将真实URL通过Scrapy-Splash传递到字典

Question

当试图通过('url‘：response.request.url)将url保存到字典中时，Scrapy从Scrapy中保存所有相同的url (http://localhost:8050/render.html)

我尝试过添加额外的参数，这些参数将传递真正的URL，但没有效果。

from scrapy import Spider
from scrapy.http import FormRequest
from scrapy.utils.response import open_in_browser
from scrapy import Request
import scrapy
from scrapy_splash import SplashRequest

class QuotesJSSpider(scrapy.Spider):
    name = 'quotesjs'
    start_urls = ('https://www.facebook.com/login',)
    custom_settings = {
        'SPLASH_URL': 'http://localhost:8050',
        'DOWNLOADER_MIDDLEWARES': {
            'scrapy_splash.SplashCookiesMiddleware': 723,
            'scrapy_splash.SplashMiddleware': 725,
            'scrapy.downloadermiddlewares.httpcompression.HttpCompressionMiddleware': 810,
        },
        'SPIDER_MIDDLEWARES': {
            'scrapy_splash.SplashDeduplicateArgsMiddleware': 100,
        },
        'DUPEFILTER_CLASS': 'scrapy_splash.SplashAwareDupeFilter',
    }

    def parse(self, response):
        token = response.xpath('//*[@id="u_0_a"]').extract_first()
        return FormRequest.from_response(response,
                                         formdata={'lgndim' : token,
                                                   'pass': 'xxx',
                                                   'email': 'xxxx'},
                                         callback=self.load_sites)

    def load_sites(self, response):
            urls = [
                'https://www.facebook.com/page1/about',
                'https://www.facebook.com/page2/about',
            ]
            for url in urls:
                yield SplashRequest(url=url, callback=self.scrape_pages)

    def scrape_pages(self, response):
        shops = {
            'company_name' : response.css('title::text').extract(),
            'url' : response.request.url,

        }

        yield shops

结果应该如下：'url‘：https://www.facebook.com/page1/about’

而不是这个：'url‘：http://localhost:8050/render.html，

Answer 1

原始请求的url可以在这里获得：response.request._original_url。

为了避免不得不访问内部属性，您还可以尝试：

• 在元中传递url：

    def load_sites(self, response):
                urls = [
                    'https://www.facebook.com/page1/about',
                    'https://www.facebook.com/page2/about',
                ]
                for url in urls:
                    yield SplashRequest(url=url, callback=self.scrape_pages, meta={'original_url': url})

    def scrape_pages(self, response)
        shops = {
                'company_name' : response.css('title::text').extract(),
                'url' : response.meta['original_url'],
        }
        yield shops

• 使用响应中的url：

    def scrape_pages(self, response):
        shops = {
            'company_name' : response.css('title::text').extract(),
            'url' : response.url,
        }