如何显示JSON对象的值? [英] How do I display values of an JSON object?

问题描述

这是我到目前为止所得到的.请阅读代码中的注释.它包含我的问题.

Here is what I got so far. Please read the comment in the code. It contains my questions.

var customer;   //global variable
function getCustomerOption(ddId){
      $.getJSON("http://localhost:8080/WebApps/DDListJASON?dd="+ddId, function(opts) {
           $('>option', dd).remove(); // Remove all the previous option of the drop down
           if(opts){  
                customer = jQuery.parseJSON(opts); //Attempt to parse the JSON Object.
           }
      });
}

function getFacilityOption(){
      //How do I display the value of "customer" here. If I use alert(customer), I got null
}

这是我的json对象的外观:{"3":"Stanley Furniture","2":"Shaw","1":"First Quality"}.我最终想要的是，如果我传递密钥3，我想找回Stanley Furniture，如果我传递Stanley Furniture，我又得到了3.由于在我的数据库中，3是customerId，而Stanley Furniture是customerName.

Here is what my json object should look like: {"3":"Stanley Furniture","2":"Shaw","1":"First Quality"}. What I ultimately want is that, if I pass in key 3, I want to get Stanley Furniture back, and if I pass in Stanley Furniture, I got a 3 back. Since 3 is the customerId and Stanley Furniture is customerName in my database.

推荐答案

如果servlet 已经返回JSON(如URL所示)，则无需在jQuery的$.getJSON()函数，但只需处理作为JSON.摆脱该jQuery.parseJSON().这会使情况变得更糟. getFacilityOption()函数应该用作$.getJSON()的回调函数，或者您需要在function(opts)(实际上是当前的回调函数)中编写其逻辑.

If the servlet already returns JSON (as the URL seem to suggest), you don't need to parse it in jQuery's $.getJSON() function, but just handle it as JSON. Get rid of that jQuery.parseJSON(). It would make things potentially more worse. The getFacilityOption() function should be used as callback function of $.getJSON() or you need to write its logic in the function(opts) (which is actually the current callback function).

{"3":"Stanley Furniture","2":"Shaw","1":"First Quality"}

...按以下方式访问时将返回斯坦利家具"

...would return "Stanley Furniture" when accessed as follows

var json = {"3":"Stanley Furniture","2":"Shaw","1":"First Quality"};
alert(json['3']);
// or
var key = '3';
alert(json[key]);

要了解有关JSON的更多信息，强烈建议阅读本文.要了解有关$.getJSON的更多信息，请查看其文档.

To learn more about JSON, I strongly recommend to go through this article. To learn more about $.getJSON, check its documentation.

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