mohamed Hamouday' Answer的扩展将填补缺失的关键

function Object_Manager(obj, Path, value, Action, strict) 
{
    try
    {
        if(Array.isArray(Path) == false)
        {
            Path = [Path];
        }

        let level = 0;
        var Return_Value;
        Path.reduce((a, b)=>{
            console.log(level,':',a, '|||',b)
            if (!strict){
              if (!(b in a)) a[b] = {}
            }


            level++;
            if (level === Path.length)
            {
                if(Action === 'Set')
                {
                    a[b] = value;
                    return value;
                }
                else if(Action === 'Get')
                {
                    Return_Value = a[b];
                }
                else if(Action === 'Unset')
                {
                    delete a[b];
                }
            } 
            else 
            {
                return a[b];
            }
        }, obj);
        return Return_Value;
    }

    catch(err)
    {
        console.error(err);
        return obj;
    }
}

例子

obja = {
  "a": {
    "b":"nom"
  }
}

// Set
path = "c.b" // Path does not exist
Object_Manager(obja,path.split('.'), 'test_new_val', 'Set', false);

// Expected Output: Object { a: Object { b: "nom" }, c: Object { b: "test_new_value" } }

/**
 * Access a deep value inside a object 
 * Works by passing a path like "foo.bar", also works with nested arrays like "foo[0][1].baz"
 * @author Victor B. https://gist.github.com/victornpb/4c7882c1b9d36292308e
 * Unit tests: http://jsfiddle.net/Victornpb/0u1qygrh/
 */
function getDeepVal(obj, path) {
    if (typeof obj === "undefined" || obj === null) return;
    path = path.split(/[\.\[\]\"\']{1,2}/);
    for (var i = 0, l = path.length; i < l; i++) {
        if (path[i] === "") continue;
        obj = obj[path[i]];
        if (typeof obj === "undefined" || obj === null) return;
    }
    return obj;
}

适用于

getDeepVal(obj,'foo.bar')
getDeepVal(obj,'foo.1.bar')
getDeepVal(obj,'foo[0].baz')
getDeepVal(obj,'foo[1][2]')
getDeepVal(obj,"foo['bar'].baz")
getDeepVal(obj,"foo['bar']['baz']")
getDeepVal(obj,"foo.bar.0.baz[1]['2']['w'].aaa[\"f\"].bb")

AngularJS

Speigg的方法非常简洁，尽管我在搜索通过字符串路径访问AngularJS $scope属性的解决方案时发现了这个回复，并进行了一些修改:

$scope.resolve = function( path, obj ) {
    return path.split('.').reduce( function( prev, curr ) {
        return prev[curr];
    }, obj || this );
}

只需要把这个函数放在你的根控制器中，并像这样使用它的任何子范围:

$scope.resolve( 'path.to.any.object.in.scope')

使用eval:

var part1name = eval("someObject.part1.name");

换行以在错误时返回undefined

function path(obj, path) {
    try {
        return eval("obj." + path);
    } catch(e) {
        return undefined;
    }
}

http://jsfiddle.net/shanimal/b3xTw/

在使用评估的权力时，请使用常识和谨慎。它有点像光剑，如果你打开它，有90%的几率你会切断一个肢体。并不是每个人都适合。

而不是尝试模拟JS语法，你将不得不花费大量的计算解析，或者只是错误/忘记一些事情，比如一堆这些答案(带.s的键，有人吗?)，只是使用一个键数组。

var part1name     = Object.get(someObject, ['part1', 'name']);
var part2quantity = Object.get(someObject, ['part2', 'qty']);
var part3name1    = Object.get(someObject, ['part3', 0, 'name']);

如果您需要使用单个字符串，只需JSONify它。 此方法的另一个改进是您可以删除/设置根级对象。

function resolve(obj, path) {
    let root = obj = [obj];
    path = [0, ...path];
    while (path.length > 1)
        obj = obj[path.shift()];
    return [obj, path[0], root];
}
Object.get = (obj, path) => {
    let [parent, key] = resolve(obj, path);
    return parent[key];
};
Object.del = (obj, path) => {
    let [parent, key, root] = resolve(obj, path);
    delete parent[key];
    return root[0];
};
Object.set = (obj, path, value) => {
    let [parent, key, root] = resolve(obj, path);
    parent[key] = value;
    return root[0];
};

其他功能演示:

对于.set(/.del()的bob =是不必要的，除非你的路径可能是空的(操作根对象)。 我证明我不克隆的对象使用史蒂夫保持原始的引用和检查bob == steve //true后，第一个.set(

简单的函数，允许字符串或数组路径。

function get(obj, path) {
  if(typeof path === 'string') path = path.split('.');

  if(path.length === 0) return obj;
  return get(obj[path[0]], path.slice(1));
}

const obj = {a: {b: {c: 'foo'}}};

console.log(get(obj, 'a.b.c')); //foo

OR

console.log(get(obj, ['a', 'b', 'c'])); //foo