If the parent is optional, types cannot be obtained

I'm trying to set the type ShowOptions as "ALL", "ACTIVE", or "DRAFT"

However, I encountered an error saying 

TS2339: Property show does not exist on type

Does anyone have suggestions on how I can extract the ShowOptions? Please note that I am unable to modify the QueryParameters type as it is generated from an OpenAPI specification.

// Snippet of TypeScript code auto-generated from an OpenAPI definition - this code cannot be modified by me!
type QueryParameters = {
  query?: {
    show: "ALL" | "ACTIVE" | "DRAFT";
  };
};

type Query = QueryParameters["query"];
type ShowOptions = Query["show"]; // TS2339: Property show does not exist on type
typescript

Answer №1

To ensure null safety, enclose Query within NonNullable<T>

// Extracted TypeScript snippet from an OpenAPI specification - I do not have authority over this code!
type QueryParameters = {
  query?: {
    show: "ALL" | "ACTIVE" | "DRAFT";
  };
};

type Query = QueryParameters["query"];
type ShowOptions = NonNullable<Query>["show"];

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