爬虫项目——xpath练手（1）

本文爬取目标网址中的电影名，介绍，及评分

前期准备：
工具：Spyder
引用的库：requests,csv,lxml里的 etree

1.前期基本处理:

url = "https://film.sohu.com/list_4_0_0_0_0_1_60.html?channeled=1200100000"
headers = {
    "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/94.0.4606.61 Safari/537.36"    
}
resp = requests.get(url,headers=headers)
resp.encoding = "utf-8"

2.对其进行x-path解析

html = etree.HTML(resp.text)

3.进行定位

选中神秘巨星的字样,右键->检查
选择右边源码区域，选择到每个影片同级的区域，这里找到

右键->复制->复制Xpath
/html/body/div[4]/div[2]/ul/li[1]
li[1]就是神秘巨星影片所包含的信息，我们去掉[1],就能寻找到所有影片的信息了

lis = html.xpath("/html/body/div[4]/div[2]/ul/li")

4.深层寻找每个信息

for li in lis:
    title = li.xpath("./div[2]/div[1]/text()")
    abstract = li.xpath("./div[2]/div[2]/text()")
    score = li.xpath("./div[1]/div[3]/span/text()")

附上完整代码

# -*- coding: utf-8 -*-
"""
Created on Sun Oct  3 17:37:09 2021

@author: yingzi

E-mail:guotaomath@163.com
"""

import requests
from lxml import etree
import csv

url = "https://film.sohu.com/list_4_0_0_0_0_1_60.html?channeled=1200100000"
headers = {
    "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/94.0.4606.61 Safari/537.36"    
}
resp = requests.get(url,headers=headers)
resp.encoding = "utf-8"

f = open("movie.csv",mode="w",encoding="utf-8")
csvwriter = csv.writer(f)

html = etree.HTML(resp.text)

lis = html.xpath("/html/body/div[4]/div[2]/ul/li")

for li in lis:
    title = li.xpath("./div[2]/div[1]/text()")
    abstract = li.xpath("./div[2]/div[2]/text()")
    score = li.xpath("./div[1]/div[3]/span/text()")
    csvwriter.writerow([title,abstract,score])
f.close()
resp.close()
print("over!!!")

评分那块,只取出数字待解决