我有一个这样的数据结构:

var someObject = {
    'part1' : {
        'name': 'Part 1',
        'size': '20',
        'qty' : '50'
    },
    'part2' : {
        'name': 'Part 2',
        'size': '15',
        'qty' : '60'
    },
    'part3' : [
        {
            'name': 'Part 3A',
            'size': '10',
            'qty' : '20'
        }, {
            'name': 'Part 3B',
            'size': '5',
            'qty' : '20'
        }, {
            'name': 'Part 3C',
            'size': '7.5',
            'qty' : '20'
        }
    ]
};

我想使用这些变量访问数据:

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

part1name应该用someObject.part1.name的值填充,即“Part 1”。part2quantity也是一样,它的容量是60。

有没有办法实现这与纯javascript或JQuery?


当前回答

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.

其他回答

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

使用对象扫描,这就变成了一行。然而,更重要的是,这个解决方案考虑性能:

在搜索期间遍历一次输入(即使查询了多个键) 解析只在init上发生一次(如果查询多个对象) 允许使用*进行扩展语法

// const objectScan = require('object-scan'); const someObject = { part1: { name: 'Part 1', size: '20', qty: '50' }, part2: { name: 'Part 2', size: '15', qty: '60' }, part3: [{ name: 'Part 3A', size: '10', qty: '20' }, { name: 'Part 3B', size: '5', qty: '20' }, { name: 'Part 3C', size: '7.5', qty: '20' }] }; const get = (haystack, needle) => objectScan([needle], { rtn: 'value', abort: true })(haystack); console.log(get(someObject, 'part1.name')); // => Part 1 console.log(get(someObject, 'part2.qty')); // => 60 console.log(get(someObject, 'part3[0].name')); // => Part 3A const getAll = (haystack, ...needles) => objectScan(needles, { reverse: false, rtn: 'entry', joined: true })(haystack); console.log(getAll(someObject, 'part1.name', 'part2.qty', 'part3[0].name')); /* => [ [ 'part1.name', 'Part 1' ], [ 'part2.qty', '60' ], [ 'part3[0].name', 'Part 3A' ] ] */ console.log(getAll(someObject, 'part1.*')); /* => [ [ 'part1.name', 'Part 1' ], [ 'part1.size', '20' ], [ 'part1.qty', '50' ] ] */ .as-console-wrapper {max-height: 100% !important; top: 0} <script src="https://bundle.run/object-scan@13.8.0"></script>

声明:我是object-scan的作者

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" } }

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。

这可以通过将逻辑分解为三个独立的函数来简化:

const isVal = a => a != null; // everything except undefined + null

const prop = prop => obj => {
    if (isVal(obj)) {
        const value = obj[prop];
        if (isVal(value)) return value;
        else return undefined;
    } else return undefined;
};

const path = paths => obj => {
    const pathList = typeof paths === 'string' ? paths.split('.') : paths;
    return pathList.reduce((value, key) => prop(key)(value), obj);
};

//usage:
const myObject = { foo: { bar: { baz: 'taco' } } };
const result = path('foo.bar')(myObject);
//results => { baz: 'taco' }

这个变体支持:

传递一个数组或字符串参数 在调用和执行期间处理未定义的值 独立测试每个功能 单独使用每个函数