我试图要求用户输入是或否，如果是，那么它将重新启动，并且将添加特征，因为他们做了 2 个选择。然而，它不允许我问是/否，而且我显然在某个地方搞砸了。如果可能的话，我如何才能最大限度地减少代码，而不是使用那个令人眼花缭乱的优化不佳的 if 语句。
public static void main(String[]args)
{
//Number 1
Scanner scan = new Scanner(System.in);
int num = 0;
int num2 = 0;
int num3 = 0;
int count = 2;
String yesno;
do
{
System.out.println("Which aircraft would you like to simulate?");
System.out.println("1. Blimp");
System.out.println("2. Helicopter");
System.out.println("3. Fighter Jet");
System.out.println("4. Space Shuttle");
num = scan.nextInt();
if(num<1 || num>4) {System.out.println("Invalid");}
}
while(num<1 || num>4);
do
{
System.out.println("What characteristics would you like? (Input one, then the other)");
System.out.println("1. Position Trim ");
System.out.println("2. Force Breakout");
System.out.println("3. Force Gradient");
System.out.println("4. Force Friction");
System.out.println("5. Damping");
System.out.println("6. Hard Stop");
num2 = scan.nextInt();
num3 = scan.nextInt();
if(num2 == 1 || num3 == 1)
{
System.out.println("The position to which a flight control returns");
}
if(num2 == 2 ||num3 == 2)
{
System.out.println("A force that returns a control to Trim. This is a constant force applied toward Trim which remains the same despite how far the control is moved (displacement) and how fast a control is moved (velocity).");
}
if(num2 == 3 || num3 == 3)
{
System.out.println("A force that returns a control to Trim, but one that varies with displacement. The farther the control is moved, the stronger the force applied toward trim.");
}
if (num2 == 4 || num3 == 4)
{
System.out.println("A constant force that is opposite to the direction of movement");
}
if(num2 == 5 || num3 == 5)
{
System.out.println("A force that is oppisite to the direction of movement. Damping varies with velocity. The faster a control is moved the stronger the force.");
}
if (num2 == 6 || num3 == 6)
{
System.out.println("A force that simulates a mechanical limit of travel. By varying the Hard Stops, the range of travel can be adjusted");
}
if(num2 < 1 || num2 > 6)
{
System.out.println("Invalid input");
}
if (yesno.equalsIgnoreCase("y")) {
do
{
System.out.println("What characteristics would you like? (Input one, then the other)");
System.out.println("1. Position Trim ");
System.out.println("2. Force Breakout");
System.out.println("3. Force Gradient");
System.out.println("4. Force Friction");
System.out.println("5. Damping");
System.out.println("6. Hard Stop");
num2 = scan.nextInt();
num3 = scan.nextInt();
if(num2 == 1 || num3 == 1)
{
System.out.println("The position to which a flight control returns");
}
if(num2 == 2 ||num3 == 2)
{
System.out.println("A force that returns a control to Trim. This is a constant force applied toward Trim which remains the same despite how far the control is moved (displacement) and how fast a control is moved (velocity).");
}
if(num2 == 3 || num3 == 3)
{
System.out.println("A force that returns a control to Trim, but one that varies with displacement. The farther the control is moved, the stronger the force applied toward trim.");
}
if (num2 == 4 || num3 == 4)
{
System.out.println("A constant force that is opposite to the direction of movement");
}
if(num2 == 5 || num3 == 5)
{
System.out.println("A force that is oppisite to the direction of movement. Damping varies with velocity. The faster a control is moved the stronger the force.");
}
if (num2 == 6 || num3 == 6)
{
System.out.println("A force that simulates a mechanical limit of travel. By varying the Hard Stops, the range of travel can be adjusted");
}
if(num2 < 1 || num2 > 6)
{
System.out.println("Invalid input");
}
System.out.println("Would you like to re-select?(Y/N)");
yesno = scan.nextLine();
count += 2;
}
while(num2 < 1 || num2 > 6 || num3 < 1 ||num3 > 6 );
}
else if (yesno.equalsIgnoreCase("n"))
{
System.out.println("You've used " + count + " characteristics");
}
}
while(num2 < 1 || num2 > 6 || num3 < 1 ||num3 > 6 );
}
}
最佳答案
您的代码可以压缩一些，但主要问题是 yesno 变量从未初始化，因此条件 if (yesno.equalsIgnoreCase("y")) 计算结果为 false 并且后续 block 从不计算，因此这些行
System.out.println("Would you like to re-select?(Y/N)");
yesno = scan.nextLine();
永远不会被评估，也永远不会询问用户。如果您保留代码不变，则应在创建代码时设置 yesno="y"。但是，就像我说的，您的代码可以简化。
关于java - 请帮忙，在 while 循环中询问某些问题时遇到问题，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/22677526/