基于这个答案，我使用ES2015提出了一个通用函数，可以解决这个问题

function validChain( object, ...keys ) {
    return keys.reduce( ( a, b ) => ( a || { } )[ b ], object ) !== undefined;
}

var test = {
  first: {
    second: {
        third: "This is not the key your are looking for"
    }
  }
}

if ( validChain( test, "first", "second", "third" ) ) {
    console.log( test.first.second.third );
}

使现代化

看起来lodash已经为您的所有嵌套属性获取需求添加了_.get。

_.get(countries, 'greece.sparta.playwright')

https://lodash.com/docs#get

上一个答案

lodash用户可能喜欢lodash.contrib，它有几种方法可以缓解这个问题。

获取路径

签名：_.getPath（obj:Object，ks:String|Array）

基于所描述的路径获取嵌套对象中任何深度处的值给出的钥匙。键可以以数组或点分隔字符串的形式给出。如果无法到达路径，则返回undefined。

var countries = {
        greece: {
            athens: {
                playwright:  "Sophocles"
            }
        }
    }
};

_.getPath(countries, "greece.athens.playwright");
// => "Sophocles"

_.getPath(countries, "greece.sparta.playwright");
// => undefined

_.getPath(countries, ["greece", "athens", "playwright"]);
// => "Sophocles"

_.getPath(countries, ["greece", "sparta", "playwright"]);
// => undefined

//Just in case is not supported or not included by your framework
//***************************************************
Array.prototype.some = function(fn, thisObj) {
  var scope = thisObj || window;
  for ( var i=0, j=this.length; i < j; ++i ) {
    if ( fn.call(scope, this[i], i, this) ) {
      return true;
    }
  }
  return false;
};
//****************************************************

function isSet (object, string) {
  if (!object) return false;
  var childs = string.split('.');
  if (childs.length > 0 ) {
    return !childs.some(function (item) {
      if (item in object) {
        object = object[item]; 
        return false;
      } else return true;
    });
  } else if (string in object) { 
    return true;
  } else return false;
}

var object = {
  data: {
    item: {
      sub_item: {
        bla: {
          here : {
            iam: true
          }
        }
      }
    }
  }
};

console.log(isSet(object,'data.item')); // true
console.log(isSet(object,'x')); // false
console.log(isSet(object,'data.sub_item')); // false
console.log(isSet(object,'data.item')); // true
console.log(isSet(object,'data.item.sub_item.bla.here.iam')); // true

您可以使用递归函数来实现这一点。即使您不知道所有嵌套的对象键名称，也可以使用此方法。

function FetchKeys(obj) {
    let objKeys = [];
    let keyValues = Object.entries(obj);
    for (let i in keyValues) {
        objKeys.push(keyValues[i][0]);
        if (typeof keyValues[i][1] == "object") {
            var keys = FetchKeys(keyValues[i][1])
            objKeys = objKeys.concat(keys);
        }
    }
    return objKeys;
}

let test = { level1: { level2: { level3: "level3" } } };
let keyToCheck = "level2";
let keys = FetchKeys(test); //Will return an array of Keys

if (keys.indexOf(keyToCheck) != -1) {
    //Key Exists logic;
}
else {
    //Key Not Found logic;
}

这有一个小模式，但在某些时候可能会让人不知所措。我建议您一次使用两个或三个嵌套。

if (!(foo.bar || {}).weep) return;
// Return if there isn't a 'foo.bar' or 'foo.bar.weep'.

正如我可能忘记提到的，你也可以进一步扩展。下面的示例显示了对嵌套foo.bar.weep.woop的检查，如果没有可用的，则返回。

if (!((foo.bar || {}).weep || {}).woop) return;
// So, return if there isn't a 'foo.bar', 'foo.bar.weep', or 'foo.bar.weep.woop'.
// More than this would be overwhelming.

另一个选项（接近这个答案）：

function resolve(root, path){
    try {
        return (new Function(
            'root', 'return root.' + path + ';'
        ))(root);
    } catch (e) {}
}

var tree = { level1: [{ key: 'value' }] };
resolve(tree, 'level1[0].key'); // "value"
resolve(tree, 'level1[1].key'); // undefined

更多信息：https://stackoverflow.com/a/18381564/1636522