使用scrapy获取url列表，然后在这些url中抓取内容 [英] Use scrapy to get list of urls, and then scrape content inside those urls

问题描述

我需要一个 Scrapy 蜘蛛来抓取以下页面(https://www.phidgets.com/?tier=1&catid=64&pcid=57) 对于每个 URL(30 个产品，所以 30 个 URL)，然后通过该 URL 进入每个产品并抓取其中的数据.

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.

我让第二部分完全按照我的意愿工作:

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)

但我不知道如何做第一部分.正如您将看到的，我有主页 (https://www.phidgets.com/?tier=1&catid=64&pcid=57) 设置为 start_url.但是我如何让它用我需要抓取的所有 30 个 url 填充 start_urls 列表?

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?

推荐答案

我目前无法测试，所以请告诉我这是否适合您，以便我可以在有任何错误时对其进行编辑.

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.

>

这里的想法是我们找到第一页中的每个链接并产生新的scrapy请求，将您的产品解析方法作为回调传递

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(),
            }

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