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关于javascript:强烈推荐2020年15道优秀的TypeScript练习题-上集

jiezi

TypeScript 是目前不得不学的内容

  • Ts 的货色其实十分十分的多，上到 tsconfig 的配置，下到写法，内容。
  • Ts 正在疯狂的迭代，进入 4.0 版本行将，外面的内容十分十分的多，能够说，入门很简略，然而要写精通，真的还是要花很多功夫。
  • 本文一共分上、下集, 欢送你关注我的公众号：【前端巅峰】, 前端、后端、源码、架构、算法、面试都有, 更有理财，心理学、开源我的项目等日常分享。

正式开始

  • 第一题, 根本 interface 应用考查, 定义一个 item 接口，合乎应用
interface item {
  name: string;
  age: number;
  occupation: string;
}

const users: item[] = [
  {
    name: 'Max Mustermann',
    age: 25,
    occupation: 'Chimney sweep',
  },
  {
    name: 'Kate Müller',
    age: 23,
    occupation: 'Astronaut',
  },
];

function logPerson(user: item) {console.log(` - ${chalk.green(user.name)}, ${user.age}`);
}

console.log(chalk.yellow('Users:'));
users.forEach(logPerson);
  • 第二题, 考查联结类型, 让 logPerson 函数不报错

interface User {
  name: string;
  age: number;
  occupation: string;
}

interface Admin {
  name: string;
  age: number;
  role: string;
}
type Person = User | Admin;
const persons: Person[] /* <- Person[] */ = [
  {
    name: 'Max Mustermann',
    age: 25,
    occupation: 'Chimney sweep',
  },
  {
    name: 'Jane Doe',
    age: 32,
    role: 'Administrator',
  },
  {
    name: 'Kate Müller',
    age: 23,
    occupation: 'Astronaut',
  },
  {
    name: 'Bruce Willis',
    age: 64,
    role: 'World saver',
  },
];

function logPerson(user: Person) {console.log(` - ${chalk.green(user.name)}, ${user.age}`);
}

persons.forEach(logPerson);
  • 第三题, 类型推断、联结类型、类型断言（此题我感觉不是最优解法, 欢送大家指出）, 让 logPerson 不报错

interface User {
  name: string;
  age: number;
  occupation: string;
}

interface Admin {
  name: string;
  age: number;
  role: string;
}

type Person = User | Admin;

const persons: Person[] = [
  {
    name: 'Max Mustermann',
    age: 25,
    occupation: 'Chimney sweep',
  },
  {
    name: 'Jane Doe',
    age: 32,
    role: 'Administrator',
  },
  {
    name: 'Kate Müller',
    age: 23,
    occupation: 'Astronaut',
  },
  {
    name: 'Bruce Willis',
    age: 64,
    role: 'World saver',
  },
];

function logPerson(person: Person) {
  let additionalInformation: string;
  if ((person as Admin).role) {additionalInformation = (person as Admin).role;
  } else {additionalInformation = (person as User).occupation;
  }
  console.log(` - ${chalk.green(person.name)}, ${person.age}, ${additionalInformation}`
  );
}

persons.forEach(logPerson);
  • 第四题，我这里同样应用了类型断言和类型推断、联结类型解题（感觉也不是最优），让 logPerson 不报错

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

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

type Person = User | Admin;

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'},
];

function isAdmin(person: Person) {return person.type === 'admin';}

function isUser(person: Person) {return person.type === 'user';}

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

console.log(chalk.yellow('Admins:'));
persons.filter(isAdmin).forEach(logPerson);

console.log();

console.log(chalk.yellow('Users:'));
persons.filter(isUser).forEach(logPerson);
  • 第五题, 我应用了索引签名解题, 保障 filterUsers 函数不报错

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

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

type Person = User | Admin;

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',
  },
];

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

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

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

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

filterUsers(persons, {age: 23,}).forEach(logPerson);
  • 第六题，考查 overloads, 我对filterPersons 独自进行了解决，解题，保障 logPerson 函数能够返回不同的类型数据
interface User {
  type: 'user';
  name: string;
  age: number;
  occupation: string;
}

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

type Person = User | Admin;

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'},
];

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

function filterPersons(persons: Person[],
  personType: 'user',
  criteria: {[fieldName: string]: number }
): User[];

function filterPersons(persons: Person[],
  personType: 'admin',
  criteria: {[fieldName: string]: number }
): Admin[];

function filterPersons(persons: Person[],
  personType: string,
  criteria: {[fieldName: string]: number }
) {
  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];
      });
    });
}

let usersOfAge23: User[] = filterPersons(persons, 'user', { age: 23});
let adminsOfAge23: Admin[] = filterPersons(persons, 'admin', { age: 23});

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

console.log();

console.log(chalk.yellow('Admins of age 23:'));
adminsOfAge23.forEach(logPerson);
  • 第七题, 考查泛型应用，依据传入参数不同, 动静返回不同类型的数据, 保障 swap 函数运行失常
interface User {
  type: 'user';
  name: string;
  age: number;
  occupation: string;
}

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

function logUser(user: User) {const pos = users.indexOf(user) + 1;
  console.log(` - #${pos} User: ${chalk.green(user.name)}, ${user.age}, ${user.occupation}`
  );
}

function logAdmin(admin: Admin) {const pos = admins.indexOf(admin) + 1;
  console.log(` - #${pos} Admin: ${chalk.green(admin.name)}, ${admin.age}, ${admin.role}`
  );
}

const admins: Admin[] = [
  {
    type: 'admin',
    name: 'Will Bruces',
    age: 30,
    role: 'Overseer',
  },
  {
    type: 'admin',
    name: 'Steve',
    age: 40,
    role: 'Steve',
  },
];

const users: User[] = [
  {
    type: 'user',
    name: 'Moses',
    age: 70,
    occupation: 'Desert guide',
  },
  {
    type: 'user',
    name: 'Superman',
    age: 28,
    occupation: 'Ordinary person',
  },
];

function swap<T, L>(v1: T, v2: L): [L, T] {return [v2, v1];
}

function test1() {console.log(chalk.yellow('test1:'));
  const [secondUser, firstAdmin] = swap(admins[0], users[1]);
  logUser(secondUser);
  logAdmin(firstAdmin);
}

function test2() {console.log(chalk.yellow('test2:'));
  const [secondAdmin, firstUser] = swap(users[0], admins[1]);
  logAdmin(secondAdmin);
  logUser(firstUser);
}

function test3() {console.log(chalk.yellow('test3:'));
  const [secondUser, firstUser] = swap(users[0], users[1]);
  logUser(secondUser);
  logUser(firstUser);
}

function test4() {console.log(chalk.yellow('test4:'));
  const [firstAdmin, secondAdmin] = swap(admins[1], admins[0]);
  logAdmin(firstAdmin);
  logAdmin(secondAdmin);
}

function test5() {console.log(chalk.yellow('test5:'));
  const [stringValue, numericValue] = swap(123, 'Hello World');
  console.log(` - String: ${stringValue}`);
  console.log(` - Numeric: ${numericValue}`);
}

[test1, test2, test3, test4, test5].forEach((test) => test());
  • 第八题，考查 Omit 和多类型 & 的应用，应用 Omit 提取 type 字段，最小代价实现了这道题

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

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


type Person = User | Admin | PowerUser;

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: 'powerUser',
        name: 'Nikki Stone',
        age: 45,
        role: 'Moderator',
        occupation: 'Cat groomer'
    }
];


type PowerUser = Omit<User, 'type'> & Omit<Admin, 'type'> & {type: 'powerUser'};


function isAdmin(person: Person): person is Admin {return person.type === 'admin';}

function isUser(person: Person): person is User {return person.type === 'user';}

function isPowerUser(person: Person): person is PowerUser {return person.type === 'powerUser';}

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

console.log(chalk.yellow('Admins:'));
persons.filter(isAdmin).forEach(logPerson);

console.log();

console.log(chalk.yellow('Users:'));
persons.filter(isUser).forEach(logPerson);

console.log();

console.log(chalk.yellow('Power users:'));
persons.filter(isPowerUser).forEach(logPerson);

写给读者

  • 后面八道题并不一定是最优解法,Ts外面货色的确多，如果你有好的解法能够公众号后盾私信我
  • 前面会补充剩下的题目，由易到难
  • 不定期补充前端架构师面试题、各种技术的系列学习题目
  • 我是 Peter 酱，架构设计过20 万 人端到端加密超级群性能的桌面 IM 软件, 我微信:CALASFxiaotan,不闲聊
  • 另外欢送珍藏我的材料网站:前端生存社区:https://qianduan.life, 能够 右下角 点个 在看 ，关注一波公众号:[ 前端巅峰]
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