Typescript: transform required properties to optional
Typescript: transform required properties to optional
How can non-optional properties be transformed to optional ones?
Here's the code:
interface Foo
bar: any // no '?', hence this prop is required
type KeysOfFoo =
[K in keyof Foo]: any
const keysOfFoo: KeysOfFoo = // No tsc error wanted here, got: "type '' is not assignable to type KeysOfFoo"
Partial
Thanks, you gave me correct direction. I just added question mark immediately after 'key' part:
[K in keyof Foo]?: any– Nurbol Alpysbayev
Sep 3 at 3:54
[K in keyof Foo]?: any
In case it's not immediately clear from the linked document,
Partial is part of TypeScript. Additionally, there is a bunch of other really useful types.– cartant
Sep 3 at 4:10
Partial
Thanks! I got it. In my case I cannot use
Partial because my value is not Foo[K], but completely different value– Nurbol Alpysbayev
Sep 3 at 4:19
Partial
Foo[K]
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Have a look at the
Partialtype: typescriptlang.org/docs/handbook/advanced-types.html– cartant
Sep 3 at 3:51