Python 爬取喜马拉雅免费相声音频
from requests import getfrom lxml import etree
import os
try:
os.mkdir("Video")
os.chdir("Video")
except:
os.chdir("Video")
def open_url(url):
headers = {'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/81.0.4044.129 Safari/537.36'}
res = get(url, headers=headers)
return res
def get_xs(res):
temp = 'https://www.ximalaya.com/xiangsheng'
html = etree.HTML(res.text)
name = html.xpath("//*[@class='general-album-list']/div[@class='content']/ul/li/div/a/span/text()")
href = /div[@class='content']/ul/li/div/a/@href")]
i = 1
result = {}
for k in zip(name, href):
result = k
i += 1
return result
def get_Videourl(nm, hf):
res = open_url(hf)
vdurl = []
temp = 'https://www.ximalaya.com/xiangsheng'
html = etree.HTML(res.text)
href = html.xpath('//*[@class="sound-list _Qp"]/ul/li/div/a/@href')
name = html.xpath('//*[@class="sound-list _Qp"]/ul/li/div/a/span/text()')
ids = for each in html.xpath('//*[@class="sound-list _Qp"]/ul/li/div/a/@href')]
# for nm, hf in zip(name, href):
# print(nm, ':', temp + hf)
for id in ids:
vdurl.append('https://www.ximalaya.com/revision/play/v1/audio?id=%s&ptype=1' % id)
return vdurl, name
def get_Video(vdurl, nm):
i = 0
for url in vdurl:
res = open_url(url).json()
for each in res:
if type(res) != int:
tempurl = res['src']
video = open_url(tempurl)
filename = f"{nm}.m4a"
print("正在下载：",filename)
print(i)
with open(filename, 'wb') as f:
f.write(video.content)
i += 1
def main():
url = 'https://www.ximalaya.com/xiangsheng/xiangsheng/mr132t2722/'
res = open_url(url)
result = get_xs(res)
for i in result:
for nm in result:
print(i, end = ' ')
print(nm)
break
choice = int(input("请选择您要听的专辑序号："))
nm, hf = result
vdurl, nm = get_Videourl(nm, hf)
get_Video(vdurl, nm)
if __name__ == "__main__":
main()
我待会把思路发出来，顺便说几个缺陷:
1,翻页没有写
2，爬取相声数量太少
这些我都会一一改进，如果有好的建议，请回复我，谢谢 沙发！
支持楼主！ 付费音频的能爬吗{:10_254:} liuzhengyuan 发表于 2020-5-1 16:48
付费音频的能爬吗
前提我得有个VIP的账号{:10_266:} {:10_256:}{:10_256:}支持一波 应该是免费的都能爬吧 wuqramy 发表于 2020-5-4 17:38
应该是免费的都能爬吧
是的
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