爬虫_豆瓣全部正在热映电影（xpath）

单纯地练习一下xpath

 1 import requests
 2 from lxml import etree
 3 
 4 
 5 def get_url(url):
 6     html = requests.get(url)
 7     return html.text
 8 
 9 
10 def parse_html(html):
11     informations = []
12     html_element = etree.HTML(html)
13     ul = html_element.xpath('//ul[@class="lists"]')[0]
14     for li in ul:
15         href = li.xpath('.//li[@class="poster"]/a/@href')
16         title = li.xpath('normalize-space(.//li[@class="stitle"]/a/@title)')  #normalize-space 去掉换行符
17         mark = li.xpath('.//span[@class="subject-rate"]/text()')
18         actor = li.xpath('@data-actors')
19         director = li.xpath('@data-director')
20     # print(etree.tostring(uls, encoding='utf-8').decode('utf-8'))
21         information = {
22             'href': href,
23             'title': title,
24             'mark': mark,
25             'actors': actor,
26             'director': director
27         }
28         informations.append(information)
29     
30     print(informations)
31 
32 
33 def main():
34     url = 'https://movie.douban.com/cinema/nowplaying/beijing/'
35     html = get_url(url)
36     parse_html(html)
37 
38 
39 if __name__ == '__main__':
40     main()

爬虫_豆瓣全部正在热映电影 （xpath）

爬虫_豆瓣全部正在热映电影（xpath）