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()