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
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] }), {})
其他回答
核心库中没有这样的内置功能,但你可以使用对象解构来实现它…
const {color, height} = sourceObject;
const newObject = {color, height};
你也可以写一个效用函数…
const cloneAndPluck = function(sourceObject, keys) {
const newObject = {};
keys.forEach((obj, key) => { newObject[key] = sourceObject[key]; });
return newObject;
};
const subset = cloneAndPluck(elmo, ["color", "height"]);
像Lodash这样的库也有_.pick()。
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'};
打印稿的解决方案:
function pick<T extends object, U extends keyof T>(
obj: T,
paths: Array<U>
): Pick<T, U> {
const ret = Object.create(null);
for (const k of paths) {
ret[k] = obj[k];
}
return ret;
}
输入信息甚至允许自动补全:
Credit to definitelytyping for U extends keyof T trick!
打印稿操场
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' }
对象数组
const aListOfObjects = [{
prop1: 50,
prop2: "Nothing",
prop3: "hello",
prop4: "What's up",
},
{
prop1: 88,
prop2: "Whatever",
prop3: "world",
prop4: "You get it",
},
]
创建一个或多个对象的子集可以通过这种方式解构对象来实现。
const sections = aListOfObjects.map(({prop1, prop2}) => ({prop1, prop2}));