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

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]];

这是我使用的解决方案:

function resolve(path, obj=self, separator='.') {
    var properties = Array.isArray(path) ? path : path.split(separator)
    return properties.reduce((prev, curr) => prev?.[curr], obj)
}

使用示例:

// accessing property path on global scope
resolve("document.body.style.width")
// or
resolve("style.width", document.body)

// accessing array indexes
// (someObject has been defined in the question)
resolve("part3.0.size", someObject) // returns '10'

// accessing non-existent properties
// returns undefined when intermediate properties are not defined:
resolve('properties.that.do.not.exist', {hello:'world'})

// accessing properties with unusual keys by changing the separator
var obj = { object: { 'a.property.name.with.periods': 42 } }
resolve('object->a.property.name.with.periods', obj, '->') // returns 42

// accessing properties with unusual keys by passing a property name array
resolve(['object', 'a.property.name.with.periods'], obj) // returns 42

限制:

不能将方括号([])用于数组下标-尽管在分隔符令牌之间指定数组下标(例如，.)可以正常工作，如上所示。

// (IE9+) Two steps var pathString = "[0]['property'].others[3].next['final']"; var obj = [{ property: { others: [1, 2, 3, { next: { final: "SUCCESS" } }] } }]; // Turn string to path array var pathArray = pathString .replace(/\[["']?([\w]+)["']?\]/g,".$1") .split(".") .splice(1); // Add object prototype method Object.prototype.path = function (path) { try { return [this].concat(path).reduce(function (f, l) { return f[l]; }); } catch (e) { console.error(e); } }; // usage console.log(obj.path(pathArray)); console.log(obj.path([0,"doesNotExist"]));

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

我的解决方案是基于@AdrianoSpadoni给出的，并解决了克隆对象的需要

function generateData(object: any, path: string, value: any): object {
  const clone = JSON.parse(JSON.stringify(object));
  path
    .split(".")
    .reduce(
    (o, p, i) => (o[p] = path.split(".").length === ++i ? value : o[p] || {}),
  clone
);
  return clone;
}