Scrapy 没有输出

【问题标题】：No output with ScrapyScrapy 没有输出
【发布时间】：2022-01-20 23:05:58
【问题描述】：

我是使用scrapy 的新手，我想抓取一个网站“https://in.seamsfriendly.com/collections/shorts” 我已经编写了代码，但它给出了回溯错误，请帮助我

import scrapy
#from ..items import SeamsfriendlyItem

class SeamsSpider(scrapy.Spider):
    name = 'seams'
    start_urls = [
        'https://in.seamsfriendly.com/collections/shorts'
        ]

    def parse(self, response):

        #items = SeamsfriendlyItem()


        #all_div_shorts = response.css('div.Grid__Cell 1/2--phone 1/2--tablet-and-up 1/3--desk')
        title : response.css("#shopify-section-collection-template a::text").extract()
        price : response.css(".Price::text").extract()
        url : response.css("img::attr(data-src)").extract()

        #items['title'] = title
        #items['url'] = url

        yield {
            'title' : title,
            'price' : price,
            'url' : url
        }

【问题讨论】：

标签： python web-scraping scrapy

【解决方案1】：

您输入的是: 而不是=：

import scrapy
#from ..items import SeamsfriendlyItem

class SeamsSpider(scrapy.Spider):
    name = 'seams'
    start_urls = [
        'https://in.seamsfriendly.com/collections/shorts'
    ]

    def parse(self, response):

        #items = SeamsfriendlyItem()


        #all_div_shorts = response.css('div.Grid__Cell 1/2--phone 1/2--tablet-and-up 1/3--desk')
        title = response.css("#shopify-section-collection-template a::text").extract()
        price = response.css(".Price::text").extract()
        url = response.css("img::attr(data-src)").extract()

        #items['title'] = title
        #items['url'] = url

        yield {
            'title' : title,
            'price' : price,
            'url' : url
        }

输出：

{'title': ['Elasticated', 'Casual', 'with Pockets', 'Printed', 'Solid', '\n    ',.............

【讨论】：