如何在Scrapy中使用SgmlLinkextractor遍历下一页？

Question

我无法遍历下一页，有点坚持使用SgmlLinkextractor规则。 请进一步导航我，我不知道我在哪里犯了错误

import scrapy
from product.items import ProductItem
from scrapy.contrib.spiders import CrawlSpider, Rule
from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor


class BurlingtoncoatfactoryDSpider(CrawlSpider):
    name = "burlingtoncoatfactory_d"
    allowed_domains = ["http://www.burlingtoncoatfactory.com"]
    start_urls = ['http://www.burlingtoncoatfactory.com/catalog/searchresults.aspx?filter=&search=south+shore+furnitures&viewMode=FortyEight#pg=1']

    rules = (
        Rule(SgmlLinkExtractor(allow=('catalog/searchresults\.aspx\?filter=&search=south+shore+furnitures&viewMode=FortyEight#pg=\d*',),),
             callback='parse', follow= True),

        )


    def parse(self, response):
        item=ProductItem()
        item['product_DetailUrl'] = [content for content in   response.xpath('//div[@id="hawkitemlist"]//div[@class="row clearfix"]//div[@class="product-image"]/a/@href').extract()]
        item['product_Rating'] = [content for content in response.xpath('//*//div[@class="product-inner clearfix"]/div[@class="review"]/span[1]/@class').extract()] 
        item['product_NumberOfReviews'] = [content for content in response.xpath('//*//div[@class="product-inner clearfix"]/div[@class="review"]/span[2]/text()').extract()] 
        yield item