如果你需要访问不同的嵌套键，而不知道它在编码时(这将是微不足道的)，你可以使用数组符号访问器:

var part1name = someObject['part1']['name'];
var part2quantity = someObject['part2']['qty'];
var part3name1 =  someObject['part3'][0]['name'];

它们等价于点符号访问器，并且可能在运行时发生变化，例如:

var part = 'part1';
var property = 'name';

var part1name = someObject[part][property];

等于

var part1name = someObject['part1']['name'];

or

var part1name = someObject.part1.name;

我希望这能解决你的问题…

EDIT

我不会使用字符串来维护某种xpath查询来访问对象值。 因为你必须调用一个函数来解析查询和检索值，我会遵循另一条路径(不是:

var part1name = function(){ return this.part1.name; }
var part2quantity = function() { return this['part2']['qty']; }
var part3name1 =  function() { return this.part3[0]['name'];}

// usage: part1name.apply(someObject);

或者，如果你不习惯apply方法

var part1name = function(obj){ return obj.part1.name; }
var part2quantity = function(obj) { return obj['part2']['qty']; }
var part3name1 =  function(obj) { return obj.part3[0]['name'];}

// usage: part1name(someObject);

函数更短，更清晰，解释器会为你检查语法错误等等。

顺便说一下，我觉得在适当的时候做一个简单的任务就足够了……

如果您想要一个能够正确检测和报告路径解析的任何问题的详细信息的解决方案，我为此编写了自己的解决方案——库路径-值。

const {resolveValue} = require('path-value');

resolveValue(someObject, 'part1.name'); //=> Part 1
resolveValue(someObject, 'part2.qty'); //=> 50
resolveValue(someObject, 'part3.0.name'); //=> Part 3A

请注意，对于索引，我们使用.0，而不是[0]，因为解析后者会增加性能损失，而.0直接在JavaScript中工作，因此非常快。

然而，完整的ES5 JavaScript语法也被支持，它只需要首先被标记化:

const {resolveValue, tokenizePath} = require('path-value');

const path = tokenizePath('part3[0].name'); //=> ['part3', '0', 'name']

resolveValue(someObject, path); //=> Part 3A

以防万一，有人在2017年或以后访问这个问题，并寻找一种容易记住的方法，这里有一篇关于在JavaScript中访问嵌套对象而不被迷惑的详细博客文章

无法读取未定义错误的属性“foo”

使用数组缩减访问嵌套对象

让我们以这个例子结构为例

const user = {
    id: 101,
    email: 'jack@dev.com',
    personalInfo: {
        name: 'Jack',
        address: [{
            line1: 'westwish st',
            line2: 'washmasher',
            city: 'wallas',
            state: 'WX'
        }]
    }
}

为了能够访问嵌套数组，您可以编写自己的数组reduce util。

const getNestedObject = (nestedObj, pathArr) => {
    return pathArr.reduce((obj, key) =>
        (obj && obj[key] !== 'undefined') ? obj[key] : undefined, nestedObj);
}

// pass in your object structure as array elements
const name = getNestedObject(user, ['personalInfo', 'name']);

// to access nested array, just pass in array index as an element the path array.
const city = getNestedObject(user, ['personalInfo', 'address', 0, 'city']);
// this will return the city from the first address item.

还有一种出色的类型处理最小库类型可以为您完成所有这些。

使用typy，代码看起来像这样

const city = t(user, 'personalInfo.address[0].city').safeObject;

免责声明:我是这个软件包的作者。

我认为你的要求是:

var part1name = someObject.part1.name;
var part2quantity = someObject.part2.qty;
var part3name1 =  someObject.part3[0].name;

你可以这样问:

var part1name = someObject["part1"]["name"];
var part2quantity = someObject["part2"]["qty"];
var part3name1 =  someObject["part3"][0]["name"];

这两种方法都可行

也许这是你自找的

var partName = "part1";
var nameStr = "name";

var part1name = someObject[partName][nameStr];

终于轮到你自找麻烦了

var partName = "part1.name";

var partBits = partName.split(".");

var part1name = someObject[partBits[0]][partBits[1]];

在这里我提供了更多的方法，这些方法在很多方面看起来都更快:

选项1:Split string on。Or [Or] Or ' Or "，颠倒过来，跳过空项。

function getValue(path, origin) {
    if (origin === void 0 || origin === null) origin = self ? self : this;
    if (typeof path !== 'string') path = '' + path;
    var parts = path.split(/\[|\]|\.|'|"/g).reverse(), name; // (why reverse? because it's usually faster to pop off the end of an array)
    while (parts.length) { name=parts.pop(); if (name) origin=origin[name]; }
    return origin;
}

选项2(最快的，除了eval):低级字符扫描(没有regex/split/等等，只是一个快速字符扫描)。 注意:该命令不支持索引引用。

function getValue(path, origin) {
    if (origin === void 0 || origin === null) origin = self ? self : this;
    if (typeof path !== 'string') path = '' + path;
    var c = '', pc, i = 0, n = path.length, name = '';
    if (n) while (i<=n) ((c = path[i++]) == '.' || c == '[' || c == ']' || c == void 0) ? (name?(origin = origin[name], name = ''):(pc=='.'||pc=='['||pc==']'&&c==']'?i=n+2:void 0),pc=c) : name += c;
    if (i==n+2) throw "Invalid path: "+path;
    return origin;
} // (around 1,000,000+/- ops/sec)

选项3:(新:选项2扩展到支持引号-有点慢，但仍然很快)

function getValue(path, origin) {
    if (origin === void 0 || origin === null) origin = self ? self : this;
    if (typeof path !== 'string') path = '' + path;
    var c, pc, i = 0, n = path.length, name = '', q;
    while (i<=n)
        ((c = path[i++]) == '.' || c == '[' || c == ']' || c == "'" || c == '"' || c == void 0) ? (c==q&&path[i]==']'?q='':q?name+=c:name?(origin?origin=origin[name]:i=n+2,name='') : (pc=='['&&(c=='"'||c=="'")?q=c:pc=='.'||pc=='['||pc==']'&&c==']'||pc=='"'||pc=="'"?i=n+2:void 0), pc=c) : name += c;
    if (i==n+2 || name) throw "Invalid path: "+path;
    return origin;
}

JSPerf： http://jsperf.com/ways-to-dereference-a-delimited-property-string/3

"eval(...)" is still king though (performance wise that is). If you have property paths directly under your control, there shouldn't be any issues with using 'eval' (especially if speed is desired). If pulling property paths "over the wire" (on the line!? lol :P), then yes, use something else to be safe. Only an idiot would say to never use "eval" at all, as there ARE good reasons when to use it. Also, "It is used in Doug Crockford's JSON parser." If the input is safe, then no problems at all. Use the right tool for the right job, that's it.

受到@webjay的回答的启发: https://stackoverflow.com/a/46008856/4110122

我做了这个函数，你可以用它来获取/设置/取消设置对象中的任何值

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

        let level = 0;
        var Return_Value;
        Path.reduce((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;
    }
}

使用它:

 // Set
 Object_Manager(Obj,[Level1,Level2,Level3],New_Value, 'Set');

 // Get
 Object_Manager(Obj,[Level1,Level2,Level3],'', 'Get');

 // Unset
 Object_Manager(Obj,[Level1,Level2,Level3],'', 'Unset');