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

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;

其他回答

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;

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'
}

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

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);

有一些很酷的Typescript选项,你可以使用https://www.typescriptlang.org/docs/handbook/utility-types.html#omittk

你的问题是:创造一个既存在“文本”又存在附件的界面。你可以这样做:

interface AllMessageProperties {
  text: string,
  attachement: string,
}

type Message = Omit<AllMessageProperties, 'text'> | Omit<AllMessageProperties, 'attachement'>;

const messageWithText : Message = {
  text: 'some text'
}

const messageWithAttachement : Message = {
  attachement: 'path-to/attachment'
}

const messageWithTextAndAttachement : Message = {
  text: 'some text',
  attachement: 'path-to/attachment'
}

// results in Typescript error
const messageWithOutTextOrAttachement : Message = {

}