在 Scrapy 中使用请求元以不同的顺序传递参数

Question

我正在尝试从包含产品列表的页面中抓取数据，目前我正在获取所有链接并抓取详细信息，但问题是产品 manufacturer/brand 仅在列表页面，而不是产品页面。我试过在回调中使用传递请求元，但制造商数据是无序传递的，导致行显示不正确的制造商。

这是示例页面：https://www.toolmania.cl/sierras-sable-561

现在是代码：

    def parse(self, response):
    """Process toolmania.cl products"""
    # define product url xpath
    XPATH_PRODUCT_LINK = "//a[@class='thumbnail product-thumbnail']/@href"
    products = response.xpath(XPATH_PRODUCT_LINK).extract()
    XPATH_PRODUCT_BRAND = ".//h4[@class='product-manufacturer']/text()"

    for product in products:
        # obtain product brand
        brand = response.xpath(XPATH_PRODUCT_BRAND).get()
        #url = product
        yield scrapy.Request(product, callback=self.parse_product, meta={'brand': brand})

    # follow pagination link
    XPATH_NEXT_PAGE = "//li[@class='page-item directional js-search-link']//a[@rel='next']/@href"
    next_page = response.xpath(XPATH_NEXT_PAGE).get()
    if next_page:
        yield scrapy.Request(url=next_page, callback=self.parse)

def parse_product(self, response):
    """Get details from single product"""
    XPATH_SINGLE_PRODUCT = "//div[@class='single-product']"

    for product in response.xpath(XPATH_SINGLE_PRODUCT):
        # define xpaths for product details
        XPATH_PRODUCT_MODEL = ".//h5[@class='product-reference-single']/text()"
        XPATH_PRODUCT_NAME = ".//h1[@class='product-name-single mb-md-4']/text()"
        XPATH_PRODUCT_PRICE = ".//div[@class='product-prices margin__bottom__20']//span[@itemprop='price']/@content"

        product_model = product.xpath(XPATH_PRODUCT_MODEL).get()
        # clean product model
        product_model = re.sub('Código de referencia: ', '', product_model)

        yield {
            'product_brand': response.meta['brand'],
            'product_model': product_model,
            'product_price': product.xpath(XPATH_PRODUCT_PRICE).extract(),
            'product_name': product.xpath(XPATH_PRODUCT_NAME).extract(),
            'product_link': response.url,
        }

Answer 1

在下面的循环中使用 product 而不是 response，并且还看到我正在使用 css 而不是 xpath

def parse(self, response):
    """Process toolmania.cl products"""
    products = response.css('div.product-list')

    for product in products:
        # use "product" instead of "response"
        brand = product.css('.product-manufacturer::text').get()
        url = product.css(".thumbnail::attr(href)").get()
        yield scrapy.Request(product, callback=self.parse_product, meta={'brand': brand})

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