Ways to inform TypeScript about my custom extension for Object.defineProperty()

Question

Ways to inform TypeScript about my custom extension for Object.defineProperty()

Currently, I am working on a Canvas using the Konva library and found that I needed to add an extension to the Konva.Line prototype. This was necessary because when using getAbsolutePosition(), you only receive relative y values of Lines:

Object.defineProperty(Konva.Line.prototype, "absY", {
    value: function()  {
        return this.getSelfRect().y + this.y()
    },
    writable: true,
    configurable: true,
});

Although this code functions correctly, in my IDE (WebStorm), the ".absY()" call always appears red because TypeScript is not aware of this type.

How can I resolve this issue within a SvelteKit Project?

I understand that this may not be considered good practice, but I have become accustomed to Kotlin 😅

I have already attempted to include

declare namespace Konva {
    interface Line {
        absY: () => number;
    }
}

in the app.d.ts file, however this did not provide a solution and the IDE still shows "Unused interface Line"

typescript sveltekit

Answer 1

Answer №1

In order to make the necessary changes, you must update the definition where it is currently specified
https://tsplay.dev/m3Vl1N

declare module "konva/lib/shapes/Line.js" {
  export interface Line {
    foo: "bar"
  }
}

import Konva from "konva"
let line = new Konva.Line()
let bar = line.foo
//  ^?
// let bar: "bar"

It is important to note that export default namespace cannot be altered. This task may not be feasible if this particular interface is not being exported elsewhere.

Answer 2

In order to make the necessary changes, you must update the definition where it is currently specified
https://tsplay.dev/m3Vl1N

declare module "konva/lib/shapes/Line.js" {
  export interface Line {
    foo: "bar"
  }
}

import Konva from "konva"
let line = new Konva.Line()
let bar = line.foo
//  ^?
// let bar: "bar"

It is important to note that export default namespace cannot be altered. This task may not be feasible if this particular interface is not being exported elsewhere.

Answer 3

Answer №2

If you're looking to customize the behavior of the Konva Line class, consider extending it with your own properties.

import Konva from 'konva';

// extend the Konva `Line` class
class MyCustomKonvaLine extends Konva.Line {
    public absY() {
        return this.getSelfRect().y + this.y()
    }
}

// example usage
const greenLine = new MyCustomKonvaLine({
    points: [5, 70, 140, 23, 250, 60, 300, 20],
    stroke: 'green',
    strokeWidth: 2,
    lineJoin: 'round',
    dash: [33, 10]
});

// output the absY value
console.log('greenLine absY equals:', greenLine.absY());

const stage = new Konva.Stage({
    container: 'container',
    width: window.innerWidth,
    height: window.innerHeight,
});

const layer = new Konva.Layer();

layer.add(greenLine);
stage.add(layer);

I have no experience with Svelte or Konva.

Answer 4

If you're looking to customize the behavior of the Konva Line class, consider extending it with your own properties.

import Konva from 'konva';

// extend the Konva `Line` class
class MyCustomKonvaLine extends Konva.Line {
    public absY() {
        return this.getSelfRect().y + this.y()
    }
}

// example usage
const greenLine = new MyCustomKonvaLine({
    points: [5, 70, 140, 23, 250, 60, 300, 20],
    stroke: 'green',
    strokeWidth: 2,
    lineJoin: 'round',
    dash: [33, 10]
});

// output the absY value
console.log('greenLine absY equals:', greenLine.absY());

const stage = new Konva.Stage({
    container: 'container',
    width: window.innerWidth,
    height: window.innerHeight,
});

const layer = new Konva.Layer();

layer.add(greenLine);
stage.add(layer);

I have no experience with Svelte or Konva.

Ways to inform TypeScript about my custom extension for Object.defineProperty()

Answer №1

Answer №2

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