简单的“二选一”的例子:

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

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);

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
  }
})

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。

你可以更深入地使用@robstarbuck解决方案创建以下类型:

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>;

然后消息类型看起来像这样

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

使用此解决方案，您可以轻松地在MessageWithText或MessageWithAttachment类型中添加更多字段，而无需在其他类型中排除它。

还可以为通用属性使用抽象类，而不是接口，以防止某人意外实现该接口。

abstract class BaseMessage {
  timestamp?: number;
  /* more general properties here */
  constructor(timestamp?: number) {
    this.timestamp = timestamp;
    /* etc. for other general properties */
  }
}
interface IMessageWithText extends BaseMessage {
  text: string;
  attachment?: never;
}
interface IMessageWithAttachment extends BaseMessage {
  text?: never;
  attachment: string;
}
type Message = IMessageWithText | IMessageWithAttachment;