在 Typescript 中解构赋值和“选择”时，类型上不存在属性

Question

I want to use the interface defined in the model as the argument of a function.

// model/interface.ts
import { Document } from 'mongoose';

export interface Post extends Document {
  location: {
    type: string;
    coordinates: number[];
  };
};

// service/post_create.ts
import { Post } from '../../model/interface/post';

const exec = async (location: Pick<Post, 'location'>) => {
  const { coordinates: [x, y] } = location;
  // -> Property 'coordinates' does not exists on type 'Pick<Post, 'location'>'
}

I've specified a type via Pick, which I've imported to use the interface as a function argument, but I can't actually find a property named "coordinates" inside the function.

What did I make a mistake?

Answer 1

Pick utility type is used to "filter" keys of the object type, so if you say

type X = Pick<Post, 'location'>
/*
It means:
X = {
  location: {
    type: string;
    coordinates: number[];
  };
}
*/

As you can see we have object type with picked properties only, in your example we picked one - location . But that means location is a property in this object and in order to access it you need to say location.location .

---

Looking at your use case, I think you want to get type of location property and not filtered object. Getting property type can be done by indexed types . Consider:

const exec = async (location: Post['location']) => {
  const { coordinates: [x, y] } = location; // works
}

Post['location'] takes type of location property of the Post type.

Answer 2

Location contains location

const exec = async (location: Pick<Post, 'location'>) => {
  const { coordinates: [x, y] } = location.location;
}