是否有方法更改*.d中定义的接口属性的类型?Ts在typescript中?
例如: x.d.ts中的接口定义为
interface A {
property: number;
}
我想在我写入的typescript文件中改变它
interface A {
property: Object;
}
甚至这个也可以
interface B extends A {
property: Object;
}
这种方法有效吗?当我试我的系统时，它不工作。只是想确认一下有没有可能?
当前回答
如果你只想修改一个现有属性的类型，而不是删除它，那么&就足够了:
// Style that accepts both number and percent(string)
type BoxStyle = {
height?: string | number,
width?: string | number,
padding?: string | number,
borderRadius?: string | number,
}
// These are both valid
const box1: BoxStyle = {height: '20%', width: '20%', padding: 0, borderRadius: 5}
const box2: BoxStyle = {height: 85, width: 85, padding: 0, borderRadius: 5}
// Override height and width to be only numbers
type BoxStyleNumeric = BoxStyle & {
height?: number,
width?: number,
}
// This is still valid
const box3: BoxStyleNumeric = {height: 85, width: 85, padding: 0, borderRadius: 5}
// This is not valid anymore
const box4: BoxStyleNumeric = {height: '20%', width: '20%', padding: 0, borderRadius: 5}
其他回答
试试这个:
type Override<T extends object, K extends { [P in keyof T]?: any }> = Omit<T, keyof K> & K;
用法:
type TransformedArticle = Override<Article, { id: string }>;
如果你只想修改一个现有属性的类型，而不是删除它，那么&就足够了:
// Style that accepts both number and percent(string)
type BoxStyle = {
height?: string | number,
width?: string | number,
padding?: string | number,
borderRadius?: string | number,
}
// These are both valid
const box1: BoxStyle = {height: '20%', width: '20%', padding: 0, borderRadius: 5}
const box2: BoxStyle = {height: 85, width: 85, padding: 0, borderRadius: 5}
// Override height and width to be only numbers
type BoxStyleNumeric = BoxStyle & {
height?: number,
width?: number,
}
// This is still valid
const box3: BoxStyleNumeric = {height: 85, width: 85, padding: 0, borderRadius: 5}
// This is not valid anymore
const box4: BoxStyleNumeric = {height: '20%', width: '20%', padding: 0, borderRadius: 5}
日期:19/3/2021。 我认为最新的typescript(4.1.2)版本支持d.ts文件中的接口覆盖。
// in test.d.ts
interface A {
a: string
}
export interface B extends A {
a: number
}
// in any ts file
import { B } from 'test.d.ts'
// this will work
const test: B = { a: 3 }
// this will not work
const test1: B = { a: "3" }
对于像我这样的懒人来说，简单的答案是:
type Overrided = Omit<YourInterface, 'overrideField'> & { overrideField: <type> };
interface Overrided extends Omit<YourInterface, 'overrideField'> {
overrideField: <type>
}
type ModifiedType = Modify<OriginalType, {
a: number;
b: number;
}>
interface ModifiedInterface extends Modify<OriginalType, {
a: number;
b: number;
}> {}
受到ZSkycat扩展的省略解决方案的启发，我想出了这个:
type Modify<T, R> =省略<T, R> & R键; //在typescript@3.5之前 type修改<T, R> =选择<T，排除<keyof T, keyof R>> & R
例子:
interface OriginalInterface {
a: string;
b: boolean;
c: number;
}
type ModifiedType = Modify<OriginalInterface , {
a: number;
b: number;
}>
// ModifiedType = { a: number; b: number; c: number; }
一步一步地:
type R0 = Omit<OriginalType, 'a' | 'b'> // { c: number; }
type R1 = R0 & {a: number, b: number } // { a: number; b: number; c: number; }
type T0 = Exclude<'a' | 'b' | 'c' , 'a' | 'b'> // 'c'
type T1 = Pick<OriginalType, T0> // { c: number; }
type T2 = T1 & {a: number, b: number } // { a: number; b: number; c: number; }
TypeScript实用工具类型
深度修改v3
interface Original {
a: {
a: string
b: { a: string }
c: string
d: string // <- keep this one
}
}
interface Overrides {
a: {
a: { a: number } // <- overwrite string with object
b: number // <- overwrite object with number
c: number // <- overwrite string with number
e: number // <- new property
}
}
type ModifiedType = ModifyDeep<Original, Overrides>
interface ModifiedInterface extends ModifyDeep<Original, Overrides> {}
const example: ModifiedType = {
a: {
a: { a: number },
b: number,
c: number,
d: string,
e: number,
}
}
在下面找到ModifyDeep。
