根据@Stephane LaFlèche的回答，我提出了另一个剧本版本。

JSFiddle演示

var obj = {"a":{"b":{"c":"Hello World"}},"resTest":"potato","success":"This path exists"};
checkForPathInObject = function(object,path,value) {
        var pathParts   = path.split("."),
            result      = false;
        // Check if required parameters are set; if not, return false
        if(!object || typeof object == 'undefined' || !path || typeof path != 'string')
            return false;
        /* Loop through object keys to find a way to the path or check for value
         * If the property does not exist, set result to false
         * If the property is an object, update @object
         * Otherwise, update result */
        for(var i=0;i<pathParts.length;i++){
            var currentPathPart = pathParts[i];
            if(!object.hasOwnProperty( currentPathPart )) {
                result = false;
            } else if (object[ currentPathPart ] && path == pathParts[i]) {
                result = pathParts[i];
                break;
            } else if(typeof object[ currentPathPart ] == 'object') {
                object = object[ currentPathPart ];
            } else {
                result = object[ currentPathPart ];
            }
        }
        /* */
        if(typeof value != 'undefined' && value == result)
            return true;
        return result;
};
// Uncomment the lines below to test the script
// alert( checkForPathInObject(obj,'a.b.c') ); // Results "Hello World"
// alert( checkForPathInObject(obj,'a.success') ); // Returns false
// alert( checkForPathInObject(obj,'resTest', 'potato') ); // Returns 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;
}

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。

怎么样

try {
   alert(test.level1.level2.level3)
} catch(e) {
 ...whatever

}

这是我使用的一个小助手函数，对我来说，它非常简单明了。希望这对一些人有帮助：）。

static issetFromIndices(param, indices, throwException = false) {
    var temp = param;

    try {
        if (!param) {
            throw "Parameter is null.";
        }

        if(!Array.isArray(indices)) {
            throw "Indices parameter must be an array.";
        }

        for (var i = 0; i < indices.length; i++) {
            var index = indices[i];
            if (typeof temp[index] === "undefined") {
                throw "'" + index + "' index is undefined.";
            }


            temp = temp[index];
        }
    } catch (e) {
        if (throwException) {
            throw new Error(e);
        } else {
            return false;
        }
    }

    return temp;
}

var person = {
    hobbies: {
        guitar: {
            type: "electric"
        }
    }
};

var indices = ["hobbies", "guitar", "type"];
var throwException = true;

try {
    var hobbyGuitarType = issetFromIndices(person, indices, throwException);
    console.log("Yay, found index: " + hobbyGuitarType);
} catch(e) {
    console.log(e);
}

您可以使用“.”分隔对象和路径

函数checkPathExist（obj，路径）{var pathArray=路径.split（“.”）for（pathArray的var i）{if（反射get（obj，i））{obj=obj[i]；}其他{return false；}}返回true；}var测试=｛level1：｛level2：｛level3：‘level3‘｝｝｝｝；console.log（'level.level.level3=>'，checkPathExist（测试，'level.level 2.level3'））；//真的console.log（'level.level.foo=>'，checkPathExist（测试，'level.level 2.foo'））；//假的