Ex 6

문제

/*

Intro:

    Filtering requirements have grown. We need to be
    able to filter any kind of Persons.

Exercise:

    Fix typing for the filterPersons so that it can filter users
    and return User[] when personType='user' and return Admin[]
    when personType='admin'. Also filterPersons should accept
    partial User/Admin type according to the personType.
    `criteria` argument should behave according to the
    `personType` argument value. `type` field is not allowed in
    the `criteria` field.

Higher difficulty bonus exercise:

    Implement a function `getObjectKeys()` which returns more
    convenient result for any argument given, so that you don't
    need to cast it.

    let criteriaKeys = Object.keys(criteria) as (keyof User)[];
    -->
    let criteriaKeys = getObjectKeys(criteria);

*/

interface User {
    type: 'user';
    name: string;
    age: number;
    occupation: string;
}

interface Admin {
    type: 'admin';
    name: string;
    age: number;
    role: string;
}

export type Person = User | Admin;

export const persons: Person[] = [
    { type: 'user', name: 'Max Mustermann', age: 25, occupation: 'Chimney sweep' },
    { type: 'admin', name: 'Jane Doe', age: 32, role: 'Administrator' },
    { type: 'user', name: 'Kate Müller', age: 23, occupation: 'Astronaut' },
    { type: 'admin', name: 'Bruce Willis', age: 64, role: 'World saver' },
    { type: 'user', name: 'Wilson', age: 23, occupation: 'Ball' },
    { type: 'admin', name: 'Agent Smith', age: 23, role: 'Anti-virus engineer' }
];

export function logPerson(person: Person) {
    console.log(
        ` - ${person.name}, ${person.age}, ${person.type === 'admin' ? person.role : person.occupation}`
    );
}

export function filterPersons(persons: Person[], personType: string, criteria: unknown): unknown[] {
    return persons
        .filter((person) => person.type === personType)
        .filter((person) => {
            let criteriaKeys = Object.keys(criteria) as (keyof Person)[];
            return criteriaKeys.every((fieldName) => {
                return person[fieldName] === criteria[fieldName];
            });
        });
}

export const usersOfAge23 = filterPersons(persons, 'user', { age: 23 });
export const adminsOfAge23 = filterPersons(persons, 'admin', { age: 23 });

console.log('Users of age 23:');
usersOfAge23.forEach(logPerson);

console.log();

console.log('Admins of age 23:');
adminsOfAge23.forEach(logPerson);

// In case you are stuck:
// https://www.typescriptlang.org/docs/handbook/2/functions.html#function-overloads

풀이

함수 filterPersons에 적절한 타입을 추가해 문제를 해결해야 한다.

코드를 살펴보면 filterPersons은 personType에 따라 criteria와 리턴되는 값이 다르다.

따라서 함수 오버로드를 활용해 적절한 타입을 축적하자.

export function filterPersons(persons: Person[], personType: "user", criteria: Partial<User>): User[]
export function filterPersons(persons: Person[], personType: "admin", criteria: Partial<Admin>): Admin[]
export function filterPersons(persons: Person[], personType: string, criteria: Partial<Person>): Person[] {
    return persons
        .filter((person) => person.type === personType)
        .filter((person) => {
            let criteriaKeys = Object.keys(criteria) as (keyof Person)[];
            return criteriaKeys.every((fieldName) => {
                return person[fieldName] === criteria[fieldName];
            });
        });
}