Consider the following interfaces:
export interface FooInterface<T, O extends FooOptions<T>> {
items: T[];
options: O;
}
interface FooOptions<T> {
doSomething: (e: T) => void;
}
Now whenever i need to do something with this I need to repeat T. For example:
const t: FooInterface<string, FooOptions<string>> = {
items: ["a","b","c"],
options: {doSomething: (e) => {
console.log(e);
}}
}
Now, repeating is boring, is it possible to create a type or interface where T should be string for both types, or redefine the interfaces somehow, so that I write something like this instead:
const t: FooInterfaceString
//or
const t: FooInterfaceOfType<string>
how about :
type FooInterfaceString = FooInterface<string, FooOptions<string>>;
also, maybe consider this if it fits your design:
export interface FooInterface<T> {
items: T[];
options: FooOptions<T>;
}
type FooInterfaceString = FooInterface<string>;
The other answers here are correct, but I want to add that you might consider using generic parameter defaults if you are usually (but not always) going to just pass in FooOptions<T>
as the type of O
:
interface FooOptions<T> {
doSomething: (e: T) => void;
}
// note how O has a default value now
export interface FooInterface<T, O extends FooOptions<T> = FooOptions<T>> {
items: T[];
options: O;
}
That lets you just leave out the O
parameter when you intend it to just be FooOptions<T>
:
const t: FooInterface<string> = {
items: ["a", "b", "c"],
options: {
doSomething: (e) => {
console.log(e);
}
}
}
And in the event that you actually want O
to be more specific than FooOptions<T>
(that's why you have it as a separate parameter, right?), you can still do it:
interface MoreOptions {
doSomething: (e: string) => void;
doNothing: () => void;
}
const u: FooInterface<string, MoreOptions> = {
items: ["d", "e", "f"],
options: {
doSomething(e) { console.log(e); },
doNothing() { console.log("lol nvm"); }
}
}
Oh, and if T
will usually be string
then you can add a default for it too and then FooInterface
without parameters will be interpreted as FooInterface<string, FooOptions<string>>
:
export interface FooInterface<T = string, O extends FooOptions<T> = FooOptions<T>> {
items: T[];
options: O;
}
const t: FooInterface = {
items: ["a", "b", "c"],
options: {
doSomething: (e) => {
console.log(e);
}
}
}
Okay, hope that helps. Good luck!
You could use a TypeScript alias :
// Declaration
type FooInterfaceOfType<T> = FooInterface<T, FooOptions<T>>;
// Usage
const t: FooInterfaceOfType<string>;
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