I'd like to write a function that takes an object parameter and captures all the remaining parameters in a variable. The goal is to allow the function to receive named parameters (as opposed to positional parameters), some of them optional, and set default values in the function. So, in pseudo code something like this:
interface IMyProps {
a: string
b?: number
}
const myfun1 = (p: {a: string, b?:number, ...rest}) => {
const {a, b = 'hello'} = p;
}
What would be the best way to achieve this in Typescript 2.0?
You can use destructuring assignment directly in the function arguments:
I think object destructing with the rest operator in TypeScript corresponds to the index signature on the type of the destructured object if you have pulled off all the explicitly named properties already. That means you would have the same restriction where the types of the rest properties would have to be at least as wide as the union of all the explicitly labeled properties. In your case, you could extend
IMyProps
with an index signature like this:since the type of
a
isstring
and the type ofb
isnumber | undefined
. You could add more stuff to the union but you couldn't make it something narrower likestring
. If that's okay with you, then the way to do destructuring would be something like this:If you inspect the types of the variables, you get something like the above.
That's probably the best you can do with typing the rest properties. Ideally you could use generics, but you can't use the rest/spread operator with generics as of now, and there is an open issue in GitHub tracking this.
Hope that helps. Good luck.