf4</hash>
<url>
<![CDATA[ ]]>
</url>
<type>mp3</type>
<size>3778335</size>
<bitrate>128</bitrate>
</p2p>
</result>

?

??????????????????????lrc????????????2829?????
?encode?decode?????????mp3??????????
http://zhangmenshiting.baidu.com/data2/music/12762845/YmRqamdua21fn6NndK6ap5WXcJlrmG1xlJhobWibmGpjk5ZtmWiZcWRjZ5lqbGyelGKWlZtubGljZ5lka2uanWSXY1qin5t1YWBmZW5ocGlhaWdnbGtqbzE$12762845.mp3?xcode=e6b69cf593ea22ac9d2b9314e565fc0caf85125f065ce3e0&mid=0.31929107437537
????????????????????????
????????lrcid?????2829
http://box.zhangmen.baidu.com/bdlrc/ ?????lrc??????,
??????????http://box.zhangmen.baidu.com/bdlrc/28/2829.lrc
???????????????????????lrcid??100????????????????????????lrcid?????????.lrc????
??lrc???????????????????????????????ok??
???????????????

Python

import os
import os.path
import re
import eyed3
import urllib2
import urllib
from urllib import urlencode
import sys 

import os
reload(sys) 
sys.setdefaultencoding('utf8')

music_path = r"E:\music"
lrc_path = r"e:\lrc"

os.remove('nolrc.txt')
os.remove('lrcxml.txt')

the_file = open('lrcxml.txt','a')
nolrc_file = open('nolrc.txt','a')

for root,dirs,files in os.walk(music_path):
    for filepath in files:
        the_path = os.path.join(root,filepath)
        if (the_path.find("mp3") != -1):
            print the_path
            the_music = eyed3.load(the_path)
            the_teg = the_music.tag._getAlbum()
            the_artist = the_music.tag._getArtist()
            the_title = the_music.tag._getTitle()
           # print the_teg
           # print the_title
           # print the_artist
            b = the_title.replace(' ','+')
           # print b
            a = the_artist.replace(' ','+')
            #print urlencode(str(b))
            if isinstance(a,unicode):
                a = a.encode('utf8')
            song_url = "http://box.zhangmen.baidu.com/x?op=12&count=1&title="+b+"" />"+a+" "

            the_file.write(song_url+'\n')
            page = urllib2.urlopen(song_url).read()
            print page
            theid = 0

            lrcid =  re.compile('<lrcid>(.*?)</lrcid>',re.S).findall(page)
            have_lrc = True
            if lrcid != []:
                theid = lrcid[0]

            else:
                nolrc_file.write(the_title+'\n')
                have_lrc = False
            print theid

            if have_lrc:
                firstid = int(theid)/100
                lrcurl = "http://box.zhangmen.baidu.com/bdlrc/"+str(firstid)+"/"+theid+".lrc"
                print lrcurl
                lrc = urllib2.urlopen(lrcurl).read()
                if(lrc.find('html')== -1):
                    lrcfile = open(lrc_path+"\\"+the_title+".lrc",'w')
                    lrcfile.writelines(lrc)
                    lrcfile.close()
                else:
                    nolrc_file.write(the_title+'\n')

the_file.close()
nolrc_file.close()
print "end!"

 

有用第一步请求所获取到底是xml格式的，所以本来想着解析xml来获取lrcid，但是在实现过程中遇到了各种问题，别的还容易，就在这一块儿浪费的时间最长，纠结未果之后，只能改用正则表达式来获取了。只能说明还是学艺不精呢。