pick<> type of array in typescript

Question

I want to construct type {name:string,value:string} from another type Todo.

Below code throws error: Property '0' does not exist on type 'Pick<Todo, "items">'.

type Todo = {
  title: string;
  items:Array<{name:string,value:string}>;
  description: string;
  completed: boolean;
}
 
type TodoItems = Pick<Todo, "items">[0]; //ERROR: Property '0' does not exist on type 'Pick<Todo, "items">'.
 
const todo: TodoItems = {
  name: "Clean room",
  value: "yes",
};

Note: Todo comes from a ts module so it's not editable.

What approach could I follow to extract {name:string,value:string} from Todo?

Mike Kokadii · Accepted Answer · 2021-10-14 14:14:47Z

4

I think you need to split Todo item into separate type as Dmitrii Zolotuhin said, it will be the rightest way. BUT if you can't do it, you can create this type

type TodoItems = Todo["items"][number]
// ...or
type TodoItems = Pick<Todo, "items">["items"][number]

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edited Oct 14, 2021 at 14:14

answered Oct 14, 2021 at 14:02

Mike Kokadii

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Comments

Tiko · Accepted Answer · 2021-10-14 19:16:19Z

2

You were almost right, try type TodoItems = Todo["items"][0];

type Todo = {
  title: string;
  items:Array<{name:string,value:string}>;
  description: string;
  completed: boolean;
}
 
type TodoItems = Todo["items"][0]; 
//...

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edited Oct 14, 2021 at 19:16

Tiko

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answered Oct 14, 2021 at 13:43

Mike

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Comments

Dmitrii Zolotuhin · Accepted Answer · 2021-10-14 13:44:56Z

1

I would do it in the other way:

interface TodoItem {
    name: string
    value: string
}

interface Todo {
    title: string
    items: TodoItem[]
    description: string
    completed: boolean
}

const todo: TodoItem = {
    name: "Clean room",
    value: "yes",
}

answered Oct 14, 2021 at 13:44

Dmitrii Zolotuhin

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1 Comment

GorvGoyl Over a year ago

thanks for alternative solution but due to some constraints I want to extract type of array from an existing type.

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