Is it possible to make a property required in a type based on the presence of another property?
Here's an example:
type Parent = {
children?: Child[];
childrenIdSequence: string[]; // This property should only be required when `children` is provided
}
To achieve your desired outcome, you can utilize a union type like this:
type Family = {
members: Member[],
memberIds: string[]
} | {
members: undefined
}
With this setup, a Family can either consist of a list of members and corresponding memberIds, or the members property can simply be undefined, in which case there may or may not be a list of memberIds. By checking the members property, you can effectively narrow down the type:
function validate(f: Family): void {
if(f.members) {
// f: { members: Member[], memberIds: string[] } - all good
console.log(f.memberIds);
} else {
// f: { members: undefined } - type error
console.log(f.memberIds);
}
}
One limitation to this approach is that the members property must be present, even when set to undefined. You'll need to explicitly define it as { members: undefined } rather than simply {}, as that's required for it to conform to the Family type.
Trying to declare the type with members?: undefined as an optional property won't work, as it disrupts the union structure and leads to the type getting needlessly narrowed to f: never in the function.
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