Question

我想循环输入（重复），直到输入5退出。例如，我输入1个展示柜1，然后再次询问输入，输入4并显示案例4.直到我输入5退出。

下面的代码只是一直问我输入，从不向我显示输出。

public static void main(String[] args){
        System.out.println("1.Soup\n2.Main Course\n3.Dessert\n4.Beverage\n5.Exit");
        //ask input from user
        System.out.println("Enter your choice: ");
        Scanner scan=new Scanner(System.in);
        int input=scan.nextInt();
        int check=input;
        //keep looping until user input 5 which is Exit
        while (input!=5){
             System.out.println("Please enter 1-5\n1.Soup\n2.Main Course\n3.Dessert\n4.Beverage\n5.Exit");
             int input1=scan.nextInt();
             check=input1;
        }
        switch (check){
            case 1 :
                System.out.println("1.Mushroom Soup\n2.Miso Soup\n3.Tomato Soup");
                break;
            case 2 :
                System.out.println("1.Chiken Katsu Don\n2.Curry Katsu Don\n3.Teriyaki Katsu Don\n4.Curry Udon");
                break;
            case 3 :
                System.out.println("1.Matcha Ice Cream\n2.Chocolate Ice Cream\n3.Oreo Chocolate\n4.Dango");
                break;
            case 4 :
                System.out.println( "1.Green Tea\n2.Matcha RedBean\n3.Chocolate MilkShake\n4.Strawberry Ice Blended");
                break;
            case 5 :
                System.exit(0);
               break;
        }
    }
}

Answer 1

switch应位于while内。只有在处理完第一个输入后才能询问另一个输入。

while (input is valid)
    get input
    process input

Answer 2

您的switch声明超出while-loop范围。

        while (input!=5)
        {
            System.out.println("Please enter 1-5\n1.Soup\n2.Main Course\n3.Dessert\n4.Beverage\n5.Exit");
            int input1=scan.nextInt();
            check=input1;
            switch (check)
            {
            case 1 :
                System.out.println("1.Mushroom Soup\n2.Miso Soup\n3.Tomato Soup");
                break;
            case 2 :
                System.out.println("1.Chiken Katsu Don\n2.Curry Katsu Don\n3.Teriyaki Katsu Don\n4.Curry Udon");
                break;
            case 3 :
                System.out.println("1.Matcha Ice Cream\n2.Chocolate Ice Cream\n3.Oreo Chocolate\n4.Dango");
                break;
            case 4 :
                System.out.println( "1.Green Tea\n2.Matcha RedBean\n3.Chocolate MilkShake\n4.Strawberry Ice Blended");
                break;
            case 5 :
                System.exit(0);
               break;
            }        
       }

编辑：另外，正如其他人在最近的答案中指出的那样，你的循环条件变量没有在循环内重置（但这是你提出的问题的偶然问题，可以说是一个不同的问题）。 p>

Answer 3

我会这样修理：

do{
    System.out.println("Please enter 1-5\n1.Soup\n2.Main Course\n3.Dessert\n4.Beverage\n5.Exit");
    int input1=scan.nextInt();
    check=input1;
    switch (check){
    case 1 :
        System.out.println("1.Mushroom Soup\n2.Miso Soup\n3.Tomato Soup");
        break;
    case 2 :
        System.out.println("1.Chiken Katsu Don\n2.Curry Katsu Don\n3.Teriyaki Katsu Don\n4.Curry Udon");
        break;
    case 3 :
        System.out.println("1.Matcha Ice Cream\n2.Chocolate Ice Cream\n3.Oreo Chocolate\n4.Dango");
        break;
    case 4 :
        System.out.println( "1.Green Tea\n2.Matcha RedBean\n3.Chocolate MilkShake\n4.Strawberry Ice Blended");
        break;
    case 5 :
        System.exit(0);
        break;
    }
}
while (input!=5);

添加Do-While循环可确保它至少显示一次。

或者你可以简单地

while (input!=5){
    System.out.println("Please enter 1-5\n1.Soup\n2.Main Course\n3.Dessert\n4.Beverage\n5.Exit");
    int input1=scan.nextInt();
    check=input1;
    switch (check){
    case 1 :
        System.out.println("1.Mushroom Soup\n2.Miso Soup\n3.Tomato Soup");
        break;
    case 2 :
        System.out.println("1.Chiken Katsu Don\n2.Curry Katsu Don\n3.Teriyaki Katsu Don\n4.Curry Udon");
        break;
    case 3 :
        System.out.println("1.Matcha Ice Cream\n2.Chocolate Ice Cream\n3.Oreo Chocolate\n4.Dango");
        break;
    case 4 :
        System.out.println( "1.Green Tea\n2.Matcha RedBean\n3.Chocolate MilkShake\n4.Strawberry Ice Blended");
        break;
    case 5 :
        System.exit(0);
        break;
    }
}

Answer 4

您需要进行两次修复。首先，将switch块移到while循环中。其次，你的while循环永远不会终止，因为你只分配input一次。

我只能想象一些混淆来自代码中包含太多变量：input，input1，check ......

在我看来，您的代码可以大大简化。考虑以下替代方案 - 我们可以使用简单的while(true)循环，因为您计划在输入5时退出VM：

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);

    while (true) {

        System.out.println("Please enter 1-5\n1.Soup\n2.Main Course\n3.Dessert\n4.Beverage\n5.Exit");

        int input = scan.nextInt();

        switch (input) {
        case 1:
            System.out.println("1.Mushroom Soup\n2.Miso Soup\n3.Tomato Soup");
            break;
        case 2:
            System.out.println("1.Chiken Katsu Don\n2.Curry Katsu Don\n3.Teriyaki Katsu Don\n4.Curry Udon");
            break;
        case 3:
            System.out.println("1.Matcha Ice Cream\n2.Chocolate Ice Cream\n3.Oreo Chocolate\n4.Dango");
            break;
        case 4:
            System.out.println("1.Green Tea\n2.Matcha RedBean\n3.Chocolate MilkShake\n4.Strawberry Ice Blended");
            break;
        case 5:
            System.exit(0);
            break;
        }
    }
}

如何在用户输入退出之前创建菜单

4 个答案: