how to sort object array in TypeScript

Question

I am new in TypeScript. I might be asking a rookie question. I have the following array of objects. I would like to sort it either by age or name. This question was asked before. I tried following after reading this answer. And this. I couldn't compile it in TypeScript.

    sortData2() {
  const testData: object[] =  [
    {
      name: 'name 1',
      age: 20,
      sex: 'M',
      height: 7.5
    },
    {
      name: 'name 1',
      age: 10,
      sex: 'M',
      height: 6.5
    },
    {
      name: 'name 3',
      age: 30,
      sex: 'F',
      height: 4.5
    }
  ];

  testData.sort( (a, b) => {
      return compare(a, b);
  });

  
  function compare(a: number | string | object, b: number | string | object) {
    // compiler error Property 'age' does not exist on type 'string | number | object'.
    return (a.age < b.age ? -1 : 1);
  }
  console.log(testData);
}

How can I sort it? Thanks in advance!

Is there a reason you're declaring compare as a function instead of defining it inline ? You should also probably remove the object[] type: typescript will infer something better — Axnyff
– Axnyff, Commented Sep 27, 2020 at 13:37
a and b wouldn't be number | string | object only object. And even then, you are better off defining an interface for your objects, so you can more accurately typecheck. — VLAZ
– VLAZ, Commented Sep 27, 2020 at 13:37
Actually, I got an array of objects from another function call. I can’t change that 3rd party api call. So I can’t define any kind of interface of my own. — masiboo
– masiboo, Commented Sep 27, 2020 at 13:41
You cannot write interface Person {name: string, age: number, sex: 'F' | 'M, height: number }? — VLAZ
– VLAZ, Commented Sep 27, 2020 at 13:45
I tried to simulate my real code and issue by this simple person like interface. My real code is far more complex than this simple interface. — masiboo
– masiboo, Commented Sep 27, 2020 at 13:58

Linda Paiste · Accepted Answer · 2020-09-27 18:57:40Z

In your compare function, you are saying that a and b can be an object, string, or number; but then you are checking a.age, which doesn't make sense unless a and b are objects which have the property age. So why are you accepting so many possible types here?

Replace this:

function compare(a: number | string | object, b: number | string | object) {
    return (a.age < b.age ? -1 : 1);
}

With this:

function compareAge(a: {age: number}, b: {age: number}): number {
    return (a.age < b.age ? -1 : 1);
}

Or better yet, this, which better handles the case where a.age and b.age are equal. An array.sort compare function should return a number less than 0 if a is less than b, greater than 0 if a is greater than b, or 0 if they are equal. The actually value of the number doesn't matter -- just the sign. So you can map to -1 or 1, but it is unnecessary. A simple subtraction allows for returning positive, negative, and 0.

function compareAge(a: {age: number}, b: {age: number}): number {
    return a.age - b.age;
}

You will still get an error when calling testData.sort( compareAge ); because you have declared a very vague type with testData: object[]. You can either remove this typing entirely and Typescript will infer the proper type, or you can declare it as a more specific type, like testData: Person[]. You will be able to sort with compareAge as long as the elements of testData (or any other array) extend {age: number}.

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