Question

我有从PHP 5.3.10的json_encode函数返回到jQuery 1.9.1的以下JSON字符串

{
   "messsages":{
      "error":["API_CONTROLLER.INVALID_LOGIN_PASSWORD"],
      "warning":[],
      "notice":[],
      "success":[]
   },
   "result":false
}

jQuery代码：

$.ajax(targetURL, {
   type: method,
   data: params,
   dataType: "json",
   success: function(resp){
      console.log(resp); //Shows object with all vars correctly displayed

      //The for loop also outputs all keys and values correctly
      for(var i in resp){
         console.log("I: "+i);
         console.log("RESP: ");
         console.log(resp[i]);
      }

      console.log("DIRECT ACCESS");
      console.log(resp.messages); //WHY undefined?
      //console.log(resp.messages[error]); //undefined
      //console.log(resp.messages[error][0]); //undefined
      //console.log(resp.messages.error); //undefined
      //console.log(resp.messages.length); //Just trying => undefined
      console.log(resp.result); //this works!
      //console.log(resp[messages]); //Just trying => undefined
   }
});

如何访问消息？

第一个console.log和循环显示消息存在，还有messages.error，messages.notice等等。但是我怎样才能访问它们？这种jQuery错误吗？

Answer 1

你的回复的消息var中有三个's'。因此它在循环中有效，但在您尝试访问未发送的“消息”时失败。

Answer 2

首先解决拼写错误（消息）然后您可以稍微重写代码：）

请在Google Chrome中使用JavaScript Developer Tools。

{
   "messages":{
      "error":["API_CONTROLLER.INVALID_LOGIN_PASSWORD"],
      "warning":[],
      "notice":[],
      "success":[]
   },
   "result":false
}

The jQuery code:

var performLogin = function(){
 return $.getJSON(targetURL, params).done(function(response){

      // TODO - do something with response

      console.log(response.messages);
      console.log(response.messages[error]);
      console.log(response.messages[error][0]);
      console.log(response.messages.error);
      console.log(response.messages.length);
      console.log(response.result);
      console.log(response[messages]);
   });
}

虽然它在那里，但无法直接访问ajax响应属性

2 个答案: