This is totally not necessary, but I want to know if types can be assigned dynamically to interfaces.
Suppose I have the following interface:
interface UserAuth {
username: string;
password: string;
auth0: SomeOtherAuthInterface;
}
Obviously, if auth0 is set, then there's no need for a password (maybe there is but let's just assume there's not) and vice versa. Is there a way to type that?
If you label the distinct cases with a distinct value of a literal member, then you can compose a discriminated union, in which different 'variants' of your type can be distinguished by the compiler using a type guard.
interface UserData {
kind:"basic"
username: string;
}
interface UserAuth {
kind:"authenticated",
username: string;
password: string;
auth0: "something";
}
type UserRecord = UserData | UserAuth
function (record:UserRecord){
if(record.kind==="basic"){
console.log(record.username)
//console.log(record.auth0) this would be an error
}
else if(record.kind==="authenticated"){
console.log(record.username)
console.log(record.password)
console.log(record.auth0)
}
}
See this page for more information https://basarat.gitbook.io/typescript/type-system/discriminated-unions
You could use the Omit utility type to choose which ones you need:
interface Auth {
username: string;
password: string;
auth0: SomeOtherAuthInterface;
}
type Auth0 = Omit<Auth, "password">
type BasicAuth = Omit<Auth, "auth0">