Honestly, just curious, on the documentation website for type-graphql, they pass an argument to the @Resolver decorator. Here is the link to the page where it occurs: https://typegraphql.com/docs/getting-started.html (under the "Resolvers" heading).
I've also included their code snippet here:
@Resolver(Recipe)
class RecipeResolver {
constructor(private recipeService: RecipeService) {}
@Query(returns => Recipe)
async recipe(@Arg("id") id: string) {
const recipe = await this.recipeService.findById(id);
if (recipe === undefined) {
throw new RecipeNotFoundError(id);
}
return recipe;
}
@Query(returns => [Recipe])
recipes(@Args() { skip, take }: RecipesArgs) {
return this.recipeService.findAll({ skip, take });
}
@Mutation(returns => Recipe)
@Authorized()
addRecipe(
@Arg("newRecipeData") newRecipeData: NewRecipeInput,
@Ctx("user") user: User,
): Promise<Recipe> {
return this.recipeService.addNew({ data: newRecipeData, user });
}
@Mutation(returns => Boolean)
@Authorized(Roles.Admin)
async removeRecipe(@Arg("id") id: string) {
try {
await this.recipeService.removeById(id);
return true;
} catch {
return false;
}
}
}
My code still works without the argument to @Resolver so I'm sure it's not that important but I'm wondering what it's used for?
CodePudding user response:
It's used in a @FieldResolver
method.
Field resolvers in TypeGraphQL are very similar to queries and mutations. Create them as a method on the resolver class, but with a few modifications. First, declare which object type fields we are resolving by providing the type to the @Resolver decorator:
@Resolver(of => Recipe) class RecipeResolver { // queries and mutations }
Then, create a class method (e.g.,
averageRating
) that will become the field resolver. The method is marked with the@FieldResolver()
decorator. Also, decorate the method parameters with the@Root
decorator in order to inject the recipe object:@Resolver(of => Recipe) class RecipeResolver { // queries and mutations @FieldResolver() averageRating(@Root() recipe: Recipe) { const ratingsSum = recipe.ratings.reduce((a, b) => a b, 0); return recipe.ratings.length ? ratingsSum / recipe.ratings.length : null; } }