haskell - Haskell wai 中间件 - 如果从第一个应用程序返回 404，如何运行应用程序？

Question

我正在尝试将两个 waiApplication一起“加入”。本质上，第一个应用程序将提供404 Not found 响应或不是404 Not found 响应的响应。我想添加第二个应用程序，如果第一种情况（第一个应用程序返回404）存在，它将尝试处理请求。

但是查看类型Application：

type Application = Request -> (Response -> IO ResponseReceived) -> IO ResponseReceived

不清楚如何检查响应？我不会只是ResponseRecived以最终结果结束 - 它似乎不包含任何响应代码。如下图所示：

xyzMiddle :: (Request -> (Response -> IO ResponseReceived) -> IO ResponseReceived) ->
             (Request -> (Response -> IO ResponseReceived) -> IO ResponseReceived)
xyzMiddle app req respond = do
  zzz <- (app req respond)
  pure zzz

我怎么能实现这个？或者，我在上面没有理解什么？

Answer 1

我忽略了这一点(Response -> IO ResponseReceived)，因为这是在IO上下文中 - 这是我们可以运行第二个应用程序的地方。这意味着，要检查Response我们可以执行以下操作：

xyzMiddle :: (Request -> (Response -> IO ResponseReceived) -> IO ResponseReceived) ->
             Request -> (Response -> IO ResponseReceived) -> IO ResponseReceived
xyzMiddle app req respond = app req (\r -> do
  print $ show $ responseStatus r
  respond r)