python – 我怎么能告诉Scrapy只抓取Xpath中的链接？

就像一个例子,我只想让一个爬行器通过将start_urls设置为https://www.airbnb.com/s?location=New+York%2C+NY&checkin=&checkout=&guests=1的AirBnB列表

而不是抓取URL中的所有链接,我只想抓取xpath内的链接// * [@ id =“results”]

目前我正在使用以下代码来抓取所有链接,我如何才能使其仅适用于抓取// * [@ id =“results”]

from scrapy.selector import HtmlXPathSelector
    from tutorial.items import DmozItem
    from scrapy.contrib.spiders import CrawlSpider,Rule
    from scrapy.contrib.linkextractors.sgml import SgmlLinkExtractor
    from scrapy.selector import HtmlXPathSelector


    class BSpider(CrawlSpider):
            name = "bt"
            #follow = True
            allowed_domains = ["mydomain.com"]
            start_urls = ["http://myurl.com/path"]
            rules =(Rule(SgmlLinkExtractor(allow = ()),callback = 'parse_item',follow=True),)


        def parse_item(self,response):
        {parse code}

正确方向的任何提示将非常感激,
谢谢！