这里有一个简单的方法来实现其中一个，但不是两个

type MenuItem =  {
  title: string;
  component: any;
  click?: never;
  icon: string;
} | {
  title: string;
  component?: never;
  click: any;
  icon: string;
}

// good
const menuItemWithComponent: MenuItem = {
  title: 'title',
  component: "my component",
  icon: "icon"
}

// good
const menuItemWithClick: MenuItem = {
  title: 'title',
  click: "my click",
  icon: "icon"
}

// compile error
const menuItemWithBoth: MenuItem = {
  title: 'title',
  click: "my click",
  component: "my click",
  icon: "icon"
}

在TypeScript 2.8中添加的Exclude类型的帮助下，提供了一种要求至少一组属性中的一个的通用方法:

type RequireAtLeastOne<T, Keys extends keyof T = keyof T> =
    Pick<T, Exclude<keyof T, Keys>> 
    & {
        [K in Keys]-?: Required<Pick<T, K>> & Partial<Pick<T, Exclude<Keys, K>>>
    }[Keys]

要求提供一个且仅提供一个的部分而非绝对的方法是:

type RequireOnlyOne<T, Keys extends keyof T = keyof T> =
    Pick<T, Exclude<keyof T, Keys>>
    & {
        [K in Keys]-?:
            Required<Pick<T, K>>
            & Partial<Record<Exclude<Keys, K>, undefined>>
    }[Keys]

下面是一个TypeScript游乐场链接，展示了两者的作用。

RequireOnlyOne的警告是TypeScript在编译时并不总是知道运行时存在的每个属性。因此，显然RequireOnlyOne不能做任何事情来防止它不知道的额外属性。我提供了一个示例，说明RequireOnlyOne如何错过playground链接末尾的内容。

使用以下示例快速概述它的工作原理:

interface MenuItem {
  title: string;
  component?: number;
  click?: number;
  icon: string;
}

type ClickOrComponent = RequireAtLeastOne<MenuItem, 'click' | 'component'>

Pick<T, Exclude<keyof T, Keys>> from RequireAtLeastOne becomes { title: string, icon: string}, which are the unchanged properties of the keys not included in 'click' | 'component' { [K in Keys]-?: Required<Pick<T, K>> & Partial<Pick<T, Exclude<Keys, K>>> }[Keys] from RequireAtLeastOne becomes { component: Required<{ component?: number }> & { click?: number }, click: Required<{ click?: number }> & { component?: number } }[Keys] Which becomes { component: { component: number, click?: number }, click: { click: number, component?: number } }['component' | 'click'] Which finally becomes {component: number, click?: number} | {click: number, component?: number} The intersection of steps 1 and 2 above { title: string, icon: string} & ({component: number, click?: number} | {click: number, component?: number}) simplifies to { title: string, icon: string, component: number, click?: number} | { title: string, icon: string, click: number, component?: number}

没有多个接口的替代方案是

export type MenuItem = {
  title: string;
  component: any;
  icon: string;
} | {
  title: string;
  click: any;
  icon: string;
};

const item: MenuItem[] = [
  { title: "", icon: "", component: {} },
  { title: "", icon: "", click: "" },
  // Shouldn't this error out because it's passing a property that is not defined
  { title: "", icon: "", click: "", component: {} },
  // Does error out :)
  { title: "", icon: "" }
];

我在如何创建一个需要设置单个属性的Partial-like中问过类似的问题

上面的内容可以简化，但它可能更容易阅读，也可能不容易

export type MenuItem = {
  title: string;
  icon: string;
} & (
 {component: any} | {click: string}
)

注意，这些都不会阻止你同时添加两者，因为TypeScript确实允许在使用AND/OR的对象上添加额外的属性，请参阅https://github.com/Microsoft/TypeScript/issues/15447

我用这个:

type RequireField<T, K extends keyof T> = T & Required<Pick<T, K>>

用法:

let a : RequireField<TypeA, "fieldA" | "fieldB">;

这使得fieldA和fieldB是必需的。

只是延伸到上面的酷答案!对于登陆这里的人，同时寻找一个需要能力的部分版本!这里有一个片段，我要采取!

PartialReq

你想要一个接口的部分，但同时需要一些字段!这是如何做到的

export type PartialReq<T, Keys extends keyof T = keyof T> =
    Pick<Partial<T>, Exclude<keyof T, Keys>>
    & {
        [K in Keys]: T[K]
    };

使用的例子

export interface CacheObj<SigType = any, ValType = any> {
    cache: Map<SigType, ValType>,
    insertionCallback: InsertionCallback<SigType, ValType> // I want this to be required
}

// ...

export class OneFlexibleCache<SigType = any, ValType = any> {
    private _cacheObj: CacheObj<SigType, ValType>;

    constructor(
        cacheObj: PartialReq<CacheObj<SigType, ValType>, 'insertionCallback'> // <-- here
                                                                           //  i used it
    ) {
        cacheObj = cacheObj || {};

        this._cacheObj = {

// ...

// _______________ usage
this._caches.set(
    cacheSignature,
    new OneFlexibleCache<InsertionSigType, InsertionValType>({
        insertionCallback // required need to be provided
    })
);

在这里你可以看到它工作得很完美

如未提供要求

更新:对于我在这里暗示的用法，一个更好的答案

我刚从医生那里找到了奥米特。

https://www.typescriptlang.org/docs/handbook/utility-types.html#omittk

我是来加的。但在此之前，我看到了一个很酷的答案。它涵盖了所有:

https://stackoverflow.com/a/48216010/7668448

看看吧!它展示了如何为所有不同版本的Typescript做这件事!为了不重复!去看看!

还有另一个解决方案:

type RequiredKeys<T, K extends keyof T> = Required<Pick<T, K>> & Omit<T, K>;

type MenuItem2 = RequiredKeys<MenuItem, "component" | "click">;