我对编程很陌生，因为这是我大学的第一个学期，没有先验知识。在使用Python之后，现在使用Java工作，我们正在开发一个算命计划。我遇到的主要问题是试图回到开关询问用户是否想再次播放，或者他们是否输入了8个案例之外的无效回复。还必须有一个while循环嵌套在另一个while循环中。
Scanner user = new Scanner(System.in);
System.out.println("Welcome to the Fortune Telling program.\n"); //Welcome message
System.out.print("Would you like me to tell your fortune? Type 1 for yes and any other number for no: "); //ask for yes or no to run
int Var0 = user.nextInt();
if (Var0 == 1)
{
System.out.print("Enter a number 1-8 and I will tell your fortune: "); //ask for number between 1-8 to find fortune or invalid
int Var1 = user.nextInt();
switch (Var1)
{
case 1: //case 1-8 fortunes
System.out.println("\nYou will become great if you believe in yourself.");
break;
case 2:
System.out.println("\nSerious trouble with bypass you.");
break;
case 3:
System.out.println("\nYou will travel to many exotic places in your lifetime.");
break;
case 4:
System.out.println("\nYour ability for accomplishment will follow with success.");
break;
case 5:
System.out.println("\nWhen fear hurts you, conquer it and defeat it!");
break;
case 6:
System.out.println("\nYou will be called in to fulfill a position of higher honor and responsibility.");
break;
case 7:
System.out.println("\nYour golden opportunity is coming shortly.");
break;
case 8:
System.out.println("\nIntegrity is doing the right thing, even when nobody is watching.");
break;
default:
System.out.print("That's not a valid number. Try again.\n"); //invalid number try to rerun for correct response
}
} //display next print only on case not default
System.out.print("Would you like another fortune? Type 1 for yes and any other number for no: "); //loop this back into 'switch'
int Var2= user.nextInt();
System.out.print("Thank you for trying the fortune telling program."); //Thank you message
user.close();
}
}
答案 0 :(得分：0)
您不需要嵌套循环。虽然循环工作正常。您还需要一个Var0，这也是保持/停止循环的条件。当用户输入非int的东西时，整个代码在try-catch块内部来解决问题。最后阻止扫描仪结束 - 无论是否有例外。
Scanner user = new Scanner(System.in);
try {
System.out.println("Welcome to the Fortune Telling program.\n");
System.out
.print("Would you like me to tell your fortune? Type 1 for yes and any other number for no: ");
int Var0 = user.nextInt();
while (Var0 == 1) {
System.out
.print("Enter a number 1-8 and I will tell your fortune: ");
int Var1 = user.nextInt();
switch (Var1) {
case 1: // case 1-8 fortunes
System.out
.println("\nYou will become great if you believe in yourself.");
break;
case 2:
System.out.println("\nSerious trouble with bypass you.");
break;
case 3:
System.out
.println("\nYou will travel to many exotic places in your lifetime.");
break;
case 4:
System.out
.println("\nYour ability for accomplishment will follow with success.");
break;
case 5:
System.out
.println("\nWhen fear hurts you, conquer it and defeat it!");
break;
case 6:
System.out
.println("\nYou will be called in to fulfill a position of higher honor and responsibility.");
break;
case 7:
System.out
.println("\nYour golden opportunity is coming shortly.");
break;
case 8:
System.out
.println("\nIntegrity is doing the right thing, even when nobody is watching.");
break;
default:
System.out.print("That's not a valid number. Try again.\n");
}
System.out
.print("Would you like another fortune? Type 1 for yes and any other number for no: ");
Var0 = user.nextInt();
}
System.out
.print("Thank you for trying the fortune telling program.");
} catch (Exception e) {
System.out.println("This is what you tell if user types something which is not a digit");
} finally {
user.close();
}
答案 1 :(得分：0)
一些概念上的帮助：
当你说“我需要返回开关......”时，这意味着你需要一个循环。
需要重复的部分是什么？你（或者更确切地说，你的教练）可能期望的行为是，在告诉财富之后，“你想让我告诉你的财富”这个问题会再次出现。这意味着你必须把它和它所需的一切（算命本身）放在一个循环中。
在这种情况下通常的构造是
你能想到程序中适合这种模式的部分吗？
现在接下来的事情是你需要一段时间。这是一个提示：程序希望用户输入值1-8。如果他输入'9'或'0'或其他什么，那么程序是否应该忽略这一点并再次询问他是否想要发财，还是应该坚持？
答案 2 :(得分：0)
尝试以下方法。你不需要var1
public static void main(String[] args)
{
// TODO Auto-generated method stub
Scanner user = new Scanner(System.in);
System.out.println("Welcome to the Fortune Telling program.\n"); //Welcome message
System.out.print("Would you like me to tell your fortune? Type 1 for yes and any other number for no: ");
int Var0 = 0;
while(Var0 != -1)
{
System.out.print("Enter a number 1-8 and I will tell your fortune or -1 to quit "); //ask for number between 1-8 to find fortune or invalid
Var0 = user.nextInt();
switch (Var0)
{
case 1: //case 1-8 fortunes
System.out.println("\nYou will become great if you believe in yourself.");
break;
case 2:
System.out.println("\nSerious trouble with bypass you.");
break;
case 3:
System.out.println("\nYou will travel to many exotic places in your lifetime.");
break;
case 4:
System.out.println("\nYour ability for accomplishment will follow with success.");
break;
case 5:
System.out.println("\nWhen fear hurts you, conquer it and defeat it!");
break;
case 6:
System.out.println("\nYou will be called in to fulfill a position of higher honor and responsibility.");
break;
case 7:
System.out.println("\nYour golden opportunity is coming shortly.");
break;
case 8:
System.out.println("\nIntegrity is doing the right thing, even when nobody is watching.");
break;
default:
System.out.print("That's not a valid number. Try again.\n"); //invalid number try to rerun for correct response
}
}
System.out.print("Thank you for trying the fortune telling program.");//Thank you message
user.close();
}
将代码包含在try catch块中，只需确保捕获使用扫描程序时可能发生的任何异常。关闭finally块中的扫描程序，如果使用java 1.7及更高版本，请使用try-with-resources更好。