抓取一个电影网站图片，图片链接没有规律，该怎么改才能运行

Question

image_url = html.xpath('//div[@class="media-wrapper"]/a/img/@data-original')
# # print(div_image)
# for j in image_url:
# response1 = requests.get(img_url1,headers=headers)
#
# # with open("./data/{}-{}.jpg".format(name_list, j), "wb") as f:
# # f.write(response1.content)

Answer 1

不知道是不是题主要的，下载图片的代码题主自加下

import requests
from lxml import etree
from openpyxl import Workbook
wb=Workbook()
ws=wb.active
ws.append(["电影名称","电影分类","所属国家","年份","导演","演员","封面图片"])
#准备url和headers
headers={"User-Agent":"Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/80.0.3987.87 Safari/537.36 SE 2.X MetaSr 1.0"}

tmpurl="https://www.kkdsa.com/vodtype/6-{}.html"#分页url模板

for i in range(1,2):#############################采集多个分页修改这里的2
    url=tmpurl.format(i)
    response=requests.get(url=url,headers=headers)
    html=etree.HTML(response.text)
    div_list=html.xpath('//div[@class="cards video-list"]/div')
    for div in div_list:
        name=div.xpath('.//div[@class="card-heading text-ellipsis"]/strong/a/text()')[0]# 剧名
        director='，'.join(div.xpath('.//div[@class="caption"]/p[1]/a/text()'))# 导演
        actor ='，'.join(div.xpath('.//div[@class="caption"]/p[2]/a/text()'))# 演员
        img=div.xpath('.//img[@class="lazy"]/@data-original')[0]
       
        # 分类 国家 年份
        arr=''.join(div.xpath('.//div[@class="card-content text-ellipsis text-muted"]//text()')).split('/')
        classify=arr[0]#分类
        country=arr[1]#国家
        year=arr[2]#年份
        print(name,classify,country,year,director,actor,img)
        ws.append([name,classify,country,year,director,actor,img])
    
wb.save("天堂电影网.xlsx")