通过字符串路径访问嵌套的JavaScript对象和数组

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

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

2021-01-20 19:30:45

其他回答

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(

2021-05-19 13:10:31

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.

2014-08-21 22:12:32

现在有一个npm模块可以做到这一点:https://github.com/erictrinh/safe-access

使用示例:

var access = require('safe-access');
access(very, 'nested.property.and.array[0]');

2014-06-13 18:56:11

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

声明:我是object-scan的作者

2021-01-20 19:30:45

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。

2012-12-20 14:13:31

通过字符串路径访问嵌套的JavaScript对象和数组

推荐文章

最新文章

标签