如何使用splash获取弹出式内容

docker run -p 8050:8050 scrapinghub/splash

function main(splash, args)
  -- 打开目标网页
  assert(splash:go(args.url))
  assert(splash:wait(2))

  -- 获取弹出窗口的句柄
  local handles = splash:get_viewport_handles()
  for _, handle in ipairs(handles) do
    if handle ~= splash:get_main_window_handle() then
      -- 切换到弹出窗口
      splash:set_viewport_handle(handle)
      local popup_content = splash:html()
      -- 处理弹出窗口的内容
      -- ...
      break
    end
  end

  return {
    html = splash:html(),
    png = splash:png(),
    har = splash:har(),
  }
end

import requests

url = 'http://localhost:8050/run'
lua_script = """
function main(splash, args)
  -- Lua脚本内容（如上所示）
end
"""

response = requests.post(url, json={
    'lua_source': lua_script,
    'url': 'http://example.com',  # 目标网页URL
})

result = response.json()
print(result['html'])  # 打印弹出窗口的HTML内容