Question

我需要实现一些接口（例如IUserAccess）。

所有这些接口都继承了包含执行方法的IBase接口，因此我只想实现一次IBase。

赞：

class IBase{
  Execute(data: any): void{
    console.log('Execute');
  };

interface IUserAccess extends IBase {
  CheckPassword(username: string, password: string): boolean;
}

class UserAccess implements IUserAccess{
  CheckPassword(username: string, password: string): boolean {
    console.log('CheckPassword');
    Execute({...});
    ...
  }

}

但是IDE说：

Class 'UserAccess' incorrectly implements interface 'IUserAccess'.
  Property 'Execute' is missing in type 'UserAccess' but required in type 'IUserAccess'.

如何解决我的问题？

Answer 1

可能的解决方案

class IBase{
  Execute(data: any): void{
    console.log('Execute');
  };
}

interface IUserAccess {
  CheckPassword(username: string, password: string): boolean;
}

class UserAccess extends IBase implements IUserAccess {
  CheckPassword(username: string, password: string) {
    console.log('CheckPassword');
    this.Execute({...});
    ...
    return true;
  }
}

如何实现在Typescript中继承类的接口？

1 个答案: