typescript extend an interface as not required

问题

I have two interfaces;

interface ISuccessResponse {
    Success: boolean;
    Message: string;
}

and

interface IAppVersion extends ISuccessResponse {
    OSVersionStatus: number;
    LatestVersion: string;
}

I would like to extend ISuccessResponse interface as Not Required; I can do it as overwrite it but is there an other option?

interface IAppVersion {
    OSVersionStatus: number;
    LatestVersion: string;
    Success?: boolean;
    Message?: string;
}

I don't want to do this.

回答1:

A bit late, but Typescript 2.1 introduced the Partial<T> type which would allow what you're asking for:

interface ISuccessResponse {
    Success: boolean;
    Message: string;
}

interface IAppVersion extends Partial<ISuccessResponse> {
    OSVersionStatus: number;
    LatestVersion: string;
}

declare const version: IAppVersion;
version.Message // Type is string | undefined

回答2:

If you want Success and Message to be optional, you can do that:

interface IAppVersion {
    OSVersionStatus: number;
    LatestVersion: string;
    Success?: boolean;
    Message?: string;
}

You can't use the extends keyword to bring in the ISuccessResponse interface, but then change the contract defined in that interface (that interface says that they are required).

回答3:

Your base interface can define properties as optional:

interface ISuccessResponse {
    Success?: boolean;
    Message?: string;
}
interface IAppVersion extends ISuccessResponse {
    OSVersionStatus: number;
    LatestVersion: string;
}
class MyTestClass implements IAppVersion {
    LatestVersion: string;
    OSVersionStatus: number;
}

来源：https://stackoverflow.com/questions/29650402/typescript-extend-an-interface-as-not-required

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