2 replies

Typescript function parameter into returnType

Here's my function:

export const useSomething = <Data extends object>(
    func: (values?: Data) => Promise<any>,
    name?: string
)

export const useSomething = <Data extends object>(
    func: (values?: Data) => Promise<any>,
    name?: string
)

It returns as below

return {} as returnType<Data,Name>

return {} as returnType<Data,Name>

I want to use the function in the below syntax:

useSomething<SomeInterface>(TestFunction, 'test')

useSomething<SomeInterface>(TestFunction, 'test')

and not

useSomething<SomeInterface,'test'>(TestFunction, 'test')

useSomething<SomeInterface,'test'>(TestFunction, 'test')

How can I create the type

Name

Name

- from the value of the parameter

name

name

- and use it in the

returnType

returnType

thing so I avoid typing

'test'

'test'

twice in the call?