Use scrapy to get list of urls, and then scrape content inside those urls

问题

I need a Scrapy spider to scrape the following page (https://www.phidgets.com/?tier=1&catid=64&pcid=57) for each URL (30 products, so 30 urls) and then go into each product via that url and scrape the data inside.

I have the second part working exactly as I want:

import scrapy

class ProductsSpider(scrapy.Spider):
    name = "products"
    start_urls = [
        'https://www.phidgets.com/?tier=1&catid=64&pcid=57',
    ]

    def parse(self, response):
        for info in response.css('div.ph-product-container'):
            yield {
                'product_name': info.css('h2.ph-product-name::text').extract_first(),
                'product_image': info.css('div.ph-product-img-ctn a').xpath('@href').extract(),
                'sku': info.css('span.ph-pid').xpath('@prod-sku').extract_first(),
                'short_description': info.css('div.ph-product-summary::text').extract_first(),
                'price': info.css('h2.ph-product-price > span.price::text').extract_first(),
                'long_description': info.css('div#product_tab_1').extract_first(),
                'specs': info.css('div#product_tab_2').extract_first(),
            }

        # next_page = response.css('div.ph-summary-entry-ctn a::attr("href")').extract_first()
        # if next_page is not None:
        #     yield response.follow(next_page, self.parse)

But I don't know how to do the first part. As you will see I have the main page (https://www.phidgets.com/?tier=1&catid=64&pcid=57) set as the start_url. But how do I get it to populate the start_urls list with all 30 urls I need crawled?

回答1:

I am not able to test at this moment, so please let me know if this works for you so I can edit it should there be any bugs.

The idea here is that we find every link in the first page and yield new scrapy requests passing your product parsing method as a callback

import scrapy
from urllib.parse import urljoin

class ProductsSpider(scrapy.Spider):
    name = "products"
    start_urls = [
        'https://www.phidgets.com/?tier=1&catid=64&pcid=57',
    ]

    def parse(self, response):
        products = response.xpath("//*[contains(@class, 'ph-summary-entry-ctn')]/a/@href").extract()
        for p in products:
            url = urljoin(response.url, p)
            yield scrapy.Request(url, callback=self.parse_product)

    def parse_product(self, response):
        for info in response.css('div.ph-product-container'):
            yield {
                'product_name': info.css('h2.ph-product-name::text').extract_first(),
                'product_image': info.css('div.ph-product-img-ctn a').xpath('@href').extract(),
                'sku': info.css('span.ph-pid').xpath('@prod-sku').extract_first(),
                'short_description': info.css('div.ph-product-summary::text').extract_first(),
                'price': info.css('h2.ph-product-price > span.price::text').extract_first(),
                'long_description': info.css('div#product_tab_1').extract_first(),
                'specs': info.css('div#product_tab_2').extract_first(),
            }

来源：https://stackoverflow.com/questions/44913798/use-scrapy-to-get-list-of-urls-and-then-scrape-content-inside-those-urls

标签

python

web-scraping

scrapy