What is the process for accessing a child within an optional field?

I am working with the following data type:

export type GetWeeklyArticleQuery = {
    __typename?: 'Query';
    weeklyArticle?: {
        __typename?: 'WeeklyArticle';
        date: any;
        tags: Array<string>;
        title: string;
        authors: Array<{ __typename?: 'Author'; name: string; slug: string }>;
        bodyContent: { __typename?: 'RichText'; html: string };
        coverImage: {
            __typename?: 'Asset';
            url: string;
            title?: string | null;
            altText?: string | null;
            description?: { __typename?: 'RichText'; html: string } | null;
            caption?: { __typename?: 'RichText'; html: string } | null;
        };
        introLine: { __typename?: 'RichText'; html: string };
        footnote: Array<{ __typename?: 'RichText'; html: string }>;
    } | null;
};

In my code, I need to access the authors field but I am unsure how to handle it without defining a custom type like Array<{ __typename?}. In my code, I attempted the following:

let authors:weeklyArticleDto['authors'];

This results in an error in my VSCode editor and in TypeScript playground:

type weeklyArticleDto = GetWeeklyArticleQuery['weeklyArticle'];
const weeklyArticle: weeklyArticleDto = data.weeklyArticle;
let authors:weeklyArticleDto['authors'];
if (weeklyArticle) {
    authors  = weeklyArticle.authors;
}

How can I resolve this issue?

Answer №1

One solution is to utilize the pre-existing NonNullable<Type> utility type.

You can declare a variable like this: let authors: NonNullable<weeklyArticleDto>['authors'];

For a live example and further exploration, check out this Playground Link

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