Typescript create an object from another object, based on an interface

Question

I want to create an ExampleInterface object from another object, but keep only those properties that ExampleInterface contains.

Is it possible without copying each key manually?

export interface ExampleInterface {
  property1: string;
  property2: string;
}

and then

const exampleObject: ExampleInterface = anotherObjectThatHasMoreProperties;

Thank u in advance.

TypeScript types don't generate runtime types, so out of the box you can't do this. Why do you need to strip extra properties at runtime? Probably the easiest solution is to make a function that manually copies the desired properties. There are also libraries designed to bring runtime type features to TS, they may help. — Aaron Beall
– Aaron Beall, Commented Jan 15, 2020 at 15:01
Possible duplicate of How can I create an object based on an interface file definition in TypeScript? — jcalz
– jcalz, Commented Jan 15, 2020 at 15:47

Rodrigo · Accepted Answer · 2020-01-15 20:57:30Z

1

A possible solution is the function above:

 function createExampleInterface(sourceObject: ExampleInterface): ExampleInterface 
 {
      const emptyExampleInterface: ExampleInterface = {
        property1: '',
        property2: ''
      };
      const interfaceProperties = Object.keys(emptyExampleInterface);
      const targetObject: ExampleInterface = Object.assign({}, sourceObject) ;

      for (let property of Object.keys(targetObject)) {    
        if (interfaceProperties.indexOf(property) < 0) {      
          delete targetObject[property];
        }
      }
      return targetObject;
 }

Example using the function:

const objA = {
  property1: 'Property 1',
  property2: 'Property 2',
  property3: 'Property 3'
}

const objB: ExampleInterface = createExampleInterface(objA);

console.log(objB);

Try it in https://stackblitz.com/edit/typescript-rjgcjp

edited Jan 15, 2020 at 20:57

answered Jan 15, 2020 at 16:02

Rodrigo

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1 Comment

Hienz Over a year ago

Although in my case I chose the 'class instead of interface and use constructor' solution, imo this answer is the correct one for my question.

dave0688 · Accepted Answer · 2020-01-15 16:08:16Z

1

I think therefore a class might be a better choice, because there you can create a constructor and give it the other object as parameter like this:

    export class ExampleDomainObject {
        constructor(obj: AnotherObjectThatHasMoreProperties) {
            this.myProp = obj.myProp;
            // here you apply all properties you need from AnotherObjectThatHasMoreProperties
        }

    }

Let me know if it helps :)

answered Jan 15, 2020 at 16:08

dave0688

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3 Comments

Hienz Over a year ago

Yes, finally I admitted that classes would fit more to my case than interface, thanks. :)

dave0688 Over a year ago

Glad it helped :)

GreenAsJade Over a year ago

Don't you still have to explicitly copy all the properties of the other object, which is what this question seeks to avoid?

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