需求:爬取每一个问政标题的内容 并且翻页
解决方案:通过Scrapy来实现需求
问政url:
http://wz.sun0769.com/political/index/index
思路:
1.找到ul标签
2.找到li标签
3.找到li标签里面的href title
4.翻页
http://wz.sun0769.com/political/index/politicsNewest 第一页
http://wz.sun0769.com/political/index/politicsNewest?id=1&page=2 第二页
http://wz.sun0769.com/political/index/politicsNewest?id=1&page=3 第三页
yg.py
import scrapy
from yangguang.items import YangguangItem
class YgSpider(scrapy.Spider):
name = 'yg'
allowed_domains = ['wz.sun0769.com']
start_urls = ['http://wz.sun0769.com/political/index/politicsNewest?id=1']
def parse(self, response):
li_list = response.xpath("//ul[@class='title-state-ul']/li")
for li in li_list:
item = YangguangItem()
item['title'] = li.xpath('./span[3]/a/text()').extract_first() # 标题
item['href'] = 'http://wz.sun0769.com' + li.xpath('./span[3]/a/@href').extract_first() # 详情页的url
# print(item)
# 向详情页发起请求并获取得响应的数据
yield scrapy.Request(
url=item['href'],
callback=self.parse_detail,
meta={
'item': item}
)
# 开始进行翻页
next_url = 'http://wz.sun0769.com' + response.xpath('//div[@class="mr-three paging-box"]/a[2]/@href').extract_first()
print(next_url)
if next_url is not None:
yield scrapy.Request(
url=next_url,
callback=self.parse
)
# http://wz.sun0769.com/political/politics/index?id=492147
def parse_detail(self, response):
item = response.meta.get('item')
# 获取详情页的数据
item['content'] = response.xpath('//div[@class="details-box"]/pre/text()').extract_first()
print(item)
yield item
pipelines.py
import re
class YangguangPipeline:
def process_item(self, item, spider):
# 处理格式问题
item['content'] = self.pares_content(item['content'])
print(item)
return item
# 定义一个方法来处理格式问题
def pares_content(self, content):
content = re.sub(r'\r\n', '', content)
return content
items.py
import scrapy
class YangguangItem(scrapy.Item):
# define the fields for your item here like:
# name = scrapy.Field()
title = scrapy.Field()
href = scrapy.Field()
content = scrapy.Field()