无法使用 Scrapy 抓取下一页内容

Question

我也想从下一页抓取内容，但它没有转到下一页。我的代码是：

import scrapy
class AggregatorSpider(scrapy.Spider):
name = 'aggregator'
allowed_domains = ['startech.com.bd/component/processor']
start_urls = ['https://startech.com.bd/component/processor']

def parse(self, response):
    processor_details = response.xpath('//*[@class="col-xs-12 col-md-4 product-layout grid"]')
    for processor in processor_details:
        name = processor.xpath('.//h4/a/text()').extract_first()
        price = processor.xpath('.//*[@class="price space-between"]/span/text()').extract_first()
        print ('\n')
        print (name)
        print (price)
        print ('\n')
    next_page_url = response.xpath('//*[@class="pagination"]/li/a/@href').extract_first()
    # absolute_next_page_url = response.urljoin(next_page_url)
    yield scrapy.Request(next_page_url)

我没有使用 urljoin，因为 next_page_url 给了我整个 url。我还尝试了 yield 函数中的 dont_filter=true 参数，这给了我一个通过第一页的无限循环。我从终端收到的消息是 [scrapy.spidermiddlewares.offsite] DEBUG: Filtered offsite request to 'www.startech.com.bd': https://www.startech.com.bd /component/processor?page=2>

Answer 1

这是因为您的allowed_domains 变量错误，请使用allowed_domains = ['www.startech.com.bd'] 而不是(see the doc)。

您还可以修改下一页选择器以避免再次进入页面：

import scrapy
class AggregatorSpider(scrapy.Spider):
    name = 'aggregator'
    allowed_domains = ['www.startech.com.bd']
    start_urls = ['https://startech.com.bd/component/processor']

    def parse(self, response):
        processor_details = response.xpath('//*[@class="col-xs-12 col-md-4 product-layout grid"]')
        for processor in processor_details:
            name = processor.xpath('.//h4/a/text()').extract_first()
            price = processor.xpath('.//*[@class="price space-between"]/span/text()').extract_first()
            yield({'name': name, 'price': price})
        next_page_url = response.css('.pagination li:last-child a::attr(href)').extract_first()
        if next_page_url:
            yield scrapy.Request(next_page_url)