TypeScript exercises 5번 문제

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: 'Administrator'
    }
];

export const isAdmin = (person: Person): person is Admin => person.type === 'admin';
export const isUser = (person: Person): person is User => person.type === 'user';

export function logPerson(person: Person) {
    let additionalInformation = '';
    if (isAdmin(person)) {
        additionalInformation = person.role;
    }
    if (isUser(person)) {
        additionalInformation = person.occupation;
    }
    console.log(` - ${person.name}, ${person.age}, ${additionalInformation}`);
}

export function filterUsers(persons: Person[], criteria: User): User[] {
    return persons.filter(isUser).filter((user) => {
        const criteriaKeys = Object.keys(criteria) as (keyof User)[];
        return criteriaKeys.every((fieldName) => {
            return user[fieldName] === criteria[fieldName];
        });
    });
}

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

filterUsers(
    persons,
    {
        age: 23
    }
).forEach(logPerson);

filterUsers 함수의 인자 criteria의 타입을 지정하는 문제. 

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: 'Administrator'
    }
];

export const isAdmin = (person: Person): person is Admin => person.type === 'admin';
export const isUser = (person: Person): person is User => person.type === 'user';

export function logPerson(person: Person) {
    let additionalInformation = '';
    if (isAdmin(person)) {
        additionalInformation = person.role;
    }
    if (isUser(person)) {
        additionalInformation = person.occupation;
    }
    console.log(` - ${person.name}, ${person.age}, ${additionalInformation}`);
}

export function filterUsers(persons: Person[], criteria: Partial<User>): User[] {
    return persons.filter(isUser).filter((user) => {
        const criteriaKeys = Object.keys(criteria) as (keyof User)[];
        return criteriaKeys.every((fieldName) => {
            return user[fieldName] === criteria[fieldName];
        });
    });
}

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

filterUsers(
    persons,
    {
        age: 23
    }
).forEach(logPerson);

유틸리티 타입인 Partial 타입을 사용해서 해결한다.

https://www.typescriptlang.org/ko/docs/handbook/utility-types.html#partialtype

Documentation - Utility Types

 

보너스 문제가 있었는데, criteria에서 type을 제외하라는 게 무슨 말인지 와닿지 않아서 해결하지 못했다. 

솔루션을 찾아보니 말 그대로 User에서 type을 제거한 타입을 요구했던 것...

이는 유틸리티 타입인 Omit을 사용해서 구현 가능하다.

export function filterUsers(persons: Person[], criteria: Partial<Omit<User, 'type'>>): User[] {
    return persons.filter(isUser).filter((user) => {
        const criteriaKeys = Object.keys(criteria) as (keyof Omit<User, 'type'>)[];
        return criteriaKeys.every((fieldName) => {
            return user[fieldName] === criteria[fieldName];
        });
    });
}

https://www.typescriptlang.org/ko/docs/handbook/utility-types.html#omittype-keys

Documentation - Utility Types