Scrapy 抓取下一頁

Question

我有這個用於scrapy框架的代碼：

# -*- 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'

但問題是我得到了 100 個結果，它不會轉到下一頁。

這里有什么問題？

Answer 1

沒有使用您的rule因為您沒有使用CrawlSpider 。

所以你必須像這樣手動創建下一頁requests ：

# -*- 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

或者像這樣使用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'

Answer 2

嘗試這個

  next_page_url = response.xpath('//a[@class="button next"]').extract_first()

  if next_page_url is not None:
        yield scrapy.Request(response.urljoin(next_page_url))