无法使用Scrapy抓取下一页内容

Question

I want to scrape the contents from the next pages too but it didn't go to the next page. My code is:

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)

I didn't use the urljoin because the next_page_url is giving me the whole url. I also tried the dont_filter=true argument in the yield function which gives me an infinite loop through the 1st page. The message I'm getting from the terminal is [scrapy.spidermiddlewares.offsite] DEBUG: Filtered offsite request to 'www.startech.com.bd': https://www.startech.com.bd/component/processor?page=2>

Answer 1

This is because your allowed_domains variable is wrong, use allowed_domains = ['www.startech.com.bd'] instead (see the doc) .

You can also modify your next page selector in order to avoid going to page one again:

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)