getValue (o, key1, key2, key3, key4, key5) {
    try {
      return o[key1][key2][key3][key4][key5]
    } catch (e) {
      return null
    }
}

ES6答案，经过彻底测试：）

const propExists = (obj, path) => {
    return !!path.split('.').reduce((obj, prop) => {
        return obj && obj[prop] ? obj[prop] : undefined;
    }, obj)
}

→请参阅具有完整测试覆盖范围的Codepen

我以以下方式使用函数。

var a = {};
a.b = {};
a.b.c = {};
a.b.c.d = "abcdabcd";

function isDefined(objectChainString) {
    try {
        var properties = objectChainString.split('.');
        var currentLevel = properties[0];
        if (currentLevel in window) {
            var consolidatedLevel = window[currentLevel];
            for (var i in properties) {
                if (i == 0) {
                    continue;
                } else {
                    consolidatedLevel = consolidatedLevel[properties[i]];
                }
            }
            if (typeof consolidatedLevel != 'undefined') {
                return true;
            } else {
                return false;
            }
        } else {
            return false;
        }
    } catch (e) {
        return false;
    }
}

// defined
console.log(checkUndefined("a.b.x.d"));
//undefined
console.log(checkUndefined("a.b.c.x"));
console.log(checkUndefined("a.b.x.d"));
console.log(checkUndefined("x.b.c.d"));

function isIn(string, object){
    var arr = string.split(".");
    var notFound = true;
    var length = arr.length;
    for (var i = 0; i < length; i++){
        var key = arr[i];
        if (!object.hasOwnProperty(key)){
            notFound = false;
            break;
        }
        if ((i + length) <= length){
            object = object[key];
        }
    }
    return notFound;
}
var musicCollection = {
    hasslehoff: {
        greatestHits : true
    }
};
console.log(isIn("hasslehoff.greatestHits", musicCollection));
console.log(isIn("hasslehoff.worseHits", musicCollection));

这里是我的基于字符串的分隔符版本。

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;
}

从这个答案开始，阐述了以下选项。两者的树相同：

var o = { a: { b: { c: 1 } } };

未定义时停止搜索

var u = undefined;
o.a ? o.a.b ? o.a.b.c : u : u // 1
o.x ? o.x.y ? o.x.y.z : u : u // undefined
(o = o.a) ? (o = o.b) ? o.c : u : u // 1

逐一确保每个级别

var $ = function (empty) {
    return function (node) {
        return node || empty;
    };
}({});

$($(o.a).b).c // 1
$($(o.x).y).z // undefined