Scrapy 抓取下一页 [英] Scrapy crawl with next page

问题描述

我有这个用于scrapy框架的代码:

I have this code for scrapy framework:

# -*- 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'

但问题是我得到了 100 个结果，它不会进入下一页.

But problem is that i get 100 results, it doesn't go to next pages.

这里出了什么问题?

推荐答案

你的 rule 没有被使用，因为你没有使用 CrawlSpider.

Your rule is not used because you don't use a CrawlSpider.

所以你必须像这样手动创建下一页requests:

So you have to create the next page requests manually like so:

# -*- 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

或者像这样使用CrawlSpider:

# -*- 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'

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