Dynamic type using a string variable as a property name

Question

I would like to be able to construct a type dynamically with a property name which is one of the specified parameters. While I can construct the actual object, I cannot seem to construct the actual type. I would like to use this type for composition alter

export function mapProp<AssignedType>(value: AssignedType, propertyName: string) {

  type ReturnType = {
    [propertyName]: value
  }; // errors on this line

  return {
    [propertyName]: value
  };
}

The error emitted is as follows:

A computed property name in a type literal must refer to an expression whose type is a literal type or a 'unique symbol' type.

Example playground: http://www.typescriptlang.org/play/#src=%0D%0Aexport%20function%20mapProp%3CAssignedType%3E(value%3A%20AssignedType%2C%20propertyName%3A%20string)%20%7B%0D%0A%0D%0A%20%20type%20ReturnType%20%3D%20%7B%0D%0A%20%20%20%20%5BpropertyName%5D%3A%20value%0D%0A%20%20%7D%3B%0D%0A%0D%0A%20%20return%20%7B%0D%0A%20%20%20%20%5BpropertyName%5D%3A%20value%0D%0A%20%20%7D%3B%0D%0A%7D

jcalz · Accepted Answer · 2018-07-14 00:27:48Z

I think the closest you're going to get is something like this:

export function mapProp<PropertyName extends string, AssignedType>(
  value: AssignedType, 
  propertyName: PropertyName
) {

  type ReturnType = {
    [K in PropertyName]: AssignedType
  }; 
  // equivalent to Record<PropertyName, AssignedType>

  return {
    [propertyName]: value
  } as ReturnType;

}

In this case you'd be using a mapped type instead of a type with an index signature. The addition of the PropertyName generic type parameter allows the narrowing of the key past string, if you pass it a string literal:

const thing = mapProp(123, "abc");
thing.abc; // number
thing.def; // error

In that case ReturnType is known to be equivalent to {abc: number}. If all you know is that the key is a string at compile time, then you get this:

declare const propName: string;
const stuff = mapProp(123, propName);
stuff.abc; // number
stuff.def; // number

Now ReturnType is equivalent to {[k: string]: number}, meaning it accepts any string key (and gives it a number value). This might not be what you want, but it's the best the compiler can do in this case.

Also note that without using a type assertion (as ReturnType), computed properties usually end up as string indexes instead of something more specific. This is currently a design limitation of TypeScript. There have been some attempts to deal with this, but nothing has made it into the language yet.

Hope that helps; good luck!

andzep · Accepted Answer · 2023-06-06 08:11:02Z

In the meantime, there are more elegant and simple ways to achieve this (see https://www.typescriptlang.org/docs/handbook/2/mapped-types.html).

So instead of using a function to return the type, we can now declare types like this:

export type MappedType<PropertyName extends string, AssignedType> = {
    [K in PropertyName]: AssignedType
};

let mapped: MappedType<'foo', boolean>;
mapped = {foo: true};  // OK
mapped = {foo: 123}; // error: Type 'number' is not assignable to type 'boolean'
mapped = {foo: true, bar: 123}; // error: 'bar' does not exist in type ...

// You can also define multiple type-properties and use them like variables
type MultipleTypes = [
    {prop: 'p_bool'; value: boolean},
    {prop: 'p_str'; value: 'abc'}
];

type FirstType = MultipleTypes[0];
let m: MappedType<FirstType['prop'], FirstType['value']>;
m = {p_bool: true}; // OK
m = {p_bool: 123};  // errror, wrong type

Or you can even create properties dynamically based on another string:

type Langs = 'en' | 'de';
export type TranslationsType<PropertyName extends string> = {
    [K in `${PropertyName}_${Langs}`]: string
};

const test: TranslationsType<'bla'> = {
    bla_de: 'Hallo',
    bla_en: 'Hello', // if left out, test would have error: Property 'bla_en' is missing in type... 
    bla_es: 'Hola',  // error 'bla_es' does not exist ....
};

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