CODEWITHSUNDEEP
typescripttypesinterface
termitavee
termitavee

Reputation: 305

How to change all interface values in typescript

If I have an function that accept a dynamic object as a param.

const fun = <IValues>(values: IValues) => // IValues = {a: '', b: ''}

I would like to use that object and a clone that has the same keys but different values:

//manage values to init valuesModified1 and valuesModified2
const originalValues: IValues = valuesModified1; // {a: 'foo', b: 'bar'}
const modifiedValues = valuesModified2; // {a: true', b: false}
return {originalValues, modifiedValues}

my problem is that modifiedValues doesn't have the correct types.

Is there a way to clone an interface but changing all it key's values?

EDIT: based on iY1NQ's answer, I just used : { [key in keyof IValues]: boolean } and errors are gone (maybe is not the best solution but it's a solution):

const fun = <IValues>(values: IValues) => // IValues = {a: '', b: ''}
    //manage values to init valuesModified1 and valuesModified2

    const originalValues: IValues = valuesModified1; // {a: 'foo', b: 'bar'}
    const modifiedValues: { [key in keyof IValues]: boolean }  = valuesModified2; // {a: true', b: false}
    return {originalValues, modifiedValues}
}

Upvotes: 0

Views: 2393

Answers (1)

iY1NQ
iY1NQ

Reputation: 2514

That changes the type of each property to a new one specified by T:

const fun = <IValues, T>(values: IValues): { [key in keyof IValues]: T } => { 
    return ...;
} 

const result = fun<{ a: string }, boolean>({ a: "a"}); // => result: { a: boolean }

Upvotes: 1

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