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?
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
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' }
其他回答
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)
两种Array.prototype.reduce:
const selectable = {a: null, b: null};
const v = {a: true, b: 'yes', c: 4};
const r = Object.keys(selectable).reduce((a, b) => {
return (a[b] = v[b]), a;
}, {});
console.log(r);
这个答案使用了神奇的逗号运算符: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comma_Operator
如果你想要更花哨,这个更紧凑:
const r = Object.keys(selectable).reduce((a, b) => (a[b] = v[b], a), {});
把所有这些放到一个可重用的函数中:
const getSelectable = function (selectable, original) {
return Object.keys(selectable).reduce((a, b) => (a[b] = original[b], a), {})
};
const r = getSelectable(selectable, v);
console.log(r);
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' }
你可以使用逗号操作符
const elmo = {
color: 'red',
annoying: true,
height: 'unknown',
meta: { one: '1', two: '2'}
};
const subset = ({color , height} = elmo , {color , height});
// {color: 'red', height: 'unknown'}
如果你正在使用ES6,有一种非常简洁的方法来使用解构来做到这一点。析构允许您轻松地使用扩展添加到对象,但它也允许您以相同的方式创建子集对象。
const object = {
a: 'a',
b: 'b',
c: 'c',
d: 'd',
}
// Remove "c" and "d" fields from original object:
const {c, d, ...partialObject} = object;
const subset = {c, d};
console.log(partialObject) // => { a: 'a', b: 'b'}
console.log(subset) // => { c: 'c', d: 'd'};