Question 72380 in Javascript Question posted in
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Type for all Omits of class/interface/type – Javascript

If I have a type:

type MyType = {
    ...
    propertyA: string;
    propertyB: number;
    ...
}

how can I type this in another type but allow all versions of it with properties omitted? Such as:

type MyOtherType = {
    propertyC: MyType | Omit<MyType, ANYTHING>;
}

I have tried Omit<MyType, any> but that removes the types of other properties on the object.

What is the correct way to do this?

Tags:

2

Answers


  1. 0

    I believe you means make all of them optional (Partial), you can try to use Partial<T> instead:

    type MyType = {
    ...
    propertyA: string;
    propertyB: number;
    ...
}

type MyOtherType = {
    propertyC: MyType | Partial<MyType>;
}
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  2. 0

    If i understood correctly. You might use the Partial utility type to make all props of MyType optional and then use Omit to remove any property you want from it.

    type MyOtherType = {
    propertyC: MyType | Omit<Partial<MyType>, 'propertyA'>;
}

    It is an example. So I removed propertyA from MyType. You can remove other property with replacing ‘propertyA’ with the name of that property.

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