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"





Have a look at the Partial type: typescriptlang.org/docs/handbook/advanced-types.html
– cartant
Sep 3 at 3:51



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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