Question

所以我试图制作一个像收银员一样的程序，该程序不断增加取决于用户购买产品的数量，但是我的程序不会增加价值，但它会不断替换它

班级：

idx=any(cell2mat(cellfun(@isempty,raw,'UniformOutput',false)),2)
raw(idx,:)=[]
idx=any(cell2mat(cellfun(@isnan,raw,'UniformOutput',false)),2)
raw(idx,:)=[]

主类：

public class Class {
    private double drink, pdrink;
                public void SetDrink (double drink)
                {
                    this.drink = drink;
                }
                public void SetPDrink (double pdrink)
                {
                    this.pdrink = pdrink;
                }

            public double getDrink()
            {
                if (drink==1)
                    drink=500;
                else if (drink==2)
                    drink=1000;
                else if (drink==3)
                    drink=3000;
                else
                    drink=0;


                return drink;
            }


    public double getPDrink()
            {
    double total = getDrink();
            return total;
        }
}

我正在考虑使用“ do while”，但是我不确定如何做到这一点

编辑：我正在使用oop，我正在尝试使用switch on class，但没有成功

Answer 1

我正在使用扫描仪，您可以使用任何其他库从控制台读取输入。不能固定您的业务逻辑，例如打印“总”价值。

    char mi;
    Scanner sc = new Scanner(System.in);
    double total = 0.0;
    do {
        System.out.println("do you want to buy drinks?");
        System.out.println("1. A (500)");
        System.out.println("2. B (1000)");
        System.out.println("3. C (1500)");
        System.out.println("0 to not choose");
        System.out.println("Please choose you drink");
        //p.SetDrink(sc.nextDouble()); // not working unknown p in OP, so commented
        char drink = sc.next().charAt(0);
        System.out.println(drink); // just printing user's choice for drink
        switch (drink) {
                case 'A':
                    total += 500;
                    break;
                case 'B':
                    total += 1000;
                    break;
                case 'C':
                    total += 1500;
                    break;
                default:
                    total += 0;
            }

        System.out.println("do you want to buy more?");

        mi = sc.next().charAt(0);
    } while (mi=='y');
    System.out.println("total : " + total);

输出：

do you want to buy drinks?
1. A (500)
2. B (1000)
3. C (1500)
0 to not choose
Please choose you drink
C
C
do you want to buy more?
y
do you want to buy drinks?
1. A (500)
2. B (1000)
3. C (1500)
0 to not choose
Please choose you drink
A
A
do you want to buy more?
n
total : 2000.0

循环数学方程

1 个答案: