Question

我正在尝试创建一个人们欠对方钱的应用程序。让人说他的朋友购买电影票，每张票的成本是20美元。现在他的每个朋友都欠那个人20美元。

我确实做了一个对象，但我认为这不是正确的方法。我想知道是否有更好的方法可以做到。

例如 person1 to person2：20美元。 person3 to person2：20美元。 person2 to person1：$ 0。 person2 to person3：$ 0。

我已经接近这样了

    var person1 = new User(1, "person1", [{"name": "person2", "amount": 0}, {"name": "person3, "amount": 0}]);
var person2 = new User(2, "person2", [{"name": "person1", "amount": 0}, {"name": "person3, "amount": 0}]);
var person3 = new User(3, "person3", [{"name": "person1", "amount": 0}, {"name": "person2", "amount": 0}]);

Answer 1

我不建议将＆＃34;交易＆＃34;对用户本身。分开保存。也许就是这样。

＆＃13;

// just an quick example using typescript

class User {
  public readonly id: string;
  public readonly name: string;
  
  constructor(name?: string) {
    this.name = name || 'unknown';
    this.id = Math.random(); // just an example;
  }
}

class Transaction {
  private target: string;

  public readonly date: number;
  public readonly origin: string;
  public readonly amount: number;
  
  constructor(origin: User, target: User, amount: number) {
    this.date = Date.now();
    this.origin = origin.id;
    this.target = target.id;
    this.amount = amount;
  }
  
  public changeTarget(user: User) {
    this.target = user;
  }
}

class TransactionManager {
  private static instance: TransactionManager;
  private transactions: Transaction[] = [];
  private users: { [id: string]: User } = {};
  
  public getInstance() {
    if (!TransactionManager.instance) {
        TransactionManager.instance = new TransactionManager();
    }
    return TransactionManager.instance;
  }
  
  public createUser(name) {
    let user = new User(name);
    this.users[user.id] = user;
    return user;
  }
  
  public lendMoney(from: User, to: User, amount: number) {
    let transaction = new Transaction(from, to, amount);
    this.transactions = this.transactions.concat(transaction);
    return transaction;
  }
  
  public getUserDept(user: User) {
    if(!this.users[user.id]) {
      throw new Error('user not found');
    }
    return this.transactions.reduce((dept: number, t: Transaction) => {
      return t.origin === user.id ? dept + t.amount : dept;
    }, 0);
  }
}

＆＃13;

我的简单钱记录应用程序的对象

1 个答案: