通过JavaScript迭代/解析JSON对象

Question

我遇到了jQuery/Ajax/JSON的问题.我正在使用jQuery ajax这样的帖子......

$.ajax({
  type: "POST",
  dataType: "json",
  url: "someurl.com",
  data: "cmd="+escape(me.cmd)+"&q="+q+"&"+me.args,
  success: function(objJSON){
    blah blah...
  }
});

我的理解是这将返回一个JavaScript JSON对象？ajax帖子产生的文本就是这个(我相信这是有效的JSON)...

{
  "student":{
    "id": 456,
    "full_name": "GOOBER, ANGELA",
    "user_id": "2733245678",
    "stin": "2733212346"
  },
  "student":{
    "id": 123,
    "full_name": "BOB, STEVE",
    "user_id": "abc213",
    "stin": "9040923411"
  }
}

我似乎无法弄清楚如何解析jQuery ajax帖子返回的JSON对象...基本上我想循环并让每个学生返回一个div像这样......

$("<div id=\"" + student.id + "\">" + student.full_name + " (" + student.user_id + " - " + student.stin + ")</div>")

我似乎无法弄清楚如何做到这一点......

谢谢!

Answer 1

您的JSON对象不正确,因为它具有多个具有相同名称的属性.你应该返回一组"学生"对象.

[
   {
     "id": 456,
     "full_name": "GOOBER ANGELA",
     "user_id": "2733245678",
     "stin": "2733212346"
   },
   {
     "id": 123,
     "full_name": "BOB, STEVE",
     "user_id": "abc213",
     "stin": "9040923411"
   }
]

然后你可以迭代它:

 for (var i = 0, len = objJSON.length; i < len; ++i) {
     var student = objJSON[i];
     $("<div id=\"" + student.id + "\">" + student.full_name + " (" + student.user_id + " - " + student.stin + ")</div>")...
 }