如何在 response.css 中正确定义并在 scrapy 中产生

Question

我是 Scrapy 的新手，有一件事我试了两天但还是没有成功。我正在练习提取 https://sofifa.com/. I adopted the code sample from https://docs.scrapy.org/ 中列出的足球运动员的信息并按如下方式进行编辑。我练习提取的信息是OVA

有谁知道我应该如何在下面的代码中正确定义“span.something...”的元素？

非常感谢，詹姆斯

import scrapy
class ToScrapeCSSSpider(scrapy.Spider):
    name = "player-css"
    start_urls = [
        'https://sofifa.com/players?type=all&tm%5B0%5D=1&r=210024&set=true',
    ]

    **def parse(self, response):
        for playerInfor in response.css("div.card"):
            yield {**
                **'OVA': playerInfor.css("span.bp3-tag p::bp3-tag p").extract()**
            }

        next_page_url = response.css("li.next > a::attr(href)").extract_first()
        if next_page_url is not None:
            yield scrapy.Request(response.urljoin(next_page_url))

Answer 1

使用此 css 选择器 response.css("tbody.list") 而不是 response.css("div.card")

因为 response.css("tbody.list") 数据很容易提取，但是当我使用 response.css("div.card") 结果是一些具有预期输出的空列表。

for playerInfor in response.css("tbody.list"):
     print( playerInfor.css('td.col.col-oa.col-sort span::text').getall())

输出

['87', '84', '84', '82', '80', '80', '80', '80', '79', '79', '79', '79', '79', '78', '77', '77', '77', '76', '76', '76', '75', '75', '74', '74', '73', '72', '72', '70', '62', '62', '60', '58', '56']

另一种方法

def parse(self, response):
        mydata =response.css('tbody.list td.col.col-oa.col-sort span::text').extract()
        yield {
            "OVA":mydata
        }

#mydata 的输出

['87', '84', '84', '82', '80', '80', '80', '80', '79', '79', '79', '79', '79', '78', '77', '77', '77', '76', '76', '76', '75', '75', '74', '74', '73', '72', '72', '70', '62', '62', '60', '58', '56']

如何在 response.css 中正确定义并在 scrapy 中产生

How to correctly define in response.css and yield in scrapy

scrapy

输出