javascript - 如何访问 JSON 对象属性

Question

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, {
    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 = new ContractorTestQuestion(qid, q, m, t, tqQuesPos);

我想做的是 cq.question 并能够获取存储在该字段中的值。

我不知道我做错了什么

Answer 1

您不能仅通过将 contractorTestQuestion 添加到私有(private)变量来设置它的属性。

设置 this 的属性，例如:

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JSON无关。这是一种数据格式。