Is it possible to Omit native type like booleans or numbers?

[silversonicaxel]

I would like to know if it is possible to have a scenario like this one:

type A = {
  a: string;
  b: number;
  c: boolean;
  d: boolean;
  e: boolean;
};

type B = Omit<A, boolean>;

where type B looks like this

type B = {
  a: string;
  b: number;
};

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[tommy-mitchell]

ConditionalExcept should work:

type B = ConditionalExcept<A, boolean>;

[silversonicaxel]

Amazing, last part of question is, what if one boolean is optional?
Because now, considering this example:

type A = {
  a: string;
  b: number;
  c: boolean;
  d?: boolean;
  e: boolean;
};

type B = ConditionalExcept <A, boolean>;

type B looks like this:

type B = {
  a: string;
  b: number;
  d?: boolean;
};

[tommy-mitchell]

d?: boolean is the same as d: boolean | undefined (for most cases). One solution is to use the built-in Required utility type (TS Playground):

type B = ConditionalExcept<Required<A>, boolean>;