Typescript enforce object destructuring

Question

I'm trying to write a class that uses its generic type as the object to be destructured to enforce only those fields to be used.

Is it possible to use typescript to indicate that an object must be destructured, and work with its fields?

class A<T> {
  doSomething ( { destructured object }: T) {
    // work with destructured T...
  }
}

For example, id like an object with this interface to be inserted in a database:

interface AnInterface { a: number; b: string; }

So I create this generic class

Class CRUDLService<T> {
  create( { destructured object }: T ) {
    // Insert object with the fields of T only, not any other
  }
}

So I can create a generic service, for example:

const anInterfaceService = new CRUDLService<AnInterface>();

This way I could try to ensure that whenever anInterfaceService.create is called, only the right fields are being used.

The way I'm doing it right now doesn't take advantage of typescript, instead when you create these generic classes, you need to specify an array of strings that represent the fields being extracted from the object for the operation. ie:

const createFields = ['a', 'b'];

I'm not following what the question is. Can you maybe provide two examples, one of what should work and one for what should not work? — Ingo Bürk
– Ingo Bürk, Commented Feb 19, 2021 at 14:14
A destructured object couldn't have a type <T> - because once you destructure it you have individual variables — Randy Casburn
– Randy Casburn, Commented Feb 19, 2021 at 14:17
How can you enforce an object of generic type to have specific properties? You can try to create an interface like: interface IMinimumProperties { property1: string, [key: string]: any }. This way you can enforce some properties — Rajesh Dixit
– Rajesh Dixit, Commented Feb 19, 2021 at 14:17
"Is it possible to use typescript to indicate that an object must be destructured," that's not a type. That's enforcing an arbitrary rule on the codebase that isn't within the purview of TS. Destructuring is just syntax sugar - instead of doing obj.a and obj.b you can shorten this to a and b but it's just an alternative, not something code should rely on. — VLAZ
– VLAZ, Commented Feb 19, 2021 at 14:47

Sandeep · Accepted Answer · 2021-02-19 14:27:20Z

1

You can ofcourse do that but what is T here, T is just a type without any property, so what will you destructure here. You can define T as a variant of some type and use the object destructuring,

Here is a simple example,

interface SomeProps {
  name: string;
}

class A<T extends SomeProps> {
  doSomething({ name }: T) {
    console.log(name);
  }
}

By the way, it does work if you try to destructure your T from the question. It is simply empty object, I dont see any use case here.

So, this is perfectly valid syntax as well,

class A<T> {
  doSomething({}: T) {

  }
}

edited Feb 19, 2021 at 14:27

answered Feb 19, 2021 at 14:18

Sandeep

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Comments

Julián · Accepted Answer · 2021-02-19 15:41:13Z

0

I don't think its possible to do this. Types will have to be explicitly defined for each case.

answered Feb 19, 2021 at 15:41

Julián

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