How to define the type of a function in TypeScript that sorts an array of objects on one property?

Question

I wrote a simple function in TypeScript that sorts an array of objects based on one of their property. The code looks like:

export const mySortFunction = (sortKey: string, invert: boolean) => {
  return (a: any, b: any) => {
    if (a[sortKey] < b[sortKey]) {
      return invert ? 1 : -1;
    } else if (a[sortKey] > b[sortKey]) {
      return invert ? -1 : 1;
    }
    return 0;
  }
}

Then, if I have a type Person, with name and lastname properties, I can sort the list of persons with a call like persons.sort(mySortFunction('name', true)) or persons.sort(mySortFunction('lastname', false)).

The function is working, but I'm not really happy with the typing here. Basically, I want to have something like:

export const mySortFunction = <T>(sortKey: string, invert: boolean) => {
  return (a: T, b: T) => {
    ...
  }
}

and indicates to TypeScript that T should extends a type that has a key which matches the value of sortKey...

How can I set a good typing for my function signature?

Nikita Madeev · Accepted Answer · 2020-06-19 07:19:36Z

1

You can use keyof: Playground

export const mySortFunction = <T>(sortKey: keyof T, invert: boolean) => {
    return (a: T, b: T) => {
        if (a[sortKey] < b[sortKey]) {
            return invert ? 1 : -1;
        } else if (a[sortKey] > b[sortKey]) {
            return invert ? -1 : 1;
        }
        return 0;
    };
};

type Person = {
    name: string;
    lastname: string;
};

const persons: Person[] = [
    { name: '1', lastname: '1' },
    { name: '2', lastname: '2' },
];

persons.sort(mySortFunction('lastname', true)); // OK
persons.sort(mySortFunction('test', true)); // Argument of type '"test"' is not assignable to parameter of type '"lastname" | "name"'

answered Jun 19, 2020 at 7:19

Nikita Madeev

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1 Comment

Romain Linsolas Over a year ago

Thanks, I knew that the solution was quite simple indeed.

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