Question

以下代码包含一个开关盒，它执行五次询问顾客的五个冰淇淋选项，但条件是每次客户必须选择不同的冰淇淋，如果他再次选择相同的冰淇淋，它应该显示错误有人可以帮助我解决上述逻辑，或者建议一个比切换案例更好的方法。

class Icecreams {

    public static void main( String[] args ) throws Exception {
        for( int i = 1; i <= 5; i++ ) {
            System.out.println("Select an Icecream:\n1.Strawberry\n2.Vanilla\n3.Chocolate\n4.Butterscotch\n5.Black current\n6.Exit");
            DataInputStream in = new DataInputStream(System.in);
            int n = Integer.parseInt(in.readLine());

            switch( n ) {
                case 1:
                    System.out.println("STRAWBERRY-Rs.50");
                    break;

                case 2:
                    System.out.println("VANILLA-Rs.40");
                    break;

                case 3:
                    System.out.println("CHOCOLATE-Rs.60");
                    break;

                case 4:
                    System.out.println("BUTTERSCOTCH-Rs.70");
                    break;

                case 5:
                    System.out.println("BLACKCURRENT-Rs.40");
                    break;

                case 6:
                    System.out.println("Exit");
                    System.exit(0);
                    break;
                default:
                    System.out.println("Select from the available choice");
                    break;
            }

        }
    }
}

Answer 1

你可以这样做：把可能的冰淇淋放在List中，即ArrayList，每次选择冰淇淋时，你都要从{{1}中取出相应的冰淇淋。 }。

您也可以使用相同的ArrayList向用户显示选项菜单。

使用ArrayList的好处是，如果将来冰淇淋的选择发生变化，您无需修改程序。根据经验：在代码中放置抽象，在数据中加入详细信息。（Dave Thomas，Andrew Hunt http://pragmatictips.com/38）

之后，您甚至可以修改程序以从文件中读取冰淇淋，因此您只需在选择冰淇淋时更改文本文件。

Answer 2

就像我在评论中所说，你需要保存以前的选择。所以它可能看起来像这样：

    public static void main( String[] args ) throws Exception {
        int previousChoice = 0;
        for( int i = 1; i <= 5; i++ ) {
            System.out.println("Select an Icecream:\n1.Strawberry\n2.Vanilla\n3.Chocolate\n4.Butterscotch\n5.Black current\n6.Exit");
            DataInputStream in = new DataInputStream(System.in);
            int n = Integer.parseInt(in.readLine());
            while (previousChoice == n) {
                System.out.println("Can't select the same ice-cream twice in a row, try again.");
                n = Integer.parseInt(in.readLine());
            }
            previousChoice = n; // save previous choice

            switch( n ) {

免责声明：这只适用于您需要避免连续两次选择相同的冰淇淋。如果您需要确保选择总是唯一，那么解决方案会有所不同，但最好在您的问题中澄清它。

Answer 3

您可以将此代码扩展到您的问题。此解决方案简单易懂。您应该很容易理解改变的位置。

import java.util.ArrayList;
import java.util.Scanner;

public class MainClass {

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

        int n = 0, selection;
        ArrayList<Integer> selectionsSoFar = new ArrayList<Integer>();

        while (n < 5) {
            System.out.println("Enter Option:\n");
            selection = keyboard.nextInt();
            if (selectionsSoFar.contains(selection) || selection <= 0 || selection > 2) {
                System.out.println("Error");
            } else {
                switch (selection) {
                case 1:
                    System.out.println("1");
                    selectionsSoFar.add(selection);
                    n++;
                    break;
                case 2:
                    System.out.println("2");
                    selectionsSoFar.add(selection);
                    n++;
                    break;
                }

            }

        }
        keyboard = null;
    }

}

Answer 4

import java.util.Scanner;

public class practice4 {
    public static void main(String[] args) {

        String I = "Icecream Rs. 10.";
        String S = "Sandwitch Rs.5";
        String C = "Cake Rs.15";
        String Ch = "Chocolate Rs.5";
        String L = "Lemon water Rs.12";
        int limit = 6;

        System.out.println(
                "Select your choice from below options of 1 - 5 \n" + 
        "1 :" + I + "\n2 :" + S + "\n3 :" + C + "\n4 :" + Ch + "\n5 :" + L);
        for (int i = 1; i <= limit; i++) {

            Scanner input = new Scanner(System.in);

            int choice = input.nextInt();

            switch (choice) {

            case 1:
                System.out.println("You have chosen a item " + I);
                break;

            case 2:
                System.out.println("You have chosen a item " + S);
                break;
            case 3:
                System.out.println("You have chosen a item " + C);
                break;
            case 4:
                System.out.println("You have chosen a item " + Ch);
                break;
            case 5:
                System.out.println("You have chosen a item " + L);
                break;
            default:
                System.out.println("Select only from the available choice of 1 - 5");
                break;

            }

        }

    }
}

在java中切换案例并选择案例条件

4 个答案: