下面是一个非常简单的实用程序类型，我使用它来创建一个新类型，允许多个接口/类型中的一个被称为InterfacesUnion:

export type InterfacesUnion<Interfaces> = BuildUniqueInterfaces<UnionToIntersection<Interfaces>, Interfaces>;

type UnionToIntersection<U> = (U extends unknown ? (k: U) => void : never) extends (k: infer I) => void ? I : never;

export type BuildUniqueInterfaces<CompleteInterface, Interfaces> = Interfaces extends object
  ? AssignNever<CompleteInterface, Interfaces>
  : never;

type AssignNever<T, K> = K & {[B in Exclude<keyof T, keyof K>]?: never};

它可以这样使用:

type NewType = InterfacesUnion<AttachmentMessageNoContext | TextMessageNoContext>

它通过接受接口/类型的联合来运行，构建一个包含它们所有属性的单一接口，并返回相同的接口/类型的联合，其中每个接口都包含其他接口拥有而它们没有的附加属性，这些属性被设置为可选的never ([propertyName]?:永远不要)。

有例子/解释的游乐场

不使用扩展

使用这里描述的XOR: https://stackoverflow.com/a/53229567/8954109

// Create a type that requires properties `a` and `z`, and one of `b` or `c`
type Without<T, U> = { [P in Exclude<keyof T, keyof U>]?: never };
type XOR<T, U> = (T | U) extends object ? (Without<T, U> & U) | (Without<U, T> & T) : T | U;

interface Az {
  a: number;
  z: number;
}

interface B {
  b: number;
}

interface C {
  c: number;
}

type XorBC = XOR<B, C>;
type AndAzXorBC = Az & XorBC;
type MyData = AndAzXorBC;

const ok1: MyData = { a: 0, z: 1, b: 2 };
const ok2: MyData = { a: 0, z: 1, c: 2 };
const badBothBC: MyData = {
  a: 0, z: 1,
  b: 2,
  c: 3
};
const badNoBC: MyData = { a: 0, z: 1 };
const badNoZ: MyData = { a: 0, b: 2 };

对无效类型产生以下错误:

src/App.tsx:30:7 - error TS2322: Type '{ a: number; z: number; b: number; c: number; }' is not assignable to type 'AndAzXorBC'.
  Type '{ a: number; z: number; b: number; c: number; }' is not assignable to type 'Az & Without<C, B> & B'.
    Type '{ a: number; z: number; b: number; c: number; }' is not assignable to type 'Without<C, B>'.
      Types of property 'c' are incompatible.
        Type 'number' is not assignable to type 'undefined'.

30 const badBothBC: MyData = {

src/App.tsx:35:7 - error TS2322: Type '{ a: number; z: number; }' is not assignable to type 'AndAzXorBC'.
  Type '{ a: number; z: number; }' is not assignable to type 'Az & Without<C, B> & B'.
    Property 'b' is missing in type '{ a: number; z: number; }' but required in type 'B'.

35 const badNoBC: MyData = { a: 0, z: 1 };
         ~~~~~~~

  src/App.tsx:17:3
    17   b: number;
         ~
    'b' is declared here.

src/App.tsx:36:7 - error TS2322: Type '{ a: number; b: number; }' is not assignable to type 'AndAzXorBC'.
  Type '{ a: number; b: number; }' is not assignable to type 'Az & Without<C, B> & B'.
    Property 'z' is missing in type '{ a: number; b: number; }' but required in type 'Az'.

36 const badNoZ: MyData = { a: 0, b: 2 };
         ~~~~~~

  src/App.tsx:13:3
    13   z: number;
         ~
    'z' is declared here.

简单的“二选一”的例子:

type Props =
  | { factor: Factor; ratings?: never }
  | { ratings: Rating[]; factor?: never }

我进一步压缩了@Voskanyan David的解决方案，得到了我个人认为非常简洁的解决方案:

你仍然需要在某个地方定义Only<>和Either<>

type Only<T, U> = {
    [P in keyof T]: T[P];
} & {
    [P in keyof U]?: never;
};

type Either<T, U> = Only<T, U> | Only<U, T>

之后，可以在没有任何中间接口/类型的情况下定义它:

type Message {
    type?: string;
    channel?: string;
    user?: string;
    // ...etc
} & Either<{message: string}, {attachment: Attachment}>

谢谢@ryan-cavanaugh，让我找到了正确的方向。

我有一个类似的情况，但是是数组类型。在语法上有点困难，所以我把它放在这里供以后参考:

interface BaseRule {
  optionalProp?: number
}

interface RuleA extends BaseRule {
  requiredPropA: string
}

interface RuleB extends BaseRule {
  requiredPropB: string
}

type SpecialRules = Array<RuleA | RuleB>

// or

type SpecialRules = (RuleA | RuleB)[]

// or (in the strict linted project I'm in):

type SpecialRule = RuleA | RuleB
type SpecialRules = SpecialRule[]

更新:

请注意，稍后在代码中使用声明的变量时，可能仍然会收到警告。然后可以使用(变量作为类型)语法。 例子:

const myRules: SpecialRules = [
  {
    optionalProp: 123,
    requiredPropA: 'This object is of type RuleA'
  },
  {
    requiredPropB: 'This object is of type RuleB'
  }
]

myRules.map((rule) => {
  if ((rule as RuleA).requiredPropA) {
    // do stuff
  } else {
    // do other stuff
  }
})

你可以使用联合类型来做到这一点:

interface MessageBasics {
  timestamp?: number;
  /* more general properties here */
}
interface MessageWithText extends MessageBasics {
  text: string;
}
interface MessageWithAttachment extends MessageBasics {
  attachment: Attachment;
}
type Message = MessageWithText | MessageWithAttachment;

如果你想同时允许文本和附件，你可以这样写

type Message = MessageWithText | MessageWithAttachment | (MessageWithText & MessageWithAttachment);