scrapy-splash返回自己的标题,而不是网站的原始标题

Question

我使用scrapy-splash来构建我的蜘蛛.现在我需要的是维护会话,所以我使用scrapy.downloadermiddlewares.cookies.CookiesMiddleware并处理set-cookie标头.我知道它处理s​​et-cookie标头,因为我设置了COOKIES_DEBUG = True,这导致CookeMiddleware关于set-cookie标头的打印输出.

问题是:当我还在图片中添加Splash时,set-cookie打印输出消失,实际上我得到的响应标题是{'Date':['Sun,2016年9月25日12:09:55 GMT'],' Content-Type':['text/html; charset = utf-8'],'Server':['TwistedWeb/16.1.1']}这与使用TwistedWeb的splash渲染引擎有关.

是否有任何指令告诉飞溅也给我原始的响应标题？

Answer 1

要获得原始响应标头,您可以编写Splash Lua脚本 ; 请参阅scrapy-splash README中的示例:

使用Lua脚本获取HTML响应,并将cookie,标题,正文和方法设置为正确的值; lua_source参数值缓存在Splash服务器上,不会随每个请求一起发送(它需要Splash 2.1+):

import scrapy
from scrapy_splash import SplashRequest

script = """
function main(splash)
  splash:init_cookies(splash.args.cookies)
  assert(splash:go{
    splash.args.url,
    headers=splash.args.headers,
    http_method=splash.args.http_method,
    body=splash.args.body,
    })
  assert(splash:wait(0.5))

  local entries = splash:history()
  local last_response = entries[#entries].response
  return {
    url = splash:url(),
    headers = last_response.headers,
    http_status = last_response.status,
    cookies = splash:get_cookies(),
    html = splash:html(),
  }
end
"""

class MySpider(scrapy.Spider):


    # ...
        yield SplashRequest(url, self.parse_result,
            endpoint='execute',
            cache_args=['lua_source'],
            args={'lua_source': script},
            headers={'X-My-Header': 'value'},
        )

    def parse_result(self, response):
        # here response.body contains result HTML;
        # response.headers are filled with headers from last
        # web page loaded to Splash;
        # cookies from all responses and from JavaScript are collected
        # and put into Set-Cookie response header, so that Scrapy
        # can remember them.

scrapy-splash还提供用于cookie处理的内置帮助程序; 一旦按照自述文件中的描述配置了 scrapy-splash,它们就会在此示例中启用.