Take a look at this solution. It addresses a similar scenario.
When TypeScript resolves union types, it identifies the best common type to use - which is the type that is common among all elements in the union. This approach ensures safety.
Let's analyze your union:
interface FruitBox {
name: string
desc: {
'orange': number;
'banana': number;
}
}
interface IceBox {
name: string
}
interface Vegabox {
name: string
desc: {
'tomato': number;
'potato': number;
}
}
type UnionBox = FruitBox | Vegabox | IceBox;
type AllowedKeys = keyof UnionBox // name
By defining AllowedKeys, you are restricted to only using the name property from the UnionBox. This limitation exists because name is present in each part of the union.
To handle this situation, consider implementing the StrictUnion helper function.
// credits goes to https://stackoverflow.com/questions/65805600/type-union-not-checking-for-excess-properties#answer-65805753
type UnionKeys<T> = T extends T ? keyof T : never;
type StrictUnionHelper<T, TAll> =
T extends any
? T & Partial<Record<Exclude<UnionKeys<TAll>, keyof T>, never>> : never;
type StrictUnion<T> = StrictUnionHelper<T, T>
Entire code snippet:
interface FruitBox {
name: string
desc: {
'orange': number;
'banana': number;
}
}
interface IceBox {
name: string
}
interface Vegabox {
name: string
desc: {
'tomato': number;
'potato': number;
}
}
type UnionKeys<T> = T extends T ? keyof T : never;
type StrictUnionHelper<T, TAll> =
T extends any
? T & Partial<Record<Exclude<UnionKeys<TAll>, keyof T>, never>> : never;
type StrictUnion<T> = StrictUnionHelper<T, T>
type UnionBox = StrictUnion<FruitBox | Vegabox | IceBox>;
type AllowedKeys = keyof UnionBox // name
type Ship = Record<string, (description: UnionBox['desc'] | UnionBox['name']) => void>; // ok
const record: Ship = {
foo: (elem) => { }
}
Playground