Specify type from keyof type typescript

Question

I have a function

newText<T, U extends keyof T>(name: keyof T, value: T[U]){...}

If I call it with T looking like { one: Function, two: string } the types should looks like 'one' | 'two' for one and Function | string for value.

This is super close to what I want but when I specify a name I want the value to be the type of that key. So if I do newText('one', I would want to see Function as my only option for value and doing newText('two', () =>{}) should throw an error.

Is this possible in typescript 2.6?

Full use case with solution

type params = {
  util?: Function;
  utilName?: string;
  other?: string;
};

type utilParams = 'util' | 'utilName';

class field {
  name: string;
  required: boolean;
  value: any;
  params: params;
  constructor(name: string, required: boolean, value: any, params: params) {
    this.name = name;
    this.required = required;
    this.value = value;
    this.params = params;
  }
}

type coreOptional<T, U extends keyof T> = {
  required?: boolean;
  value?: T[U];
};

class test<T> {
  test<U extends keyof T>(name: keyof T, optional: coreOptional<T, U> & Partial<Pick<params, utilParams>>) {
    let onlyParams: Partial<Pick<params, utilParams>> = { ...optional };
    return new field(name, optional.required || false, optional.value, onlyParams);
  }
}

let myTest = new test<{ one: Function; two: Function }>();

myTest.test('one', { value: '' }); //works!
myTest.test('one', { value: false}); //errors like it should.

use U instead of keyof T for the parameter? eg: newText<T, U extends keyof T>(name: U, value: T[U]){...}? — CRice
– CRice, Commented Nov 13, 2017 at 23:32

basarat · Accepted Answer · 2017-11-13 23:45:26Z

1

So if I do newText('one', I would want to see Function as my only option for value and doing newText('two', () =>{}) should throw an error.

Simple:

type T = { one: Function, two: string };
declare function newText<KEY extends keyof T>(name: KEY, value: T[KEY]): any;

newText('one', () => null); // Okay
newText('two', () => null); // Error

Cheers 🍻🌹

edited Nov 13, 2017 at 23:45

answered Nov 13, 2017 at 23:38

basarat

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1 Comment

Devcon Over a year ago

Yep that is what I thought. My actual use case is more complicated because I am wrapping my class in some helper functions (I wanted to stay compatible with old usage so I couldn't change the types). doing this fixed my more complicated version. I added the full use case to the question.

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