Given a discriminant, get type of a property for corresponding member of discriminated union

Question

I have the following discriminated union, where the property id acquires a different type based on the value of the discriminant:

type TableKind = 'administration' | 'data'

type AdministrationTableId = 'Modules' | 'Users' | 'Roles'
type DataTableId = number

type GenericIdentifier<K extends TableKind, ID> = { kind: K, id: ID }

type AdministrationTableIdentifier = GenericIdentifier<'administration', AdministrationTableId>
type DataTableIdentifier = GenericIdentifier<'data', DataTableId>

type TableIdentifier = AdministrationTableIdentifier | DataTableIdentifier

I would like to make a generic type that takes the discriminant and returns the type of id property of the corresponding union member:

type GetTableIdType<K extends TableKind> = ???

type AdminId = GetTableIdType<'administration'> // returns AdministrationTableID
type DataId = GetTableIdType<'data'> // returns DataTableId

Is this possible? How could this be achieved?

Answer 1

Just use Extract

type TableKind = 'administration' | 'data'

type AdministrationTableId = 'Modules' | 'Users' | 'Roles'
type DataTableId = number

type GenericIdentifier<K extends TableKind, ID> = { kind: K, id: ID }

type AdministrationTableIdentifier = GenericIdentifier<'administration', AdministrationTableId>

type DataTableIdentifier = GenericIdentifier<'data', DataTableId>

type TableIdentifier = AdministrationTableIdentifier | DataTableIdentifier


type GetTableIdType<K extends TableKind> = Extract<TableIdentifier, { kind: K }>['id']

type AdminId = GetTableIdType<'administration'> // returns AdministrationTableID
type DataId = GetTableIdType<'data'> // returns DataTableId

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