使用xpath爬取猫眼电影排行榜

最近在学习xpath,在网上找资料的时候,发现一个新手经常拿来练手的项目,爬取猫眼电影前一百名排行的信息,很多都是跟崔庆才的很雷同,基本照抄.这里就用xpath自己写了一个程序,同样也是爬取猫眼电影,获取的信息是一样的,这里提供一个另外的解法.

说实话,对于网页信息的匹配,还是推荐用xpath,虽然正则确实也能达到效果,但是语句过于繁琐,一不注意就匹配不出东西,特别对于新手,本身就不熟悉正则表达式,错了都找不出来,容易劝退.正则我一般用于在处理文件,简直神器.

下面贴代码.

import requests
from requests.exceptions import RequestException
from lxml import etree
import csv
import re


def get_page(url):
    """
        获取网页的源代码
    :param url:
    :return:
    """
    try:
        headers = {
            'User-Agent': 'Mozilla / 5.0(X11;Linuxx86_64) AppleWebKit / 537.36(KHTML, likeGecko) Chrome / '
                          '76.0.3809.100Safari / 537.36',

        }
        response = requests.get(url, headers=headers)
        if response.status_code == 200:
            return response.text
        return None
    except RequestException:
        return None


def parse_page(text):
    """
        解析网页源代码
    :param text:
    :return:
    """
    html = etree.HTML(text)
    movie_name = html.xpath("//p[@class='name']/a/text()")
    actor = html.xpath("//p[@class='star']/text()")
    actor = list(map(lambda item: re.sub('\s+', '', item), actor))
    time = html.xpath("//p[@class='releasetime']/text()")
    grade1 = html.xpath("//p[@class='score']/i[@class='integer']/text()")
    grade2 = html.xpath("//p[@class='score']/i[@class='fraction']/text()")
    new = [grade1[i] + grade2[i] for i in range(min(len(grade1), len(grade2)))]
    ranking = html.xpath("///dd/i/text()")
    return zip(ranking, movie_name, actor, time, new)


def change_page(number):
    """
        翻页
    :param number:
    :return:
    """
    base_url = 'https://maoyan.com/board/4'
    url = base_url + '?offset=%s' % number
    return url


def save_to_csv(result, filename):
    """
        保存
    :param result:
    :param filename:
    :return:
    """
    with open('%s' % filename, 'a') as csvfile:
        writer = csv.writer(csvfile, dialect='excel')
        writer.writerow(result)


def main():
    """
    主函数
    :return:
    """
    for i in range(0, 100, 10):
        url = change_page(i)
        text = get_page(url)
        result = parse_page(text)
        for j in result:
            save_to_csv(j, filename='message.csv')


if __name__ == '__main__':
    main()
posted @ 2019-10-27 03:14  lattesea  阅读(608)  评论(0编辑  收藏  举报