How can i define interface in x.y form in typescript?
Question
I want to extend my lib's prototype, and the lib is writing by JavaScript.
Just like I have a module X , and a Y class under it.
What I want is to extends Y by:
X.Y.prototype.method = function() { ... }
This will work in pure JavaScript, but in typescript, it throw error. It seems I need to add interface for the Y module by:
interface X.Y {
method(): any
}
However, it throw the following error:
error TS1005: '{' expected.
error TS1005: ';' expected.
I have no idea about this... Can anyone help me ? Thanks !
Update
Here is a minimal demo:
// index.html
<!doctype html>
<html>
<head>
<script src="./x.js"></script>
</head>
<body>
<script src="./app.js"></script>
</body>
</html>
// x.js
var x = {
y: function() { }
}
// x.d.ts
declare module x {
export class y {}
}
// app.ts
interface x.y {
test: () => void
}
x.y.prototype.test = function() {
}
1 answers
solution1
1 2018-01-01 17:17:35
Probably something like this will help
// let's pretend this is our original lib
const X = function () { };
type ExtendedProto = {
new (): {
test: (arg1: string) => void;
}
};
const Y = X as typeof X & ExtendedProto;
Y.prototype.test = function(arg1: string) {
// console.log('test');
}
const y = new Y();
y.test('1');
or you could creating an index.d.ts file with approximately the following
// index.d.ts
declare module 'x-y-z' {
export class X {}
}
// .ts file
import { X } from 'x-y-z';
class Y extends X {
test() { console.log('test'); }
}
const y = new Y();
y.test();
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