无法用jQuery解析JSON，返回[object Object]

Question

我怀疑我的问题是由于我的JSON字符串的结构。它似乎在JSON对象中有一个JSON对象。

这是我的JSON格式：

[ 
{"subject":{"title":"java","id":"1","desc":"Basic java programming"}, 
{"subject":{"title":"objective c","id":"2","desc":"Introduction to objective c"} 
} 

这是我的jQuery代码：

var items = []; 

    $.getJSON('theurl', function(data) { 

     $.each(data, function(key, subject) { 
      alert(subject); //returning me "[object Object]" 
      $('#tempresult').append('<p>'+ subject +'</p>'); //returning me "[object Object]" 
     }); 
    }); 

Answer 1

您已经发布了一些畸形的JSON。我要去你的元素被正确关闭的假设：

[ 
    {"subject":{"title":"java","id":"1","desc":"Basic java programming"}}, 
    {"subject":{"title":"objective c","id":"2","desc":"Introduction to objective c"}} 
] 

它看起来像你想$('#tempresult').append('<p>'+ subject.subject.desc +'</p>');

data[0]是这样的对象：

{"subject":{"title":"java","id":"1","desc":"Basic java programming"}} 

是这个对象：

{"subject":{"title":"objective c","id":"2","desc":"Introduction to objective c"}} 

data[0].subject是此目的：

{"title":"java","id":"1","desc":"Basic java programming"} 

data[0].subject.desc是这样的：

"Basic java programming" 

Answer 2

你的示例JSON是有点成形差（缺少}）。

否则，试试这个：

$.each(data, function(key, subject) { 
    alert(subject.subject.title); 
}); 

http://jsfiddle.net/7xMJP/

Answer 3

你的主题变量是一个JSON对象。 您必须指定对象的属性来获取值

alert(subject.subject.id); alert(subject.subject.title); Alert(subject.subject.desc);