这可能是一个奇怪的问题,但我很好奇是否有可能创建一个需要其中一个属性的接口。

所以,例如……

interface Message {
    text: string;
    attachment: Attachment;
    timestamp?: number;
    // ...etc
}

interface Attachment {...}

在上面的情况下,我想确保文本或附件都存在。


我现在就是这么做的。我觉得它有点啰嗦(为slack输入botkit)。

interface Message {
    type?: string;
    channel?: string;
    user?: string;
    text?: string;
    attachments?: Slack.Attachment[];
    ts?: string;
    team?: string;
    event?: string;
    match?: [string, {index: number}, {input: string}];
}

interface AttachmentMessageNoContext extends Message {
    channel: string;
    attachments: Slack.Attachment[];
}

interface TextMessageNoContext extends Message {
    channel: string;
    text: string;
}

当前回答

我在为我的案例寻找答案时偶然发现了这个线索(要么是propA,要么是prob,要么都不是)。Ryan Fujiwara的回答几乎做到了,但我已经丢失了一些支票。

我的解决方案:

interface Base {   
  baseProp: string; 
}

interface ComponentWithPropA extends Base {
  propA: string;
  propB?: never;
}

interface ComponentWithPropB extends Base {
  propB: string;
  propA?: never;
} 

interface ComponentWithoutProps extends Base {
  propA?: never;
  propB?: never;
}

type ComponentProps = ComponentWithPropA | ComponentWithPropB | ComponentWithoutProps;

这个解决方案使所有检查保持原样。也许有人会发现这很有用:)

其他回答

好吧,经过一段时间的尝试和错误,我发现答案并没有像我的用例预期的那样工作。所以为了防止其他人有同样的问题,我想我应该分享我是如何让它工作的。我的界面是这样的:

export interface MainProps {
  prop1?: string;
  prop2?: string;
  prop3: string;
}

我正在寻找的是一个类型定义,它会说我们既不能定义prop1也不能定义prop2。我们可以定义prop1而不是prop2。最后有prop2,但prop1没有。下面是我找到的解决方案。

interface MainBase {
  prop3: string;
}

interface MainWithProp1 {
  prop1: string;
}

interface MainWithProp2 {
  prop2: string;
}

export type MainProps = MainBase | (MainBase & MainWithProp1) | (MainBase & MainWithProp2);

这工作完美,除了一个警告是,当我试图引用prop1或prop2在另一个文件中,我一直得到一个属性不存在TS错误。下面是我解决这个问题的方法:

import {MainProps} from 'location/MainProps';

const namedFunction = (props: MainProps) => {
    if('prop1' in props){
      doSomethingWith(props.prop1);
    } else if ('prop2' in props){
      doSomethingWith(props.prop2);
    } else {
      // neither prop1 nor prop2 are defined
    }
 }

我只是想分享一下,因为如果我遇到了这种奇怪的事情,那么其他人可能也会遇到。

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;

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。

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'
}

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.