[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：

import requests
from lxml import etree
import random


def main():
    headers = {
        'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64)'
    }
    ajax_headers = {
        'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64)',
        'Host': 'www.pearvideo.com',
        'Referer': '',
        'X-Requested-With': 'XMLHttpRequest'}
    url = 'https://www.pearvideo.com/category_5'
    base_url = 'https://www.pearvideo.com/'
    r = requests.get(url=url, headers=headers)
    html = etree.HTML(r.text)
    videos = html.xpath('//li[@class="categoryem"]/div/a/@href')
    for video in videos:
        mrd = random.random()
        video_page = f'https://www.pearvideo.com/videoStatus.jsp?contId={video[-7:]}&mrd={mrd}'
        ajax_headers['Referer'] = f'{base_url}{video}'
        r = requests.get(video_page, headers=ajax_headers)
        video_url = r.json()['videoInfo']['videos']['srcUrl']
        print(video, video_url)


if __name__ == '__main__':
    main()

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：

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

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：

import requests
from lxml import etree
import random


def main():
    headers = {
        'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64)'
    }
    ajax_headers = {
        'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64)',
        'Host': 'www.pearvideo.com',
        'Referer': '',
        'X-Requested-With': 'XMLHttpRequest'}
    url = 'https://www.pearvideo.com/category_5'
    base_url = 'https://www.pearvideo.com/'
    r = requests.get(url=url, headers=headers)
    html = etree.HTML(r.text)
    videos = html.xpath('//li[@class="categoryem"]/div/a/@href')
    for video in videos:
        mrd = random.random()
        video_page = f'https://www.pearvideo.com/videoStatus.jsp?contId={video[-7:]}&mrd={mrd}'
        ajax_headers['Referer'] = f'{base_url}{video}'
        r = requests.get(video_page, headers=ajax_headers)
        video_url = r.json()['videoInfo']['videos']['srcUrl']
        print(video, video_url)


if __name__ == '__main__':
    main()