关于这个问题，我一直在苦苦寻找一条出路：（我可能展示的功能不起作用，是错误的，但它是更多的过程，我感到困惑）
我试图让我的蜘蛛得到“标准棚”页面上所有产品的价格。这是包含产品的页面的链接：https://www.charnleys.co.uk/product-category/gardening/garden-accessories/garden-furniture/sheds/standard-sheds/
但是，如果您单击产品链接，您会看到路径更改为“charnleys.co.uk/shop/shed-product-name“，因此我的爬行器无法跟踪。
我想做的是收集“standard-sheds”页面上的URL，将它们附加到一个数组中并进行迭代，然后让我的蜘蛛进入这些URL并收集价格。但是，我不确定如何让我的蜘蛛遍历URL数组。我将列出我当前创建的函数。
任何帮助都是非常感谢的。

from gc import callbacks
import scrapy
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors import LinkExtractor

urls = []

class CharnleySpider(CrawlSpider):
    name = 'crawler'
    allowed_domains = ['charnleys.co.uk']
    start_urls = ['https://www.charnleys.co.uk']

# https://www.charnleys.co.uk/product-category/gardening/garden-accessories/garden-furniture/sheds/standard-sheds/

# https://www.charnleys.co.uk/shop/bentley-supreme-apex/

        rules = (
        Rule(LinkExtractor(allow='product-category/gardening/garden-accessories/garden- 
       furniture/sheds', deny='sheds')),
        Rule(LinkExtractor(allow='standard-sheds'), callback='collect_urls')
        )

    def collect_urls(self, response):
        for elements in response.css('div.product-image'):
            urls.append(elements.css('div.product-image a::attr(href)').get())

    def html_return_price_strings(self, response): 

        #Searches through html of webpage and returns all string with "£" attatched.
        all_html = response.css('html').get()
        for line in all_html.split('\n'):
            for word in line.split():
                if word.startswith('£'):
                    print (word)

    def parse_product(self, response, html_return_price_strings): 
         yield {
             'name' : response.css('h2.product_title::text').get(),
             'price' : html_return_price_strings()

         }