Utilizing Typescript to implement an interface's properties

After declaring an interface as shown below

interface Base {
    required: string;
}

I proceeded to implement the interface in a class like this

class MyClass implements Base{
    method(): void {
        console.log(this.required);
    }
}

However, I encountered the following error message:

severity: 'Error' message: 'Class 'MyClass' incorrectly implements interface 'Base'. Property 'required' is missing in type 'MyClass'.' at: '5,7' source: 'ts'

severity: 'Error' message: 'Property 'required' does not exist on type 'MyClass'.' at: '7,26' source: 'ts'

To resolve this issue, I found that by declaring required: string; again within the class, the errors were eliminated

interface Base {
    required: string;
}

class MyClass implements Base{
 required: string;

    method(): void {
      this.required="ddd";
        console.log(this.required);
        // you can access HTMLElement
    }
}

var ss=new MyClass();
ss.method();

Answer №1

To avoid declaring requried: string twice, you can utilize a class instead of an interface for the Base and extend it rather than implement.

class Base {
    required: string;
}

class MyClass extends Base{
    method(): void {
      this.required="ddd";
        console.log(this.required);
        // you can access HTMLElement
    }
}

Try it out on the playground.

Answer №2

Understanding how interfaces function is crucial in programming. When a property is declared in an interface, it must also be defined in the implementing class. If you wish to utilize a required property without redefining it, consider creating a new class and extending it.

Answer №3

Your mistake has been pointed out correctly. In the case where your class implements an interface, it is necessary to implement all the specified properties and methods. If there are certain properties or methods that you do not wish to implement, you can label them as optional by adding a ? symbol.

interface Base {
    required: string;
    someProperty?: string; // Notice the `?` indicating optional
}

When implementing the interface, you have the flexibility to omit the someProperty if desired:

class MyClass implements Base{
required: string;

method(): void {
this.required="ddd";
console.log(this.required);
// Accessing HTMLElement is possible
}
}

Interfaces are not only for implementation purposes; they can also serve as types. For example, defining an interface:

interface Base {
required: string;
}

Allows you to create objects which adhere to that specific interface:

const obj: Base = { };

However, attempting to assign an object of type Base without providing all required properties will result in an error. Therefore, complete initialization is necessary:

const obj: Base = { required: 'Yes' };

This approach enhances code robustness and provides strong typing, even for objects that don't require a dedicated class but must conform to a particular structure.

For instance:

An interface is defined as follows:

interface Name {
name: string
}

With corresponding classes:

class Car implements Name {
name: string;
engine: string
constructor(name: string, engine: string){
this.name = name;
this.engine = engine;
}
}

class Person implements Name {
name: string;
surname: string;

constructor(name: string, surname: string){
this.name = name;
this.surname = surname;
}
}

var arr: Name = [new Car('Car', 'JZ'), new Person('Name', 'Surname')];

In this scenario, arr represents an array of type Name. Thus, accessing arr[0] and calling

.engine</code will result in an error because <code>Name
does not include an engine property. However, the existence of a name property is guaranteed for every object within the array due to the mandatory nature of the name property in the Name type.

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