Question

我必须从此IMDb页面：https://www.imdb.com/list/ls055386972/抓取所有电影。

我的方法是首先抓取<a href="/title/tt0068646/?ref_=ttls_li_tt"的所有值，即提取/title/tt0068646/?ref_=ttls_li_tt部分，然后添加“ https://www.imdb.com”以准备电影的完整URL，即https://www.imdb.com/title/tt0068646/?ref_=ttls_li_tt。但是，每当我给出response.xpath('//h3[@class]/a[@href]').extract()时，它就会提取想要的部分以及电影标题：[u'<a href="/title/tt0068646/?ref_=ttls_li_tt">The Godfather</a>', u'<a href="/title/tt0108052/?ref_=ttls_li_tt">Schindler\'s List</a>......]'我只想要"/title/tt0068646/?ref_=ttls_li_tt"部分。

如何进行？

Answer 1

import requests
from bs4 import BeautifulSoup

page = requests.get("https://www.imdb.com/list/ls055386972/")
soup = BeautifulSoup(page.content, 'html.parser')

movies = soup.findAll('h3', attrs={'class' : 'lister-item-header'})
for movie in movies:
    print(movie.a['href'])

输出：

/title/tt0068646/?ref_=ttls_li_tt
/title/tt0108052/?ref_=ttls_li_tt
/title/tt0050083/?ref_=ttls_li_tt
/title/tt0118799/?ref_=ttls_li_tt
.
.
.
.
/title/tt0088763/?ref_=ttls_li_tt
/title/tt0266543/?ref_=ttls_li_tt

Answer 2

我建议您使用requests-html来获取所有超链接，并删除不符合您的条件的超链接。您甚至可以使用r.html.absolute_links

获取绝对网址

from requests_html import HTMLSession
session = HTMLSession()
r = session.get('https://www.imdb.com/list/ls055386972/')
links = r.html.links
for i in range(len(links)):
    if not links[i].startswith('/title/'):
        del links[i]
print(links)

Answer 3

这是工作代码，请尝试：

class MoviesSpider():

  name = 'movies' #name of the spider
  allowed_domains = ['imdb.com']
  start_url = 'http://imdb.com/list/ls055386972/'

  def __init__(self):
      super(MoviesSpider, self).__init__()

  def start_requests(self):
      yield Request(self.start_url, callback=self.parse, headers=self.headers)

  def parse(self, response):
      #events = response.xpath('//*[@property="url"]/@href').extract()
      links = response.xpath('//h3[@class]/a/@href').extract()

      final_links = []

      for link in links:
          final_link = 'http://www.imdb.com' + link
          final_links.append(final_link)

      for final_link in final_links:
          absolute_url = response.urljoin(final_link)
          yield Request(absolute_url, callback = self.parse_movies)

          #process next page url
          #next_page_url = response.xpath('//a[text() = "Next"]/@href').extract_first()
          #absolute_next_page_url = response.urljoin(next_page_url)
          #yield Request(absolute_next_page_url)

  def parse_movies(self, response):

      title  = response.xpath('//div[@class = "title_wrapper"]/h1[@class]/text()').extract_first()

      yield{
                'title': title,
      }

Python + Web抓取+ scrapy：如何从IMDb页面获取所有电影的链接？

3 个答案: