Summary: Generating Types for Raw Return Types in TypeORM
Hello, I'm working with TypeORM and have a question about handling raw return types. When using getRawOne and getRawMany functions, the return includes strings composed of class names and properties, which are concatenated with an underscore ('_').
For example:
export class TestUser { id: number; name: string; verified: boolean;}const userRepository = getRepository(TestUser);// getRawOne is not support type. just return any type. I want to disclose this return type.const loaedRawTestUser = await userRepository.createQueryBuilder('testUser') .addSelect('COUNT(*()', 'myAliasForAllUserCount') .getRawOne();/** * { * testUser_id: number; * testUser_name: string; * testUser_verified: boolean; * } */console.log(loaedRawUser);The issue comprises three parts:
- Converting camelCase to snake_case.
- Inferring the class name for the composition of the return string.
- Exact value matching for the converted properties (Entity key).
The first part has been resolved.
I've implemented a camelCase to snake_case type inferer as follows:
export type CamelCaseToSnakeCase< T extends string, Joiner extends '' | '_' = ''> = T extends `${infer Character}${infer Rest}` ? Character extends Uppercase<Character> ? `${Joiner}${Lowercase<Character>}${CamelCaseToSnakeCase<Rest, '_'>}` : `${Character}${CamelCaseToSnakeCase<Rest, '_'>}` : '';Is there something way?
Additional Requirements if it can be.
When TestUser class has method like this:
export class TestUser { id: number; name: string; verified: boolean; getNameIfVerified(): { return this.verified ? this.name : 'not verified' }}const typeormRawColumTestSample: Partial<TypeORMRawColumns<TestUser, 'TestUser'>> = { testUser_id: 1, // when testUser_id: '1' should be error. testUser__get_name_if_verified // should be error.}How can I exclude methods from a Type in TypeScript