如果其他人需要一个通用的实用程序类型来做到这一点，我提出了以下解决方案:

/**
 * Returns object T, but with T[K] overridden to type U.
 * @example
 * type MyObject = { a: number, b: string }
 * OverrideProperty<MyObject, "a", string> // returns { a: string, b: string }
 */
export type OverrideProperty<T, K extends keyof T, U> = Omit<T, K> & { [P in keyof Pick<T, K>]: U };

我需要这个，因为在我的例子中，覆盖的关键是一个泛型本身。

如果没有准备好省略，请参阅从类型中排除属性。

稍微扩展一下@zSkycat的回答，您可以创建一个泛型，它接受两种对象类型，并返回一个合并的类型，其中第二个对象类型的成员覆盖第一个对象类型的成员。

type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>
type Merge<M, N> = Omit<M, Extract<keyof M, keyof N>> & N;

interface A {
    name: string;
    color?: string;
}

// redefine name to be string | number
type B = Merge<A, {
    name: string | number;
    favorite?: boolean;
}>;

let one: A = {
    name: 'asdf',
    color: 'blue'
};

// A can become B because the types are all compatible
let two: B = one;

let three: B = {
    name: 1
};

three.name = 'Bee';
three.favorite = true;
three.color = 'green';

// B cannot become A because the type of name (string | number) isn't compatible
// with A even though the value is a string
// Error: Type {...} is not assignable to type A
let four: A = three;

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}

我使用一种方法，首先过滤字段，然后组合它们。

从类型中排除属性

interface A {
    x: string
}

export type B = Omit<A, 'x'> & { x: number };

的接口:

interface A {
    x: string
}

interface B extends Omit<A, 'x'> {
  x: number
}

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 }>;

覆盖接口的两个或多个属性的解决方案:

接口原始{ 答:字符串; b:字符串; c:字符串; ｝ interface Modified extends省略<Original， 'a' | 'b'> { 一个吗?:字符串;//设置为可选 b:布尔;//设置为布尔值 d:数量;//添加另一个属性 ｝

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