相关疑难解决方法(0)

免责声明:我对Scrapy很新.

明确地提出我的问题:如何从页面上的链接检索Item属性并将结果返回到同一个Item中？

给出以下示例蜘蛛:

class SiteSpider(Spider):
    site_loader = SiteLoader
    ...
    def parse(self, response):
        item = Place()
        sel = Selector(response)
        bl = self.site_loader(item=item, selector=sel)
        bl.add_value('domain', self.parent_domain)
        bl.add_value('origin', response.url)
        for place_property in item.fields:
            parse_xpath = self.template.get(place_property)

            # parse_xpath will look like either:
            # '//path/to/property/text()'
            # or
            # {'url': '//a[@id="Location"]/@href', 
            #  'xpath': '//div[@class="directions"]/span[@class="address"]/text()'}
            if isinstance(parse_xpath, dict):  # place_property is at a URL
                url = sel.xpath(parse_xpath['url_elem']).extract()
                yield Request(url, callback=self.get_url_property,
                              meta={'loader': bl, 'parse_xpath': parse_xpath,
                                    'place_property': place_property})
            else:  # parse_xpath is just an xpath; process normally
                bl.add_xpath(place_property, …