Cowboy and jiffy websocket_handle events

Question

我使用cowboy和jiffy (Json库)为我的websocket..我可以在我的websocket_handle中接收来自客户端的数据，现在我有类似于此的代码

websocket_handle({text, Msg}, Req, Handle) ->
    Message = jiffy:decode(Msg),
    case Message of 　

pattern1 ->...........pattern2 - - ->..........最终,{ok, Res, State}

然而，我想在句柄级别进行模式匹配…

  websocket_handle({text, pattern1}, Req, State),
  websocket_handle({text, pattern2}, Req, State)

问题是我不能在websocket_handle参数中使用jiffy:decode ..所以我不能写

 websocket_handle({text, jiffy:decode(Msg), Req, State})

如何让它工作??

Answer 1

如果你喜欢函数签名中的模式匹配而不是case表达式，你可以将解码的消息"转发"到另一个"内部"函数。

websocket_handle({text, Msg}, Req, Handle) ->
    Message = jiffy:decode(Msg),
    websocket_handle_({test, Message}, Req, Handle).
websocket_handle_({text, Pattern1}, Req, Handle) ->
    {ok, Req, State}; 
websocket_handle_({text, Pattern2}, Req, Handle) ->
    {reply, OutFrame | [OutFrame], Req, State};
websocket_handle_({text, PatternN}, Req, Handle) ->
    {ok, Req, State}.

Answer 2

所以改成

websocket_handle({text, Msg}, Req, Handle) ->
    websocket_handle_({text, jiffy:decode(Msg)}, Req, Handle).
websocket_handle_({text, Pattern1}, Req, Handle) ->
    %% do stuff,
   {ok, Req, State};
websocket_handle_({text, Pattern2}, Req, Handle) ->
    %% do stuff,
     {reply, [OutFrame], Req, State};
websocket_handle_({text, PatternN}, Req, Handle) ->
    %% do stuff,
     {reply, OutFrame, Req, State}.