Question

我试图使用条件类型，但是它没有按预期工作。我期望abc的类型为数字，但它返回一个字符串。对此的任何帮助将不胜感激。

class TableIdClass {
    public tableId?: string;

    constructor(props: TableIdClass) {
        const { tableId } = props;
        this.tableId = `${Math.random()}`;

    }
}

export class TableBase extends TableIdClass {

    public createDate?: Date;

    constructor(props: TableBase) {
        super(props)
        const { createDate } = props;
        this.createDate = (createDate) ? createDate : new Date();
    }
}

export class EntityBase extends TableBase {
    public entityId?: string;
    public entityName?: string;
    public isActive?: boolean;

    constructor(props: EntityBase) {
        super(props)
        const { entityId, entityName, isActive } = props;
        this.entityId = entityId;
        this.entityName = (entityName) ? entityName : '';
        this.isActive = (typeof isActive === 'undefined') ? true : isActive;
    }
};

class SomeClass extends TableIdClass {

    constructor(prop: SomeClass) {
        super(prop)

    }
}

type abc = SomeClass extends EntityBase ? string : number; // Returns string.

Answer 1

您应该向EntityBase添加一些非可选属性，例如

class EntityBase {

    public isEntityBase = undefined
    ...
}

这是因为TypeScript使用结构子类型，换句话说，它通过查看对象的结构（属性名称）来检查对象是否实现了某些接口。

为了不污染EntityBase的API，可以使用Symbols：

const isEntityBase = Symbol()

class EntityBase {

    public [isEntityBase] = undefined


}

打字稿扩展名。条件类型

1 个答案: