从scrapy-splash中的第一次单击结果中单击调用另一个页面

Question

我正在废弃一个网站，其中进行了多个ajax调用并显示了内容。首先，我通过单击主页中的选项导航到页面。从生成的HTML中，我必须再次单击并导航到另一个page.如何使用scrapy-splash进行此操作？

这是我的代码

import json
import scrapy
from scrapy_splash import SplashRequest

class SA(scrapy.Spider):
    name = "sa"
    start_urls = ["http://ted.europa.eu/TED/browse/browseByBO.do"]
    def start_requests(self):
        script = """
        function main(splash)
            local url = splash.args.url
            assert(splash:go(url)) #goes to the home url
            assert(splash:wait(1))
            assert(splash:runjs("$('#link-kUK').click()")) #clicks an option in the page

            assert(splash:runjs("$('#filterTree > ul > li:nth-child(1) > a').click()")) #this click has to be done from the resulting html(i am struck with this part) and this click open a new page from where i have to scrap details
            return {
                html = splash:html()
            }
        end
        """
        for url in self.start_urls:
            yield scrapy.Request(url, self.parse_result, meta={
                'splash': {
                    'args': {'lua_source': script},
                    'endpoint': 'execute',
                }
        })

    def parse_result(self, response):
        print("response")