我想用Jquery 获取 在线 json api 应该怎样写 谢谢大神位求有效方法

Question

远程服务器 json 是这样的。

{
"success": 1,
"datas": [
{
"issue": "762903",
"kjcode": "20",
"kjcodes": "8,5,7",
"srccode": "04,08,12,20,21,23,24,26,39,41,42,53,54,56,61,63,75,78,79,80",
"time": "2016-06-04 15:50:00"
},
{
"issue": "762902",
"kjcode": "19",
"kjcodes": "8,3,8",
"srccode": "13,23,28,29,30,35,38,41,46,49,54,55,63,65,67,68,72,73,75,76",
"time": "2016-06-04 15:45:00"
},
{
"issue": "762901",
"kjcode": "19",
"kjcodes": "9,3,7",
"srccode": "02,03,04,06,09,15,17,20,24,28,31,33,36,41,45,47,60,68,75,77",
"time": "2016-06-04 15:40:00"
},
{
"issue": "762900",
"kjcode": "7",
"kjcodes": "0,2,5",
"srccode": "01,09,12,13,17,18,19,20,27,32,44,50,52,53,55,59,60,66,69,75",
"time": "2016-06-04 15:35:00"
},
{
"issue": "762899",
"kjcode": "17",
"kjcodes": "6,8,3",
"srccode": "01,07,09,10,13,16,20,26,27,44,48,53,54,57,61,63,64,74,77,79",
"time": "2016-06-04 15:30:00"
}
]
}

这是我写的，但是获取不到。

$.ajax({ dataType: 'jsonp', type: "GET", jsonp: "callback", url: 'https://www.cha28.cn/api/xy28.json', success: function (data) { //处理data数据 // $('body').append("kjcode: " + data); alert("aa"); } });

Answer 1

要用yql转为json数据，ajax不能直接请求跨域页面。JSONP跨域访问在线代理API

 <script src="http://ajax.aspnetcdn.com/ajax/jQuery/jquery-1.4.2.min.js"></script>
<script type="text/javascript" src="http://www.coding123.net/rardownload/20130106/20130106170832648.js"></script>
<script>
    $.ajax({
        url: 'http://query.yahooapis.com/v1/public/yql',
        dataType: 'jsonp',
        data: {
            q: "select * from json where url=\"https://www.cha28.cn/api/xy28.json\"",
            format: "json"
        },
        success: function (d) {
            alert(JSON.stringify(d))//远程json数据放在query.results下
        }
    });</script>