如何使用jQuery从JSON获取属性值

我使用Newtonsoft DLL从xml文件生成json。 从下面我将如何将地址详细信息放入列表中（如果示例中有更多内容）并将它们写入下拉列表我有以下有效的json（已检查onjsonlint）：

{ "?xml": { "@version": "1.0", "@encoding": "utf-8" }, "Root": { "Information": { "Error": { "ErrorNo": "0", "ErrorMsg": null }, "Address": { "Address": [ { "@AddressID": "14961943", "@Sequence": "1", "@Description": "Some Company Name, Some Building, 10 Some Street, Some County, Some City" } ] } } } } 

Answer 1

试试这个：

 var json = // that object above var addresses = json.Root.Information.Address.Address; for (var i = 0; i < addresses.length; i++) { var $option = $("").val(addresses[i]["@AddressID"]).text(addresses[i]["@Description"]); $("#mySelect").append($option); } 

示例小提琴

Answer 2

不使用jQuery的解决方案：

 var select = document.getElementById('selectID'); var addresses = json.Root.Information.Address.Address; for(var i = 0, l = addresses.length; i < l; i++) { var o = document.createElement('option'); o.value = addresses[i]['@AddressID']; o.innerHTML = addresses[i]['@Description']; select.appendChild(o); }