开关/案例Java返回主菜单

Question

我正在为我的Java类编写一个程序，我需要制作一个类似于输出的收据。我必须让用户从3项中选择。选择项目后，他们选择数量，我乘以价格给他们一个总数。用户需要挑选所有3个项目，但可以按任意顺序挑选它们。我正在尝试使用Switch/Case方法来帮助我将变量分配给用户选择的内容。这是我目前所做的：

System.out.println("Enter name of First item"); 
System.out.println(" 1. Gum, 2. Soda, 3. Chips"); 
firstItem = keyboard.nextInt(); 
switch (firstItem) { 
    case 1: 
     System.out.println("Gum"); 
     System.out.println(gumPrice); 
     System.out.println("How many packs of gum would you like to purchase?"); 
     gumBought = keyboard.nextInt(); 
     break; 
    case 2: 
     System.out.println("Soda"); 
     System.out.println(sodaPrice); 
     System.out.println("How many cups of soda would you like to purchase?"); 
     sodaBought = keyboard.nextInt(); 
     break; 
    case 3: 
     System.out.println("Chips"); 
     System.out.println(chipsPrice); 
     System.out.println("How many bags of chips would you like to purchase?"); 
     chipsBought = keyboard.nextInt(); 
     break; 
    default: 
     System.out.println("Unknown Entry"); 
     break; 
} 

然后它开始第二个项目开关/案例。我需要知道如何让它回到主要问题（口香糖/苏打/薯片），如果用户在1-3之外选择一个数字或者选择相同的数字/项目两次。我怎样才能有效地做到这一点？

此外，我可以使用字符串而不是Int来允许用户输入“Gum”而不是“1”？谢谢。

Answer 1

• 给用户一个选择与 '9' 退出
• 把switch语句中while()循环

这应该工作：

public static void main(String[] args) { 

    double gumPrice, sodaPrice, chipsPrice; 

    gumPrice = 1.25; 
    sodaPrice = 4.00; 
    chipsPrice = 3.25; 

    while(true){ 
    Scanner keyboard = new Scanner(System.in); 
    System.out.println("Enter name of First item"); 
    System.out.println(" 1. Gum, 2. Soda, 3. Chips, 9. EXIT"); 
    int firstItem = keyboard.nextInt(); 

    switch (firstItem){ 
     case 1: 
      System.out.println("Gum"); 
      System.out.println(gumPrice); 
      System.out.println("How many packs of gum would you like to purchase?"); 
      int gumBought = keyboard.nextInt(); 
      break; 
     case 2: 
      System.out.println("Soda"); 
      System.out.println(sodaPrice); 
      System.out.println("How many cups of soda would you like to purchase?"); 
      int sodaBought = keyboard.nextInt(); 
      break; 
     case 3: 
      System.out.println("Chips"); 
      System.out.println(chipsPrice); 
      System.out.println("How many bags of chips would you like to purchase?"); 
      int chipsBought = keyboard.nextInt(); 
      break; 
     case 9: 
      return; 
     } 
    } 
} 

Answer 2

我很犹豫给一个完整的工作例子，所以我会给你提供片段，让你在正确的轨道上使用Strings。

首先，如果您想重复算法直到满足条件，您可以使用loop为您提供所搜索的内容。

while (condition) { 
    //do something 
} 

由于Java 7发布时，switch-case可以用字符串中使用。你只需要改变你的输入类型来匹配你希望是什么：

System.out.print("Enter name of your item [gum] [soda] [chips]: "); 
String choice = keyboard.next().toLowerCase(); 

然后，你将不得不修改switch-case语法。

switch (choice) {//now String 
    case "gum": 
    case "soda": 
    case "chips": 
    default: 
} 

希望这会让你开始。