scrapy 采集列表页和内容页问题

This topic created in 4313 days ago, the information mentioned may be changed or developed.

打算在列表页采集缩略图和标题，内容页采集分类和标签。
不过一直采集不成功。请问该如何写这个spider。

from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
from scrapy.selector import Selector

from my.items import MyItem
import re
from scrapy.http import Request
from scrapy.selector import Selector
from scrapy.selector import HtmlXPathSelector

class MySpider(CrawlSpider):
name = 'xxx'
allowed_domains = ['xxxx.com']
start_urls = ['http://xxxxxx.com']

def parse(self, response):
    item = MyItem()
    sel = Selector(response)
    videos = sel.xpath('//ul[@class="listThumbs"]/li')
    for v in videos:
        item['img']=v.xpath('a[@class="thumb"]/img/@src').extract()[0]
        item['title'] = v.xpath('a[@class="title"]/text()').extract()[0]
        item['url'] = v.xpath('a[@class="thumb"]/@href').extract()[0]
        yield item
def parse_page(self,response):
    item=MyItem()
    hxs=Selector(response)
    cate=hxs.xpath('//div[@class="multiTag"]/ul/li')
    for c in cate:
        item['category']=c.xpath('a/text()').extract()[0]
        yield item

import

item

extract

2 replies • 2015-01-19 04:03:23 +08:00

huangguoji

Jan 14, 2015

def parse(self, response):
item = MyItem()
sel = Selector(response)
videos = sel.xpath('//ul[@class="listThumbs"]/li')
for v in videos:
item['img']=v.xpath('a[@class="thumb"]/img/@src').extract()[0]
item['title'] = v.xpath('a[@class="title"]/text()').extract()[0]
item['url'] = v.xpath('a[@class="thumb"]/@href').extract()[0]
yield Request(item['url'],callback=self.parse_page,meta={"item":item})
def parse_page(self,response):
item= response.meta["item"]
hxs=Selector(response)
cate=hxs.xpath('//div[@class="multiTag"]/ul/li')
for c in cate:
item['category']=c.xpath('a/text()').extract()[0]
yield item

willme105

Jan 19, 2015

@huangguoji 这样做还是有问题，列表页显示的内容都一样，而且都是最后一条信息。但内容页获取的信息是对的。请问改怎么办