作者：Peter谭金杰
转发链接：https://segmentfault.com/a/1190000023295719
TypeScript是目前不得不学的内容
- Ts的东西其实非常非常的多，上到tsconfig的配置，下到写法，内容。
- Ts正在疯狂的迭代，进入4.0版本即将，里面的内容非常非常的多，可以说，入门很简单，但是要写精通，真的还是要花很多功夫。
- 本文一共分上、下集,前端、后端、源码、架构、算法、面试都有,更有理财，心理学、开源项目等日常分享。
正式开始
- 第一题,基本interface使用考察,定义一个item接口，符合使用
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,考察联合类型,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、联合类型、类型断言（此题我觉得不是最优解法,欢迎大家指出）,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，我这里同样使用了类型断言和类型推断、联合类型解题（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,我使用了索引签名解题,保证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，考察overloads,我对filterPersons单独进行了处理，解题，保证logPerson函数可以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}, ${
person.type === 'admin' ? person.role : person.occupation
}`
);
}
function filterPersons(
persons: Person[],
personType: 'user',
criteria: { [fieldName: string]: number }
): User[];
function filterPersons(
persons: Person[],
personType: 'admin',
criteria: { [fieldName: string]: number }
): Admin[];
function filterPersons(
persons: Person[],
personType: string,
criteria: { [fieldName: string]: number }
) {
return persons
.filter((person) => person.type === personType)
.filter((person) => {
let criteriaKeys = Object.keys(criteria) as (keyof Person)[];
return criteriaKeys.every((fieldName) => {
return person[fieldName] === criteria[fieldName];
});
});
}
let usersOfAge23: User[] = filterPersons(persons, 'user', { age: 23 });
let adminsOfAge23: Admin[] = filterPersons(persons, 'admin', { age: 23 });
console.log(chalk.yellow('Users of age 23:'));
usersOfAge23.forEach(logPerson);
console.log();
console.log(chalk.yellow('Admins of age 23:'));
adminsOfAge23.forEach(logPerson);- 第七题,考察泛型使用，根据传入参数不同,动态返回不同类型的数据,保证swap函数运行正常
interface User {
type: 'user';
name: string;
age: number;
occupation: string;
}
interface Admin {
type: 'admin';
name: string;
age: number;
role: string;
}
function logUser(user: User) {
const pos = users.indexOf(user) + 1;
console.log(
` - #${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: ${numericValue}`);
}
[test1, test2, test3, test4, test5].forEach((test) => test());
- 第八题，考察Omit和多类型&的使用，使用Omit提取type字段，最小代价完成了这道题
interface User {
type: 'user';
name: string;
age: number;
occupation: string;
}
interface Admin {
type: 'admin';
name: string;
age: number;
role: string;
}
type Person = User | Admin | PowerUser;
const persons: Person[] = [
{ type: 'user', name: 'Max Mustermann', age: 25, occupation: 'Chimney sweep' },
{ type: 'admin', name: 'Jane Doe', age: 32, role: 'Administrator' },
{ type: 'user', name: 'Kate Müller', age: 23, occupation: 'Astronaut' },
{ type: 'admin', name: 'Bruce Willis', age: 64, role: 'World saver' },
{
type: 'powerUser',
name: 'Nikki Stone',
age: 45,
role: 'Moderator',
occupation: 'Cat groomer'
}
];
type PowerUser = Omit<User, 'type'> & Omit<Admin, 'type'> & {type: 'powerUser'};
function isAdmin(person: Person): person is Admin {
return person.type === 'admin';
}
function isUser(person: Person): person is User {
return person.type === 'user';
}
function isPowerUser(person: Person): person is PowerUser {
return person.type === 'powerUser';
}
function logPerson(person: Person) {
let additionalInformation: string = '';
if (isAdmin(person)) {
additionalInformation = person.role;
}
if (isUser(person)) {
additionalInformation = person.occupation;
}
if (isPowerUser(person)) {
additionalInformation = `${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
- 前面八道题并不一定是最优解法,Ts里面东西确实多，如果你有好的解法可以头条私信我或者下方评论
- 后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
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作者：Peter谭金杰
转发链接：https://segmentfault.com/a/1190000023295719