How to implement an interface which inherits an class in Typescript?

Question

I need to implement some count of interfaces (IUserAccess for example).

All of these interfaces inherit the IBase interface which contains an Execute method, so I want to implement IBase just one time.

Like this:

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({...});
    ...
  }

}

But IDE says:

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

How I can resolve my issue?

It seems when you extend class with interface, you loose method implementations, so you will have to implement Execute in UserAccess class. — Mamphir
– Mamphir, Commented Apr 9, 2019 at 8:12
But implement of Execute in UserAccess class is bad solution for me because I need to implement more than 20 interfaces each of them have the same Execute method and I dont want to make a copypast of Execute body for each interface implementation — hdnn
– hdnn, Commented Apr 9, 2019 at 8:23
I guess you can skip interface and just extend IBase like class UserAccess extends IBase and access Execute like this.Execute({...}) — Mamphir
– Mamphir, Commented Apr 9, 2019 at 8:28

Mamphir · Accepted Answer · 2019-04-09 08:32:41Z

2

Possible solution

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;
  }
}

answered Apr 9, 2019 at 8:32

Mamphir

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