我正在尝试制作一个索引，如果有人搞砸了，我希望他们能够返回到最后一个菜单或最后一个选择屏幕，或者只是重启。这可能吗？
我正在使用JCreator。这是我的第一个程序，所以我保持它非常简单;它只是一堆if-else语句。看起来像这样的东西。
编辑：如果我在程序界面中创建了一个重启按钮或者你可以称之为什么，那会更容易吗？
public class Index
{
public static void main(String[] args)
{
Scanner key = new Scanner(System.in);
System.out.println("What kind of sandwich are you looking for?");
System.out.println("1. Ham 2. Turkey 3. etc.");
int meat = key.nextInt();
if (meat == 1)
{
System.out.println("What type of cheese would you like"?);
System.out.println("1. Swiss 2. Cheddar 3. etc.");
int cheese = key.nextInt();
if (cheese == 1)
{
System.out.println("Would you like lettuce and/or tomatoes?");
System.out.println("1. Just lettuce 2. Just tomatoes 3. Both 4. Neither");
int lOrT = key.nextInt();
}
}
if (meat == 2)
{
System.out.println("What type of cheese would you like"?);
System.out.println("1. Swiss 2. Cheddar 3. etc.");
int cheese = key.nextInt();
if (cheese == 1)
{
System.out.println("Would you like lettuce and/or tomatoes?");
System.out.println("1. Just lettuce 2. Just tomatoes 3. Both 4. Neither");
int lOrT = key.nextInt();
}
}
}
}
答案 0 :(得分：1)
这不可能像你描述的那样，但这就是方法的用武之地。
方法（在某些其他语言中的函数）是基于对象的状态在特定类中公开可重复行为的东西，该状态专门属于该对象的实例。
还有static个方法，或者更正式的类方法，它们与普通方法类似，但它们通常不依赖于对象的状态来正常运行。
在您的情况下，这意味着您的代码需要进行一些清理。将所有常见的东西移到一个方法中。
让我们从生菜和西红柿开始吧。这两个街区是完全一样的。将它们移动到一个东西，以便它们可以被引用。
public static void main(String[] args) {
Scanner key = new Scanner(System.in);
System.out.println("What kind of sandwich are you looking for?");
System.out.println("1. Ham 2. Turkey 3. etc.");
int meat = key.nextInt();
if (meat == 1) {
System.out.println("What type of cheese would you like?");
System.out.println("1. Swiss 2. Cheddar 3. etc.");
int cheese = key.nextInt();
int lOrT = extractToppings(key, cheese);
}
if (meat == 2) {
System.out.println("What type of cheese would you like?");
System.out.println("1. Swiss 2. Cheddar 3. etc.");
int cheese = key.nextInt();
int lOrT = extractToppings(key, cheese);
}
}
private static int extractToppings(Scanner key, int cheese) {
if (cheese == 1) {
System.out.println("Would you like lettuce and/or tomatoes?");
System.out.println("1. Just lettuce 2. Just tomatoes 3. Both 4. Neither");
return key.nextInt();
}
return -1; // sentinel return value - what to return when the if fails
}
直截了当，不是吗？让我们把它弄清楚 - 让我们做一些更多的工作来重构你的代码。
接下来就是您的if语句 - 无论是meat == 1还是meat == 2，它们都会做同样的事情。将它们分开是没有意义的。我将在这里删除if语句，然后我们到达......
public static void main(String[] args) {
Scanner key = new Scanner(System.in);
System.out.println("What kind of sandwich are you looking for?");
System.out.println("1. Ham 2. Turkey 3. etc.");
int meat = key.nextInt();
if (meat == 1 || meat == 2) {
System.out.println("What type of cheese would you like?");
System.out.println("1. Swiss 2. Cheddar 3. etc.");
int cheese = key.nextInt();
int lOrT = extractToppings(key, cheese);
}
}
private static int extractToppings(Scanner key, int cheese) {
if (cheese == 1) {
System.out.println("Would you like lettuce and/or tomatoes?");
System.out.println("1. Just lettuce 2. Just tomatoes 3. Both 4. Neither");
return key.nextInt();
}
return -1;
}
现在我们可以轻松地将其转移到自己的方法中 - 例如extractCheese。我还会在这里做一些一般性的清理工作，让你感觉更直观。
public static void main(String[] args) {
Scanner key = new Scanner(System.in);
System.out.println("What kind of sandwich are you looking for?");
System.out.println("1. Ham 2. Turkey 3. etc.");
int meat = key.nextInt();
int cheese = extractCheese(key, meat);
int toppings = extractToppings(key, cheese);
}
private static int extractCheese(Scanner key, int meat) {
if (meat == 1 || meat == 2) {
System.out.println("What type of cheese would you like?");
System.out.println("1. Swiss 2. Cheddar 3. etc.");
return key.nextInt();
}
return -1;
}
private static int extractToppings(Scanner key, int cheese) {
if (cheese == 1) {
System.out.println("Would you like lettuce and/or tomatoes?");
System.out.println("1. Just lettuce 2. Just tomatoes 3. Both 4. Neither");
return key.nextInt();
}
return -1;
}
你可以合法地将关于肉类的问题转移到他们自己的方法中......
public static void main(String[] args) {
Scanner key = new Scanner(System.in);
int meat = extractMeat(key);
int cheese = extractCheese(key, meat);
int toppings = extractToppings(key, cheese);
}
private static int extractMeat(Scanner key) {
System.out.println("What kind of sandwich are you looking for?");
System.out.println("1. Ham 2. Turkey 3. etc.");
return key.nextInt();
}
private static int extractCheese(Scanner key, int meat) {
if (meat == 1 || meat == 2) {
System.out.println("What type of cheese would you like?");
System.out.println("1. Swiss 2. Cheddar 3. etc.");
return key.nextInt();
}
return -1;
}
private static int extractToppings(Scanner key, int cheese) {
if (cheese == 1) {
System.out.println("Would you like lettuce and/or tomatoes?");
System.out.println("1. Just lettuce 2. Just tomatoes 3. Both 4. Neither");
return key.nextInt();
}
return -1;
}
编写能够返回可重复结果的东西是直截了当的。
现在变得更加棘手 - 使用while循环来检索有效信息。我将把其他人留给你实施，但这是一般的要点
这是一个片段。
int meat ;
do{
meat = extractMeat(key);
if(meat < 1 || meat > 3) {
System.out.println("Please enter valid meat.");
}
} while (meat < 1 || meat > 3);
一旦meat在其范围内，您的程序将继续执行。我把剩下的实现作为练习留给读者。
答案 1 :(得分：0)
我会更改为循环我将您的逻辑更改为Sudo Code的方法，这在一层运行，您需要n层吗？如果是这样的话，你可以用嵌套循环来扩展这个逻辑，这样它们就可以回到最后一个选择，就像你在这里看到的另一个生菜循环一样。下面的Sudo代码：
int meatType = -1;
int cheeseType = -1;
bool canLeave = false;
do{
meatType = MeatMethod();
if(meatType == 1 || meatType ||2)
{
cheeseType = CheeseMethod();
if(cheeseType == 1 || cheeseType== 2)
{
canLeave = true;
}
}
}while(canLeave);
答案 2 :(得分：0)
我相信你正在寻找一个goto语句，但Java没有这样的东西 - 尽管它是一个保留字。
但是，如果我理解正确，你可以通过使用循环和休息来完成你想做的事情。例如：
while (true) {
//ask the menu questions
System.out.println("Did you mess up? Y/N");
// get input
if (answer=='Y')//they did mess up, so run loop again
{
continue; //goes back to loop's header
}
else //they didn't mess up
{
//break out of the loop
break;
}
}