Python 3.x 将所有分页链接提取到url为scrapy的页面=https://www.blablacar.in/ride-sharing/new-delhi/chandigarh/

有人能帮我提取所有分页链接到带有

url的scrapy页面吗=https://www.blablacar.in/ride-sharing/new-delhi/chandigarh/

就像我用python试过的那样 但不了解细节

我的代码如下=====================

allowed_domains = ['blablacar.in']
    start_urls = ['https://www.blablacar.in/ride-sharing/new-delhi/chandigarh/']

    def parse(self, response):
        products = response.css('.trip-search-results li')
        for p in products:
            brand = p.css('.ProfileCard-info--name::text').extract_first().strip()
            price = p.css('.description .time::attr(content)').extract_first()
            item = ProductItem()
            item['brand'] = brand
            item['price'] = price
            yield item
        nextPageLinkSelector = response.css('.js-trip-search-pagination::attr(href)').extract_first()
        if nextPageLinkSelector:
            nextPageLink = nextPageLinkSelector
            yield scrapy.Request(url=response.urljoin(nextPageLink), )

---

您只需找到下一页的链接，然后按照以下步骤操作：

def parse(self, response):
    products = response.css('.trip-search-results li')
    for p in products:
        brand = p.css('.ProfileCard-info--name::text').extract_first().strip()
        price = p.css('.description .time::attr(content)').extract_first()
        item = ProductItem()
        item['brand'] = brand
        item['price'] = price
        yield item

    # Here is the pagination following.
    for a_tag in response.css('.pagination .next:not(.disabled) a'):
        yield response.follow(a_tag, self.parse_route)

---

请尝试以下操作以查看下一页链接：

nextPageLink = response.xpath("//*[@class='pagination']//*[@class='next' and not(contains(@class,'disabled'))]/a/@href").extract_first()
if nextPageLink:
    yield response.follow(nextPageLink,callback=self.parse)

---

执行代码时，在response.css（“.trip search results li”）中获取trip的错误信息：^TabError:中制表符和空格的使用不一致indentation@rijin.p确保缩进是一致的：缩进应该只包含空格或制表符。你好像把这些混在一起了。