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

您发布了一些格式错误的JSON。我假设你的元素被正确关闭为：

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

看起来你想要$('#tempresult').append('<p>'+ subject.subject.desc +'</p>');

data[0]就是这个对象：

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

data[1]就是这个对象：

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

data[0].subject就是这个对象：

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

data[0].subject.desc就是这样：

"Basic java programming"

Answer 2

您的示例JSON有点糟糕（缺少}）。

否则，试试这个：

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

http://jsfiddle.net/7xMJP/

Answer 3

您的主题变量是json对象。您必须指定主题属性才能获得值

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

无法使用jQuery解析JSON，返回[object Object]

3 个答案: