Make some properties optional in a TypeScript type

Question

If I have a type with all required properties, how can I define another type with the same properties where some of it's properties are still required but the rest are optional?

For example I want a new type derived from SomeType where prop1 and prop2 are required but the rest are optional:

interface SomeType {
    prop1;
    prop2;
    prop3;
    ...
    propn;
}

interface NewType {
    prop1;
    prop2;
    prop3?;
    ...
    propn?;
}

score 58 · Accepted Answer · answered Oct 08 '18 at 13:29

58

You can use combination of Partial and Pick to make all properties partial and then pick only some that are required:

interface SomeType {
    prop1: string;
    prop2: string;
    prop3: string;
    propn: string;
}

type OptionalExceptFor<T, TRequired extends keyof T> = Partial<T> & Pick<T, TRequired>

type NewType = OptionalExceptFor<SomeType, 'prop1' | 'prop2'>

let o : NewType = {
    prop1: "",
    prop2: ""
}

answered Oct 08 '18 at 13:29

Titian Cernicova-Dragomir

230,986
31
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357

11

For completeness: `type RequiredExceptFor = Pick> & Partial;` with `type Diff = T extends U ? never : T;` – toancaro Mar 11 '19 at 16:01
1

There's a catch here. If the source properties are optional, say prop1 and prop2, that won't trigger any error. Wrapping Pick in Required shold do the job. `type OptionalExceptFor = Partial & Required>;` – AlexanderD Sep 22 '21 at 11:26
1

Thanks @AlexanderD. But for this one, I want only to mark optional what properties I want, so the rest remains the same. – Charles-Lévi BRIANI Dec 16 '22 at 13:45

Charles-Lévi BRIANI · Answer 2 · 2020-04-22T00:40:45.090

5

For tetter performance, in addition to Titian's answer, you must do :

type OptionalExceptFor<T, TRequired extends keyof T = keyof T> = Partial<
  Pick<T, Exclude<keyof T, TRequired>>
> &
  Required<Pick<T, TRequired>>;

edited Apr 22 '20 at 00:40

answered Apr 22 '20 at 00:35

Charles-Lévi BRIANI

316
2
9

this works for me when I tried to use the type in a spread operator into a fully required version, there Titian's above did not. – JBCP Sep 10 '22 at 05:06
In what situations would this affect performance? Would it be compile times for typescript? Or handling big objects in editor? As I guess it has no impact on js execution? – Cwista Dec 09 '22 at 19:48

score 3 · Answer 3 · answered Aug 27 '21 at 12:30

3

type onlyOneRequired = Partial<Tag> & {id: string}

In the above type, all properties of Tag are optional instead of id which is required. So you can use this method and specify the required properties.

answered Aug 27 '21 at 12:30

M Fuat

1,330
3
12
24

This solution does not work : https://www.typescriptlang.org/play?#code/JYOwLgpgTgZghgYwgAgLIQM4bgcxQbwChkTlgATALmRAFcBbAI2gBpjT7Ns9r9kEA9uAjhqGMFFA4WyKBAA2cSOQCSVZOMkgcAbQC6yAD41a8+cgC+bUsji0wACwFQxEqddII5S4EIAiShDUAZBsFoSE8hBgyOTAGAAOigCeEOQA+pxYuJjUAIJQUHDJADyEAJAACnBQYMBw8iXo2XgAfMgAZMhE5eUU1HRM0BXldo7Orlo4I14QPv6BwYEjWdxB3SPlgsKiGm7aI+Hl4a0RcYkpaZlcORjIALzI+oTnScVXq7cPTxU95Z88br8ISQXYAIgA7g5AgA3aBgyxsXpjJwuZBggBSAgcIGQfgEEDBSK23jqC0gAwgELxgQAFABKJFWCoAOjZtNelwyAMwAH5YvE3qluTc8HdqPpGYQ9BFthgBFEWfIBDgOYKudcWpgdAAGPQsnksuSKZRqelAA – cassepipe Apr 20 '23 at 13:30
My bad, it does not work with *nested* types – cassepipe Apr 20 '23 at 15:17
@cassepipe Sorry I was on a holiday I hope the answer helped you :) – M Fuat Apr 30 '23 at 14:56
@M Fuat It did thanks. Actually I believe none of the other answers can work with *nested* types anyways. Your answer should be the accepted one as it is the most readable in my opinion – cassepipe Apr 30 '23 at 15:32

score 2 · Answer 4 · answered Jun 16 '22 at 12:04

From this article:

We can derive the following type:

export type PartialPick<T, F extends keyof T> = Omit<T, F> & Partial<Pick<T, F>>;

F - the fields that should be made optional

T - the type

In your example:

type NewType = PartialPick<SomeType, 'prop3' | ... | 'propn'>

If you'd like to make some properties required instead of the partial use:

type RequiredPick<T, F extends keyof T> = Omit<T, F> & Required<Pick<T, F>>;

If NewType should be an interface and has additional fields besides the fields in SomeType, you can use extends from the type:

interface SomeType {
    prop1;
    prop2;
    prop3;
    prop4;
}
interface NewType extends PartialPick<SomeType, 'prop3' | 'prop4'> {
    prop5;
}

Make some properties optional in a TypeScript type

4 Answers4

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