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
补充另一种深奥的方法，这也很有效:
var obj = {a: 1, b:2, c:3}
var newobj = {a,c}=obj && {a,c}
// {a: 1, c:3}
但是你必须把道具名写两遍。
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' }
我也有同样的问题，并通过使用以下lib轻松解决了它:
object.pick
https://www.npmjs.com/package/object.pick
pick({a: 'a', b: 'b', c: 'c'}, ['a', 'b'])
//=> {a: 'a', b: 'b'}
object.omit
https://www.npmjs.com/package/object.omit
omit({a: 'a', b: 'b', c: 'c'}, ['a', 'c'])
//=> { b: 'b' }
这适用于我在Chrome控制台。有什么问题吗?
var { color, height } = elmo
var subelmo = { color, height }
console.log(subelmo) // {color: "red", height: "unknown"}
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] }), {})
补充另一种深奥的方法，这也很有效:
var obj = {a: 1, b:2, c:3}
var newobj = {a,c}=obj && {a,c}
// {a: 1, c:3}
但是你必须把道具名写两遍。
