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 - from the value of the parameter name- and use it in the returnType thing so I avoid typing 'test' twice in the call?

1 Reply

Sybatron•13mo ago

export const useSomething = <Data extends object>(
    func: (values?: Data) => Promise<unknown>,
    name?: string
) => {
    type returnType = { data1: Data; data2: string };
    return {
        data1:
            /*data that can satisfy Data*/
        as Data,
        data2: "test",
    } satisfies returnType;
};

export const useSomething = <Data extends object>(
    func: (values?: Data) => Promise<unknown>,
    name?: string
) => {
    type returnType = { data1: Data; data2: string };
    return {
        data1:
            /*data that can satisfy Data*/
        as Data,
        data2: "test",
    } satisfies returnType;
};

ReturnType takes only 1 argument you cant add 2 types there so just your own new type that you want to satisfy

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Typescript function parameter into returnType