Typescript: delete multiple keys looping array

Question

I have an object

const person = {
  first: 'John',
  last: 'Doe',
  id: 1,
}

I want to delete keys from person object. So in JavaScript it works

['first', 'last'].forEach((i) => {
  delete person[i]
})

Error(From editor VSCode not compiled)

Element implicitly has an 'any' type because expression of type 'string' can't be used to index type '{ first: string; last: string; id: number; }'.
  No index signature with a parameter of type 'string' was found on type '{ first: string; last: string; id: number; }'.ts(7053)

But in TypeScript it shows error. What will be the equivalent code in TypeScript. Thanks.

What is the shown error?

MoxxiManagarm
– MoxxiManagarm

2020-09-22 12:03:13 +00:00
Commented Sep 22, 2020 at 12:03 — MoxxiManagarm
– MoxxiManagarm, Commented Sep 22, 2020 at 12:03
@MoxxiManagarm I have added error. Thanks.

ARIF MAHMUD RANA
– ARIF MAHMUD RANA

2020-09-22 12:07:16 +00:00
Commented Sep 22, 2020 at 12:07 — ARIF MAHMUD RANA
– ARIF MAHMUD RANA, Commented Sep 22, 2020 at 12:07

thedude · Accepted Answer · 2020-09-22 12:11:04Z

2

Try this:

const keysToDelete: (keyof typeof person)[] = ['first', 'last'] 

keysToDelete.forEach((i) => {
   delete person[i]
})

You need to let TS know you are only using the keys of person and not just arbitrary strings

TS playground

answered Sep 22, 2020 at 12:11

thedude

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1 Comment

ARIF MAHMUD RANA Over a year ago

I will request thedude to add a little description if possible so that it helps other e.g keyof, typeof. Why you choose this solution(assigning in variable)? Use cases of keyof, typeof. One thing I like to share assigning in variable helps editor to type hint & validate data where using array directly doesn't validate mistakenly typed key.

ARIF MAHMUD RANA · Accepted Answer · 2020-09-22 14:37:01Z

2

Also this works for me.

const person = {
  first: 'John',
  last: 'Doe',
  id: 1,
};


(['first', 'last']).forEach((i) => {
  delete person[i as keyof typeof person]
})

TypeScript Playground

But one problem here it doesn't validate keys exist or not. So if someone mistakenly types a wrong key it will ignore e.g

const person = {
  first: 'John',
  last: 'Doe',
  id: 1,
};


(['first', 'lasts']).forEach((i) => {
  delete person[i as keyof typeof person]
})

answered Sep 22, 2020 at 14:37

ARIF MAHMUD RANA

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Comments

shahin mahmud · Accepted Answer · 2020-09-22 12:13:09Z

1

const person = {
  first: 'John',
  last: 'Doe',
  id: 1,
};


(['first', 'last'] as Array<keyof typeof person>).forEach((i) => {
  delete person[i]
})

you need to typecast i of forEach as key type of person object.

answered Sep 22, 2020 at 12:13

shahin mahmud

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Comments

Romain Deneau · Accepted Answer · 2020-09-23 08:19:16Z

1

Using keyof typeof person works but there is a simpler way (since TypeScript 3.4): using const assertion:

const person = {
  first: 'John',
  last: 'Doe',
  id: 1,
};

(['first', 'last'] as const).forEach(key => delete person[key]);

Indeed, ['first', 'last'] alone has type string[] but as const its type is readonly ["first", "last"], a readonly tuple of string literals which are compatible with the keys of the person object.

We also have to pay attention to the fact that the type of the person object is changed after deletes. It's now Omit<typeof person, 'first' | 'last'> i.e. just { id: number }. We can also choose a common type { id: number; first?: string; last?: string } to indicate that first and last are optional fields.

answered Sep 23, 2020 at 8:19

Romain Deneau

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1 Comment

ARIF MAHMUD RANA Over a year ago

Thanks nice explanation 🙂️👍️🙏️

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