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;
}

今天刚刚编写了这个函数，它对嵌套对象中的属性进行了深入搜索，如果找到了，则返回该属性的值。

/**
 * Performs a deep search looking for the existence of a property in a 
 * nested object. Supports namespaced search: Passing a string with
 * a parent sub-object where the property key may exist speeds up
 * search, for instance: Say you have a nested object and you know for 
 * certain the property/literal you're looking for is within a certain
 * sub-object, you can speed the search up by passing "level2Obj.targetProp"
 * @param {object} obj Object to search
 * @param {object} key Key to search for
 * @return {*} Returns the value (if any) located at the key
 */
var getPropByKey = function( obj, key ) {
    var ret = false, ns = key.split("."),
        args = arguments,
        alen = args.length;

    // Search starting with provided namespace
    if ( ns.length > 1 ) {
        obj = (libName).getPropByKey( obj, ns[0] );
        key = ns[1];
    }

    // Look for a property in the object
    if ( key in obj ) {
        return obj[key];
    } else {
        for ( var o in obj ) {
            if ( (libName).isPlainObject( obj[o] ) ) {
                ret = (libName).getPropByKey( obj[o], key );
                if ( ret === 0 || ret === undefined || ret ) {
                    return ret;
                }
            }
        }
    }

    return false;
}

CMS给出的答案也适用于空检查的以下修改

function checkNested(obj /*, level1, level2, ... levelN*/) 
      {
             var args = Array.prototype.slice.call(arguments),
             obj = args.shift();

            for (var i = 0; i < args.length; i++) 
            {
                if (obj == null || !obj.hasOwnProperty(args[i]) ) 
                {
                    return false;
                }
                obj = obj[args[i]];
            }
            return true;
    }

答案很多，但仍然是：为什么不更简单？

获取值的es5版本为：

function value(obj, keys) {
    if (obj === undefined) return obj;
    if (keys.length === 1 && obj.hasOwnProperty(keys[0])) return obj[keys[0]];
    return value(obj[keys.shift()], keys);
}

if (value(test, ['level1', 'level2', 'level3'])) {
  // do something
}

您也可以将其与value（config，['applet'，i，'height']）一起使用||42

CMS的ES6解决方案给了我这个想法。

还有一个非常紧凑的：

function ifSet(object, path) {
  return path.split('.').reduce((obj, part) => obj && obj[part], object)
}

打电话：

let a = {b:{c:{d:{e:'found!'}}}}
ifSet(a, 'b.c.d.e') == 'found!'
ifSet(a, 'a.a.a.a.a.a') == undefined

它的性能不会很好，因为它拆分了一个字符串（但增加了调用的可读性），并遍历所有内容，即使已经很明显找不到任何内容（但提高了函数本身的可读性。

至少比get快http://jsben.ch/aAtmc

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;
}