python - 得到一个带有scrapy splash的响应体

Question

我正在使用 scrapy 1.6 和 splash 3.2 我有：

import scrapy
import random
from scrapy_splash import SplashRequest
from scrapy.utils.response import open_in_browser
from scrapy.linkextractors import LinkExtractor

USER_AGENT = 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:48.0) Gecko/20100101 Firefox/48.0'

class MySpider(scrapy.Spider):


    start_urls = ["http://yahoo.com"]
    name = 'mytest'

    def start_requests(self):
        for url in self.start_urls:
            yield SplashRequest(url, self.parse, endpoint='render.html', args={'wait': 2.5},headers={'User-Agent': USER_AGENT,'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8'})

    def parse(self, response):
        # response.body is a result of render.html call; it
        # contains HTML processed by a browser.
        # from scrapy.http.response.html import HtmlResponse
        # ht = HtmlResponse('jj')
        # ht.body.replace =response
        open_in_browser(response)
        return None

问题是，当我尝试在浏览器中打开响应时，我却在记事本中打开它。

查看https://splash.readthedocs.io/en/stable/scripting-response-object.html。如何激活 response.body 以便可以在浏览器中打开响应（然后我希望能够使用浏览器开发工具获取 xpath）？

Answer 1

我得到了它的工作：

def parse(self, response):
    # response.body is a result of render.html call; it
    # contains HTML processed by a browser.
    from scrapy.http.response.html import HtmlResponse
    ht = HtmlResponse(url=response.url, body=response.body, encoding="utf-8", request=response.request)
    open_in_browser(ht)
    return None

Answer 2

open_in_browser()无法将 Splash 的响应检测为 HTML 响应。这是因为 Splash HTML 响应对象是 Scrapy 的子类，TextResponse而不是HtmlResponse（暂时）。

您可以open_in_browser()暂时以适合您的用例的方式重新实现。