Question

我正在为某人学习scrapy，因为它只返回页面上的第一项。有人能告诉我我做错了吗？

以下是我的代码：

class RuvillaSpider(Spider):

    name = "RuvillaSpider"
    allowded_domains = ["ruvilla.com"]
    start_urls = ["https://www.ruvilla.com/men/footwear.html?dir=desc&limit=45&order=news_from_date"]

    def parse(self, response):
        products = Selector(response).xpath('//div[@class="category-products"]')

        if not products:
            raise CloseSpider('RuvillaSpider: DONE, NO MORE PAGES.')

        for product in products:
            item = RuvillaItem()
            item['name'] = product.xpath('ul/li/div/div[1]/a/@title').extract()[0]
            item['link'] = product.xpath('ul/li/div/div[1]/a/@href').extract()[0]
            item['image'] = product.xpath('ul/li/div/div[1]/a/img/@src').extract()[0]
            yield item

Answer 1

您的xpath似乎只返回products变量的1个产品。

尝试：

$ scrapy shell "https://www.ruvilla.com/men/footwear.html?dir=desc&limit=45&order=news_from_date"
In[1]: response.xpath('//div[@class="category-products"]')
Out[1]: [<Selector xpath='//div[@class="category-products"]' data=u'<div class="category-products">\n<div cla'>]

所以看起来你的xpath不是针对每个单独的项目，而是针对容器中的项目。为了解决这个问题，你需要生成一个xpath来选择每个 product 容器：

def parse(self, response):
    products = Selector(response).xpath('//div[@class="category-products"]//li[contains(@class,"item")]')

    for product in products:
        item = dict()
        item['name'] = product.xpath('.//a/@title').extract_first()
        item['link'] = product.xpath('.//a/@href').extract_first()
        item['image'] = product.xpath('.//a/img/@src').extract_first()
        yield item
    next_page = response.xpath("//li[@class='current']/following-sibling::li[1]/a/@href").extract_first()
    if next_page:
        yield Request(next_page)

Answer 2

你的xpath错了。

使用此xpath：

（ '// DIV [@类= “类别的产品”] / UL /利'）

scrapy返回第一项

2 个答案: