[已解决]IndexError: list index out of range问题

gfz

import re
import requests
from lxml import etree
headers = {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/91.0.4472.164 Safari/537.36'
}
url ='https://www.pearvideo.com/category_5'  #当前页面请求
dc_text =requests.get(url=url,headers=headers).text
dc_tree =etree.HTML(dc_text)
dc_li =dc_tree.xpath('//*[@id="listvideoListUl"]/li')
#print(dc_li)
urls =[]
for li in dc_li:
    new_url ='https://www.pearvideo.com/'+li.xpath('./div/a/@href')[0]
    sp_name = li.xpath('./div/a/div[2]/text()')[0]+'.mp4'
    print(new_url,sp_name)
    sp_text =requests.get(url=new_url,headers=headers).text
    #ex ='srcUrl="(.*?)",vdoUrl'
    ex = 'srcUrl="(.*?)",vdoUrl'
    sp_url =re.findall(ex,sp_text)[0]
    print(sp_url)

报错了 line 20, in <module>
    sp_url =re.findall(ex,sp_text)[0]
IndexError: list index out of range

所以是什么问题呢？

最佳答案

月排行榜 / 总排行榜

suchocolate

2021-7-31 12:00:53

我看页面直接get html没有mp4的信息，视频url放在了ajax里。
用ajax拿到：

1. import requests
   
2. from lxml import etree
   
3. import random
   
4. 
   
5. 
   
6. def main():
   
7.     headers = {
   
8.         'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64)'
   
9.     }
   
10.     ajax_headers = {
    
11.         'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64)',
    
12.         'Host': 'www.pearvideo.com',
    
13.         'Referer': '',
    
14.         'X-Requested-With': 'XMLHttpRequest'}
    
15.     url = 'https://www.pearvideo.com/category_5'
    
16.     base_url = 'https://www.pearvideo.com/'
    
17.     r = requests.get(url=url, headers=headers)
    
18.     html = etree.HTML(r.text)
    
19.     videos = html.xpath('//li[@class="categoryem"]/div/a/@href')
    
20.     for video in videos:
    
21.         mrd = random.random()
    
22.         video_page = f'https://www.pearvideo.com/videoStatus.jsp?contId={video[-7:]}&mrd={mrd}'
    
23.         ajax_headers['Referer'] = f'{base_url}{video}'
    
24.         r = requests.get(video_page, headers=ajax_headers)
    
25.         video_url = r.json()['videoInfo']['videos']['srcUrl']
    
26.         print(video, video_url)
    
27. 
    
28. 
    
29. if __name__ == '__main__':
    
30.     main()
    

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跳转到最佳答案楼层

2012277033

看报错能知道是超过了列表范围，看那句的意思，你这里取0都报错，应该是找到的列表为空了。
这里可以加个判断：

1. sp_ulrs = re.findall(ex,sp_text)
   
2. if len(sp_ulrs):
   
3.     sp_url=sp_urls[0]

复制代码

suchocolate

我看页面直接get html没有mp4的信息，视频url放在了ajax里。
用ajax拿到：

1. import requests
   
2. from lxml import etree
   
3. import random
   
4. 
   
5. 
   
6. def main():
   
7.     headers = {
   
8.         'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64)'
   
9.     }
   
10.     ajax_headers = {
    
11.         'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64)',
    
12.         'Host': 'www.pearvideo.com',
    
13.         'Referer': '',
    
14.         'X-Requested-With': 'XMLHttpRequest'}
    
15.     url = 'https://www.pearvideo.com/category_5'
    
16.     base_url = 'https://www.pearvideo.com/'
    
17.     r = requests.get(url=url, headers=headers)
    
18.     html = etree.HTML(r.text)
    
19.     videos = html.xpath('//li[@class="categoryem"]/div/a/@href')
    
20.     for video in videos:
    
21.         mrd = random.random()
    
22.         video_page = f'https://www.pearvideo.com/videoStatus.jsp?contId={video[-7:]}&mrd={mrd}'
    
23.         ajax_headers['Referer'] = f'{base_url}{video}'
    
24.         r = requests.get(video_page, headers=ajax_headers)
    
25.         video_url = r.json()['videoInfo']['videos']['srcUrl']
    
26.         print(video, video_url)
    
27. 
    
28. 
    
29. if __name__ == '__main__':
    
30.     main()
    

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