此次的目标是爬取电影天堂最新200页的最新电影的电影名称和下载链接,电影的下载链接在二级页面,所以需要先匹配一级页面的所有链接,然后逐个请求二级页面,代码如下:
""" 爬取电影天堂2019年的电影名称和链接 """ import requests import csv from fake_useragent import UserAgent from lxml import etree import re import time import random class DianyingtiantangSpider(object): def __init__(self): self.url = ‘https://www.dytt8.net/html/gndy/dyzz/list_23_{}.html‘ # self.url2 = ‘https://www.dytt8.net/html/gndy/dyzz/20190918/59138.html‘ def get_headers(self): """ 构造请求头 :return: """ ua = UserAgent() headers = { "Accept": "text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3", "Accept-Encoding": "gzip, deflate, br", "Accept-Language": "zh-CN,zh;q=0.9", "Cache-Control": "max-age=0", "Connection": "keep-alive", "Cookie": "UM_distinctid=16bdec86bc2679-07c211dd7aebc-15231708-1fa400-16bdec86bc3464; CNZZDATA1260535040=961678368-1562805532-https%253A%252F%252Fwww.baidu.com%252F%7C1562805532", "Host": "www.dytt8.net", "If-Modified-Since": "Thu, 19 Sep 2019 00:34:23 GMT", "If-None-Match": "80d1b3fb816ed51:326", "Sec-Fetch-Mode": "navigate", "Sec-Fetch-Site": "none", "Sec-Fetch-User": "?1", "Upgrade-Insecure-Requests": "1", "User-Agent": ua.random } return headers def re_func(self, re_bds, html): """ 正则表达式解析出第一页的排名链接 :param re_bds: :param html: :return: """ pattern = re.compile(re_bds, re.S) r_list = pattern.findall(html) return r_list def parse_index_page(self, url): """ 请求解析第一页的链接 :param url: :return: """ text = requests.get(url=url, headers=self.get_headers()) text.encoding = ‘GBK‘ # print(text.text) re_bds = r‘<table width="100%".*?<td width="5%".*?<a href="(.*?)".*?ulink">.*?</table>‘ # link = re.findall(‘<table width="100%".*?<td width="5%".*?<a href="(.*?)".*?ulink">.*?</table>‘, text.text) # html = etree.HTML(text.text) # link = html.xpath("//ul/table[@class=‘tbspan‘][1]/tbody/tr[4]/td/text()") link = self.re_func(re_bds, text.text) print(link) # print(text) return link def parse_two_page(self, url): """ 请求解析第二页的电影名和下载链接 :param url: :return: """ text = requests.get(url=url, headers=self.get_headers()) # print(text.text) text.encoding = ‘GBK‘ html = etree.HTML(text.text) movie = html.xpath(‘//*[@id="header"]/div/div[3]/div[3]/div[1]/div[2]/div[1]/h1/font/text()‘) download = html.xpath(‘//tbody/tr/td/a/@href‘) # movie=re.findall("",text.text) print(movie) print(download) # print(html) return (movie[0], download[0]) def save_csv(self, result): """ 保存到csv文件 :param result: :return: """ with open(‘movie.csv‘, ‘a‘, newline=‘‘)as f: writer = csv.writer(f) writer.writerows(result) def run(self): """ 主函数 :return: """ for i in range(1, 201): url1 = self.url.format(i) list_result = [] link = self.parse_index_page(url1) for j in link: url2 = ‘https://www.dytt8.net‘ + j try: result = self.parse_two_page(url2) print(result) list_result.append(result) time.sleep(random.uniform(1, 3)) except Exception as err: print(err) self.save_csv(list_result) print("第%s页保存成功" % i) if __name__ == ‘__main__‘: spider = DianyingtiantangSpider() spider.run()
一级页面:
二级页面:
总结:
1.在爬取电影天堂的时候发现,请求一级页面的时候得到的网页源码中居然缺少了一些标签,所以刚开始总是匹配不到东西,xpath有时候并非万能,特别是遇到这种页面缺失的.这时候就不得不拿出杀手锏正则表达式了,只要页面出现的东西,正则就能拿到,当然正则有时候会匹配出一下重复值,这时候只要用到python中的集合函数set()处理一下就行.
2.还遇到了另一个问题,就是电影天堂的编码是gb2312,所以刚开始请求得到的网页中的中文是乱码,一番查询之后发现只要添加一下代码就能成功解决
text = requests.get(url=url, headers=self.get_headers()) text.encoding = ‘GBK‘
原文地址:https://www.cnblogs.com/lattesea/p/11746482.html
时间: 2024-12-08 15:50:42