在Javascript中获取深层对象的所有键

Question

我有以下对象:

var abc = {
    1: "Raggruppamento a 1",
    2: "Raggruppamento a 2",
    3: "Raggruppamento a 3",
    4: "Raggruppamento a 4",
    count: '3',
    counter: {
        count: '3',
    },
    5: {
        test: "Raggruppamento a 1",

        tester: {
            name: "Ross"
        }
    }
};

我想检索以下结果:

• ABC [1]
• ABC [2]
• ABC [3]
• ABC [4]
• abc.count
• abc.counter.count
• ABC [5]
• ABC [5].测试
• ABC [5] .tester
• ABC [5] .tester.name

可能在插件的帮助下使用nodejs吗？

Answer 1

您可以通过递归遍历对象来执行此操作:

function getDeepKeys(obj) {
    var keys = [];
    for(var key in obj) {
        keys.push(key);
        if(typeof obj[key] === "object") {
            var subkeys = getDeepKeys(obj[key]);
            keys = keys.concat(subkeys.map(function(subkey) {
                return key + "." + subkey;
            }));
        }
    }
    return keys;
}

getDeepKeys(abc)在问题中运行对象将返回以下数组:

["1", "2", "3", "4", "5", "5.test", "5.tester", "5.tester.name", "count", "counter", "counter.count"]

Answer 2

我知道这个帖子有点旧了...

此代码涵盖了 JSON 对象格式的所有条件，例如对象、对象数组、嵌套数组对象、带有数组对象的嵌套对象等。

getDeepKeys = function (obj) {
  var keys = [];
    for(var key in obj) {
        if(typeof obj[key] === "object" && !Array.isArray(obj[key])) {
            var subkeys = getDeepKeys(obj[key]);
            keys = keys.concat(subkeys.map(function(subkey) {
                return key + "." + subkey;
            }));
        } else if(Array.isArray(obj[key])) {
            for(var i=0;i<obj[key].length;i++){
               var subkeys = getDeepKeys(obj[key][i]);
               keys = keys.concat(subkeys.map(function(subkey) {
                return key + "[" + i + "]" + "." + subkey;
               }));
            }
        } else {
          keys.push(key);
        }
    }
    return keys;
}

Answer 3

较小的版本，没有副作用，函数体中只有 1 行：

function objectDeepKeys(obj){
  return Object.keys(obj).filter(key => obj[key] instanceof Object).map(key => objectDeepKeys(obj[key]).map(k => `${key}.${k}`)).reduce((x, y) => x.concat(y), Object.keys(obj))
}

function objectDeepKeys(obj){
  return Object.keys(obj).filter(key => obj[key] instanceof Object).map(key => objectDeepKeys(obj[key]).map(k => `${key}.${k}`)).reduce((x, y) => x.concat(y), Object.keys(obj))
}