把form表单中的元素值封装成json function toJSON(form)//form是要进行封装的form表单对象,dom对象,可以通过document.mainForm获得,mainForm是你的表单id { var objJSON = {}; for(var i = 0; i < form.elements.length; i++) { var element = form.elements[i]; if(element.type == 'text' || element.type
1.一种为使用eval()函数 var jsonObj=eval("("+data+")"); 2.使用Function对象来进行返回解析 var jsonstr='{"name":"CJ","age":18}'; var jsonObj =(new Function("","return "+jsonstr))(); 3.JSON.parse()方法
$.fn.serializeObject = function () { var o = {}; var a = this.serializeArray(); $.each(a, function () { if (o[this.name]) { if (!o[this.name].push) { o[this.name] = [o[this.name]]; } o[this.name].push(this.value || ''); } else { o[this.name] = this.v
使用eval函数来解析 <script> var data="{root: [{name:'1',value:'0'},{name:'6101',value:'北京市'},{name:'6102',value:'天津市'},{name:'6103',value:'上海市'},{name:'6104',value:'重庆市'},{name:'6105',value:'渭南市'}, {name:'6110',value:'商洛市'} ]}"; data = eval('('+d
1 对象 public class Person { public string Name { get; set; } public int Age { get; set; } public DateTime Birth { get; set; } public string BirthDay { get; set; } } 2 封装成json格式 Person person = new Person(); person.Name = "张小晓"; person.Age = ; per