Is there a way to utilize an interface with both optional and required properties to generate a new interface that excludes all optional properties?
For instance, if we consider the following interface:
interface FirstInterface {
name: string,
surname?: string,
}
How can we derive a new interface without any optional properties from this initial interface:
interface SecondInterface {
name: string,
}
Here is a clever solution:
type KeysRequired<T> = { [K in keyof T]-?: undefined extends T[K] ? never : K }[keyof T]
type PropertiesRequired<T> = Pick<T, KeysRequired<T>>
The solution works by creating a type that maps the keys of an interface to determine which ones are required, and then uses the Pick utility to extract only those required properties from the original interface.
You can implement this like so:
type NewInterface = PropertiesRequired<OriginalInterface>If you're considering using TypeScript, be sure to explore the collection of Utility Types
type newType = Pick<oldType, "property">
// type newType = Omit<oldType, "attribute"> serves the same purpose
By the way, remember to name interfaces in PascalCase convention
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