Exclude property from type

How to Exclude a Property from a Type in TypeScript

So, you want to exclude a single property from a type in TypeScript? 🤔 No worries! I've got you covered! In this blog post, I'll walk you through the common issues and provide easy solutions to help you achieve your goal.

The Problem

Let's start by understanding the problem at hand. You have an interface called XYZ which has three properties: x, y, and z. However, you want to exclude the z property and create a new type called XY that only contains the x and y properties.

The Solution

To exclude a property from a type in TypeScript, you can use the Omit utility type. This utility type allows you to create a new type by omitting specific properties from an existing type.

In your case, you can define the XY type by using the Omit utility type as follows:

type XY = Omit<XYZ, 'z'>;

This creates a new type called XY that excludes the 'z' property from the XYZ type. Now, the XY type will only have the 'x' and 'y' properties.

Example

Here's an example to help you visualize how it works:

interface XYZ {
  x: number;
  y: number;
  z: number;
}

type XY = Omit<XYZ, 'z'>;

const xy: XY = {
  x: 1,
  y: 2,
};

In the example above, we first define the XYZ interface with three properties: x, y, and z. Then, we use the Omit utility type to create the XY type by excluding the 'z' property from XYZ. Finally, we create a variable xy of type XY and assign values to the x and y properties.

Conclusion

You've learned how to easily exclude a property from a type in TypeScript using the Omit utility type. Now you can confidently create new types that only contain the properties you need, without any hassle.

So go ahead and give it a try! Start excluding properties and make your types more refined and efficient. Happy coding! 🚀

Have any questions or want to share your thoughts? Leave a comment below and let's start a conversation!