不使用变量名即可打印嵌套的JSON

一尘不染

不使用变量名即可打印嵌套的JSON

json

Web服务返回以下嵌套的json对象：

{"age":"21-24","gender":"Male","location":"San Francisco, CA","influencer score":"70-79","interests":{"Entertainment":{"Celebrities":{"Megan Fox":{},"Michael Jackson":{}},},"Social Networks & Online Communities":{"Web Personalization": {},"Journals & Personal Sites": {},},"Sports":{"Basketball":{}},},"education":"Completed Graduate School","occupation":"Professional/Technical","children":"No","household_income":"75k-100k","marital_status":"Single","home_owner_status":"Rent"}

我只想遍历此对象而不指定属性名称，我尝试了以下代码：

for (var data in json_data) {
    alert("Key:" + data + " Values:" + json_data[data]);
}

但是，如果它是嵌套值，则将值打印为[object Object]，是否有任何方法可以使迭代更深入嵌套值？

阅读 219

2020-07-27

共1个答案

一尘不染

试试这个：

function iter(obj) {
  for (var key in obj) {
    if (typeof(obj[key]) == 'object') {
      iter(obj[key]);
    } else {
      alert("Key: " + key + " Values: " + obj[key]);
    }
  }
}

BB：添加了+以防止错误。

2020-07-27