Scrapy爬行与下一页

Question

我有scrapy框架的代码:

# -*- coding: utf-8 -*-
import scrapy
from scrapy.contrib.spiders import Rule
from scrapy.linkextractors import LinkExtractor
from lxml import html

class Scrapy1Spider(scrapy.Spider):
    name = "scrapy1"
    allowed_domains = ["sfbay.craigslist.org"]
    start_urls = (
        'http://sfbay.craigslist.org/search/npo',
    )

    rules = (Rule(LinkExtractor(allow=(), restrict_xpaths=('//a[@class="button next"]',)), callback="parse", follow= True),)

    def parse(self, response):
        site = html.fromstring(response.body_as_unicode())
        titles = site.xpath('//div[@class="content"]/p[@class="row"]')
        print len(titles), 'AAAA'

但问题是,我得到100个结果,它不会转到下一页.

这有什么不对？

Answer 1

你rule没有被使用,因为你没有使用CrawlSpider.

所以你必须requests手动创建下一页,如下所示:

# -*- coding: utf-8 -*-
import scrapy
from scrapy.contrib.spiders import Rule
from scrapy.linkextractors import LinkExtractor
from lxml import html

class Scrapy1Spider(scrapy.Spider):
    name = "craiglist"
    allowed_domains = ["sfbay.craigslist.org"]
    start_urls = (
        'http://sfbay.craigslist.org/search/npo',
    )

    Rules = (Rule(LinkExtractor(allow=(), restrict_xpaths=('//a[@class="button next"]',)), callback="parse", follow= True),)

    def parse(self, response):
        site = html.fromstring(response.body_as_unicode())
        titles = site.xpath('//div[@class="content"]/p[@class="row"]')
        print len(titles), 'AAAA'

        # follow next page links
        next_page = response.xpath('.//a[@class="button next"]/@href').extract()
        if next_page:
            next_href = next_page[0]
            next_page_url = 'http://sfbay.craigslist.org' + next_href
            request = scrapy.Request(url=next_page_url)
            yield request

或者使用CrawlSpider如下:

# -*- coding: utf-8 -*-
import scrapy
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors import LinkExtractor
from lxml import html

class Scrapy1Spider(CrawlSpider):
    name = "craiglist"
    allowed_domains = ["sfbay.craigslist.org"]
    start_urls = (
        'http://sfbay.craigslist.org/search/npo',
    )

    rules = (Rule(LinkExtractor(allow=(), restrict_xpaths=('//a[@class="button next"]',)), callback="parse_page", follow= True),)

    def parse_page(self, response):
        site = html.fromstring(response.body_as_unicode())
        titles = site.xpath('//div[@class="content"]/p[@class="row"]')
        print len(titles), 'AAAA'