假设我有一个对象:
elmo = {
color: 'red',
annoying: true,
height: 'unknown',
meta: { one: '1', two: '2'}
};
我想用它的属性子集创建一个新对象。
// pseudo code
subset = elmo.slice('color', 'height')
//=> { color: 'red', height: 'unknown' }
我怎样才能做到呢?
当前回答
动态解决方案
['color', 'height'].reduce((a,b) => (a[b]=elmo[b],a), {})
让子集= (obj、钥匙)= > keys.reduce ((a, b) = > ([b] = obj [b], a), {}); / /测试 让elmo = { 颜色:红色, 讨厌:没错, 高度:“未知”, Meta: {1: '1'， 2: '2'} }; Console.log(子集(elmo， ['color'， 'height']));
其他回答
值得注意的是，Zod模式在默认情况下会删除未知属性。如果您已经在使用Zod，那么它很可能适合您的开发过程。
https://github.com/colinhacks/zod
import { z } from "zod";
// muppet schema
const muppet = z.object({
color: z.string(),
annoying: z.boolean(),
height: z.string(),
meta: z.object({ one: z.string(), two: z.string() }),
});
// TypeScript type if you want it
type TMuppet = z.infer<typeof muppet>;
// elmo example
const elmo: TMuppet = {
color: "red",
annoying: true,
height: "unknown",
meta: { one: "1", two: "2" },
};
// get a subset of the schema (another schema) if you want
const subset = muppet.pick({ color: true, height: true });
// parsing removes unknown properties by default
subset.parse(elmo); // { color: 'red', height: 'unknown' }
我建议看看Lodash;它有很多实用函数。
例如，pick()就是你要找的东西:
var subset = _.pick(elmo, ['color', 'height']);
小提琴
只是另一种方式……
var elmo = {
color: 'red',
annoying: true,
height: 'unknown',
meta: { one: '1', two: '2'}
}
var subset = [elmo].map(x => ({
color: x.color,
height: x.height
}))[0]
你可以在Objects =)数组中使用这个函数
我想在这里提到一个非常好的策展:
pick-es2019.js
Object.fromEntries(
Object.entries(obj)
.filter(([key]) => ['whitelisted', 'keys'].includes(key))
);
pick-es2017.js
Object.entries(obj)
.filter(([key]) => ['whitelisted', 'keys'].includes(key))
.reduce((obj, [key, val]) => Object.assign(obj, { [key]: val }), {});
pick-es2015.js
Object.keys(obj)
.filter((key) => ['whitelisted', 'keys'].indexOf(key) >= 0)
.reduce((newObj, key) => Object.assign(newObj, { [key]: obj[key] }), {})
omit-es2019.js
Object.fromEntries(
Object.entries(obj)
.filter(([key]) => !['blacklisted', 'keys'].includes(key))
);
omit-es2017.js
Object.entries(obj)
.filter(([key]) => !['blacklisted', 'keys'].includes(key))
.reduce((obj, [key, val]) => Object.assign(obj, { [key]: val }), {});
omit-es2015.js
Object.keys(obj)
.filter((key) => ['blacklisted', 'keys'].indexOf(key) < 0)
.reduce((newObj, key) => Object.assign(newObj, { [key]: obj[key] }), {})
我发现最简单的方法，它不会创建不必要的变量，是一个函数，你可以调用，工作原理与lodash相同，如下所示:
pick(obj, keys){
return Object.assign({}, ...keys.map(key => ({ [key]: obj[key] })))
}
例如:
pick(obj, keys){
return Object.assign({}, ...keys.map(key => ({ [key]: obj[key] })))
}
const obj = {a:1, b:2, c:3, d:4}
const keys = ['a', 'c', 'f']
const picked = pick(obj,keys)
console.log(picked)
Pick = (obj, keys) => { 返回对象。分配({},…键。Map (key => ({ (例子):obj(例子) }))) ｝ Const obj = { 答:1, b: 2 c: 3, d: 4 ｝ Const keys = ['a'， 'c'， 'f'] Const selected = pick(obj, keys) console.log(选)
