Scrapy: Following pagination link to scrape data [

Question

This question already has an answer here:

• Scrapy: scraping data from Pagination 2 answers

I am trying to scrape data from a page and continue scraping following the pagination link.

The page I am trying to scrape is --> here

# -*- coding: utf-8 -*-
import scrapy


class AlibabaSpider(scrapy.Spider):
    name = 'alibaba'
    allowed_domains = ['alibaba.com']
    start_urls = ['https://www.alibaba.com/catalog/agricultural-growing-media_cid144?page=1']

def parse(self, response):
    for products in response.xpath('//div[contains(@class, "m-gallery-product-item-wrap")]'):
        item = {
            'product_name': products.xpath('.//h2/a/@title').extract_first(),
            'price': products.xpath('.//div[@class="price"]/b/text()').extract_first('').strip(),
            'min_order': products.xpath('.//div[@class="min-order"]/b/text()').extract_first(),
            'company_name': products.xpath('.//div[@class="stitle util-ellipsis"]/a/@title').extract_first(),
            'prod_detail_link': products.xpath('.//div[@class="item-img-inner"]/a/@href').extract_first(),
            'response_rate': products.xpath('.//i[@class="ui2-icon ui2-icon-skip"]/text()').extract_first('').strip(),
            #'image_url': products.xpath('.//div[@class=""]/').extract_first(),
         }
        yield item

    #Follow the paginatin link
    next_page_url = response.xpath('//link[@rel="next"]/@href').extract_first()
    if next_page_url:
        yield scrapy.Request(url=next_page_url, callback=self.parse)

Problem

• The code is not able to follow the pagination link.

How can you help

• Modify the code to follow the pagination link.

Answer 1

It doesn't work because url isn't valid. If you want to keep using scrapy.Request, you could use:

next_page_url = response.xpath('//link[@rel="next"]/@href').extract_first()
if next_page_url:
    next_page_url = response.urljoin(next_page_url)
    yield scrapy.Request(url=next_page_url, callback=self.parse)

A shorter solution:

next_page_url = response.xpath('//link[@rel="next"]/@href').extract_first()
if next_page_url:
    yield response.follow(next_page_url)

Answer 2

To get your code working, you need to fix the broken link by using response.follow() or something similar. Try the below approach.

import scrapy

class AlibabaSpider(scrapy.Spider):
    name = 'alibaba'
    allowed_domains = ['alibaba.com']
    start_urls = ['https://www.alibaba.com/catalog/agricultural-growing-media_cid144?page=1']

    def parse(self, response):
        for products in response.xpath('//div[contains(@class, "m-gallery-product-item-wrap")]'):
            item = {
            'product_name': products.xpath('.//h2/a/@title').extract_first(),
            'price': products.xpath('.//div[@class="price"]/b/text()').extract_first('').strip(),
            'min_order': products.xpath('.//div[@class="min-order"]/b/text()').extract_first(),
            'company_name': products.xpath('.//div[@class="stitle util-ellipsis"]/a/@title').extract_first(),
            'prod_detail_link': products.xpath('.//div[@class="item-img-inner"]/a/@href').extract_first(),
            'response_rate': products.xpath('.//i[@class="ui2-icon ui2-icon-skip"]/text()').extract_first('').strip(),
            #'image_url': products.xpath('.//div[@class=""]/').extract_first(),
            }
            yield item

        #Follow the paginatin link
        next_page_url = response.xpath('//link[@rel="next"]/@href').extract_first()
        if next_page_url:
            yield response.follow(url=next_page_url, callback=self.parse)

Your pasted code was badly indented. I've fixed that as well.