python - 使用 scrapy 在多页上递归获取链接

Question

我正在使用我在网上找到的以下代码来递归地抓取多个页面上的链接。它应该递归地返回所有页面上我需要的所有链接。但我最终最多只获得 100 个链接。任何建议都会有帮助。

class MySpider(CrawlSpider):
    name = "craigs"
    allowed_domains = ["craigslist.org"]
    start_urls = ["http://seattle.craigslist.org/search/jjj?is_parttime=1"]   

    rules = (Rule (SgmlLinkExtractor(allow=("index\d00\.html", ),restrict_xpaths=('//a[@class="button next"]',))
    , callback="parse_items", follow= True),
    )

    def parse_items(self, response):
        hxs = HtmlXPathSelector(response)
        titles = hxs.select('//span[@class="pl"]')
        items = []
        for titles in titles:
            item = CraigslistSampleItem()
            item ["title"] = titles.select("a/text()").extract()
            item ["link"] = titles.select("a/@href").extract()           
            items.append(item)     
        return(items)

Answer 1

只需删除 allow=("index\d00\.html", ) 即可让它解析 next 链接:

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