您可以为所需的条件创建一些接口，并将它们以如下类型连接起来:

interface SolidPart {
    name: string;
    surname: string;
    action: 'add' | 'edit' | 'delete';
    id?: number;
}
interface WithId {
    action: 'edit' | 'delete';
    id: number;
}
interface WithoutId {
    action: 'add';
    id?: number;
}

export type Entity = SolidPart & (WithId | WithoutId);

const item: Entity = { // valid
    name: 'John',
    surname: 'Doe',
    action: 'add'
}
const item: Entity = { // not valid, id required for action === 'edit'
    name: 'John',
    surname: 'Doe',
    action: 'edit'
}

到目前为止还没有人提到过，但我认为，无论谁偶然看到这一页，都可以考虑使用受歧视的工会。如果我正确理解OP代码的意图，那么它可能会像这样转换。

interface Attachment {}

interface MessageBase {
    type?: string;
    user?: string;
    ts?: string;
    team?: string;
    event?: string;
    match?: [string, {index: number}, {input: string}];
}

interface AttachmentMessageNoContext extends MessageBase {
    kind: 'withAttachments',
    channel: string;
    attachments: Attachment[];
}

interface TextMessageNoContext extends MessageBase {
    kind: 'justText', 
    channel: string;
    text: string;
}

type Message = TextMessageNoContext | AttachmentMessageNoContext

const textMessage: Message = {
  kind: 'justText',
  channel: 'foo',
  text: "whats up???" 
}

const messageWithAttachment: Message = {
  kind: 'withAttachments',
  channel: 'foo',
  attachments: []
}

现在Message接口需要附件或文本，这取决于kind属性。

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}>

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);

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.