[SOLVED] How to create a type that requires some properties but tolerates any additional properties?

Issue

I have a util function whose first parameter param must include (simplified for example) the properties param.X and param.Y, but it doesn’t matter what other properties it has. For instance it can also have param.Z but that’s not used by the util function and can be ignored.

How can I type the param such that it enforces this constraint without complaining something like ParamType does not include property Z?

Solution

Intersect the object type with Record<PropertyKey, unknown>.

const fn = (obj: { x: string; y: string } & Record<PropertyKey, unknown>) => {
};

fn({x: 'x'}) // Fails
fn({x: 'x', y: 'y'}) // Passes
fn({x: 'x', y: 'y', z: 'z'}) // Passes

Answered By – CertainPerformance

Answer Checked By – Jay B. (BugsFixing Admin)