Javascript 如何访问JSON对象属性
我无法访问我的JSON对象属性。给定这个构造函数Javascript 如何访问JSON对象属性,javascript,json,Javascript,Json,我无法访问我的JSON对象属性。给定这个构造函数 contractorTestQuestion = function (id, question, multianswer, textanswer, order) { var cntrQuestion; this.initialize(id, question, multianswer, textanswer, order); } $.extend(contractorTestQuestion.prototype, { i
contractorTestQuestion = function (id, question, multianswer, textanswer, order) {
var cntrQuestion;
this.initialize(id, question, multianswer, textanswer, order);
}
$.extend(contractorTestQuestion.prototype, {
initialize: function (_i, _q, _m, _t, _o) {
cntrQuestion = {
"questid" : _i,
"question": _q,
"multianswer": _m,
"textanswer": _t,
"order": _o,
"answers": [],
"change": 'none'
}
},
addAnswer: function (a) {
answers.push(a);
},
getAnswer: function (idx) {
return this.answers[idx];
}
});
然后我在代码中创建它,如下所示:
var cq=新合同测试问题(qid、q、m、t、tqQuesPos)
我想做的是cq.question并能够得到存储在该字段中的值
我无法找出我做错了什么您不能通过将
contractorTestQuestion
添加到私有变量来设置其属性
设置此
上的属性,如下所示:
initialize: function (_i, _q, _m, _t, _o) {
this.questid = _i;
this.question = _q;
this.multianswer = _m;
this.textanswer = _t;
this.order = _o;
this.answers = [];
this.change = 'none';
},
此外,这与数据格式无关。我看到的是使用jQuery的extend方法吗?可能还想用jquery标记这个问题。我看到的是JSON吗?犯错误可能要删除对它的引用。