When working with a union type in TypeScript, it is important to note that the property may

Question

When working with a union type in TypeScript, it is important to note that the property may

Imagine a scenario where I am dealing with two specific interfaces:

interface Box {
    x: number
    y: number
}

and

interface ColouredBox {
    x: number
    y: number
    colour: string
}

For the sake of this discussion, let's assume that the interfaces cannot be modified.

When it comes to creating objects, the code I typically use looks something like this:

let a: Box|ColouredBox = {x: 1, y: 2}
if( something ){
    a.colour = "Blue"  // Compilation Error
}

Attempting to assign a value to a.colour = "Blue" results in the following error:

Error:(24, 26) TS2339: Property 'colour' does not exist on type 'Box'.

What is causing this error? Is this a limitation of the TypeScript compiler? Are there any alternative approaches I could take without completely rebuilding the object?

typescript union-types

Answer 1

Answer №1

To check for the existence of a property, you can utilize the in type guard:

if ("color" in obj) {
    obj.color = "Blue" // This will work
}

Using this method, you can narrow down to the specific union type ColoredBox based on whether the property color exists. Keep in mind that you can only access common properties like x and y of Box | ColoredBox if they haven't been narrowed down before.

Check out this live code sample for a demonstration.

Answer 2

To check for the existence of a property, you can utilize the in type guard:

if ("color" in obj) {
    obj.color = "Blue" // This will work
}

Using this method, you can narrow down to the specific union type ColoredBox based on whether the property color exists. Keep in mind that you can only access common properties like x and y of Box | ColoredBox if they haven't been narrowed down before.

Check out this live code sample for a demonstration.

Answer 3

Answer №2

Optimizing with Partial Types

Instead of utilizing a union type, an alternative approach is to make use of the Partial type

let a: Partial<ColouredBox> = {x: 1, y: 2}

By using Partial, all properties of the ColoredBox become optional.

For a hands-on demonstration, check out the live example here.

Update:

Simplify with Elimination of an Interface

To make only the color part of the interface optional, consider decreasing the number of interfaces.

interface Box {
  x: number;
  y: number;
  color?: string;
}

To check if you're working with a colored box:

if (box.color) { ... }

Explicit Type Casting

If utilizing both interfaces is necessary, you can specifically specify a as a ColouredBox using the as keyword, which is known as Type Casting.

let a: Box | ColouredBox = { x: 1, y: 2 };
(a as ColouredBox).colour = "Blue";

Answer 4

Optimizing with Partial Types

Instead of utilizing a union type, an alternative approach is to make use of the Partial type

let a: Partial<ColouredBox> = {x: 1, y: 2}

By using Partial, all properties of the ColoredBox become optional.

For a hands-on demonstration, check out the live example here.

Update:

Simplify with Elimination of an Interface

To make only the color part of the interface optional, consider decreasing the number of interfaces.

interface Box {
  x: number;
  y: number;
  color?: string;
}

To check if you're working with a colored box:

if (box.color) { ... }

Explicit Type Casting

If utilizing both interfaces is necessary, you can specifically specify a as a ColouredBox using the as keyword, which is known as Type Casting.

let a: Box | ColouredBox = { x: 1, y: 2 };
(a as ColouredBox).colour = "Blue";

When working with a union type in TypeScript, it is important to note that the property may

Answer №1

Answer №2

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