Typescript: return a value depends on generic type-CodePudding

I need to make this logic work.

const defaultNumberOption = {
  label: 'Select',
  value: 0,
};

const defaultStringOption = {
  label: 'Select',
  value: '',
};

const getDefaultOption = <ValueType extends string | number>(): IOption => {
  // if ValueType is 'string' - return defaultStringOption 
  // if ValueType is 'number' - return defaultNumberOption 
}

I tried to use different approaches, but with each of them had the same problem: I can't find a way to make a structure with types in condition, but with values in expression.

If there is a way to do it - it would be great.

CodePudding user response：

enum FixedLabel {Select}

const defaultNumberOption = {
 label:  FixedLabel.Select,
 value: 0,
};

const defaultStringOption = {
   label:  FixedLabel.Select,
   value: '',
};
type DefaultNumber = { label: FixedLabel, value: number};
type DefaultString = { label: FixedLabel, value: string};
type IOption = DefaultNumber | DefaultString;

const getDefaultOption = (v : number | string): IOption => {
 // if ValueType is 'string' - return defaultStringOption 
 // if ValueType is 'number' - return defaultNumberOption 
 if(typeof v === 'string'){
     return defaultNumberOption;
  } else {
   return defaultStringOption;
 }  
}

You can try this might work for you can also use string instead of enum if you are not sepcific to labels.

CodePudding user response：

You could use conditional types to achieve this.

interface DefaultNumberOption {
    label: 'Select',
    value: number,
}

interface DefaultStringOption {
    label: 'Select',
    value: string,
}

type DefaultOption<T extends number | string> = T extends string
    ? DefaultStringOption
    : DefaultNumberOption

const getDefaultOption = <T extends number | string>(value: T): DefaultOption<T> => {
    if (typeof value === 'string') {
        return {
            label: 'Select',
            value: ''
        } as DefaultOption<T>;
    }
    return {
        label: 'Select',
        value: 0
    } as DefaultOption<T>;
}

let x = getDefaultOption(''); // x will have type DefaultStringOption
let y = getDefaultOption(0); // y will have type DefaultNumberOption

Here is a link to a TS playground if you want to play around with the solution.

CodePudding user response：

TypeScript loses the typing information provided in <T extends string | number> when building to JS, which is why it does not allow you to use that.

In order to obtain the type, you may want to provide it as a parameter in the function.

const getDefaultOption = <T extends string | number>(optionValue: T): IOption => {
    return typeof optionValue === "string" ? defaultStringOption : defaultNumberOption;
}

It is also possible to add it to the return type instead:

const getDefaultOption = <T extends string | number>(getOptionType: T): IOption<string | number> => {
    if (typeof getOptionType === "string") return defaultStringOption;
    else return defaultNumberOption;
}

Note that IOption cannot contain <T> as TS would require the return type to be string | number instead of string or number.

Here's my TS playground