在不使用变量名的情况下打印嵌套的JSON

Question

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"}

我只是想在不指定属性名称的情况下遍历此对象,我尝试了以下代码:

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

但是如果它是一个嵌套值,它会将值打印为[object Object],是否有任何方法可以更深入地迭代嵌套值？

Answer 1

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+以防止错误.