为了更清楚，请使用menuChoice == 2协助我。
我已经到了这样的程度，如果你输入负值，它会提示你再次输入，直到它为正，然后计算结果很好。但是，当我现在只输入正值时，它不计算任何东西。我已经尝试了一段时间，但我无法理解。
我需要做什么？
package finalExam;
//this is required for JOptionPane to work
import javax.swing.JOptionPane;
public class Geometry {
public static void main(String[] args) {
boolean valid = false;
int menuChoice;
do {
// create a menu and display it to the user
// then ask the user to choose an option
String menu = "1) Calculate the area of a circle\n"
+ "2) Calculate the area of a rectangle\n"
+ "3) Calculate the area of a triangle\n"
+ "4) Quit\n"
+ "Please enter your choice: (1, 2, 3, or 4)";
menuChoice = Integer.parseInt(JOptionPane.showInputDialog(menu));
if(menuChoice == 1)
{
String unknownRadius = JOptionPane.showInputDialog("What is the radius of the circle?");
if(Double.parseDouble(unknownRadius) < 0){
do{
JOptionPane.showMessageDialog(null, "Please enter positive numbers only.");
unknownRadius = JOptionPane.showInputDialog("What is the radius of the circle?");
}
while(Double.parseDouble(unknownRadius) < 0);
double knownRadius = Double.parseDouble(unknownRadius);
double circleArea = Math.pow(knownRadius, 2) * 3.14159;
JOptionPane.showMessageDialog(null, "The area of the circle is " + circleArea);
}
else if(Double.parseDouble(unknownRadius) > 0) {
double knownRadius = Double.parseDouble(unknownRadius);
double circleArea = Math.pow(knownRadius, 2) * 3.14159;
JOptionPane.showMessageDialog(null, "The area of the circle is " + circleArea);
valid = true;
}
} else if(menuChoice == 2){
String unknownLength = JOptionPane.showInputDialog("What is the length of the rectangle?");
if(Double.parseDouble(unknownLength) < 0){
do{
JOptionPane.showMessageDialog(null, "Please enter positive numbers only.");
unknownLength = JOptionPane.showInputDialog("What is the length of the rectangle?");
}
while(Double.parseDouble(unknownLength) < 0);
double knownLength = Double.parseDouble(unknownLength);
String unknownWidth = JOptionPane.showInputDialog("What is the width of the rectangle?");
if(Double.parseDouble(unknownWidth) < 0){
do{
JOptionPane.showMessageDialog(null, "Please enter positive numbers only.");
unknownWidth = JOptionPane.showInputDialog("What is the width of the rectangle?");
}
while(Double.parseDouble(unknownWidth) < 0);
double knownWidth = Double.parseDouble(unknownWidth);
double rectangleArea = knownLength * knownWidth;
JOptionPane.showMessageDialog(null, "The area of the rectangle is " + rectangleArea);
}
else if(Double.parseDouble(unknownLength) > 0){
knownLength = Double.parseDouble(unknownLength);
unknownWidth = JOptionPane.showInputDialog("What is the width of the rectangle?");
if(Double.parseDouble(unknownWidth) > 0) {
double knownWidth = Double.parseDouble(unknownWidth);
double rectangleArea = knownLength * knownWidth;
JOptionPane.showMessageDialog(null, "The area of the rectangle is " + rectangleArea);
valid = true;
}
}
}
} else if(menuChoice == 3){
String unknownBase = JOptionPane.showInputDialog("What is the base length of the triangle?");
if(Double.parseDouble(unknownBase) > 0){
double knownBase = Double.parseDouble(unknownBase);
String unknownHeight = JOptionPane.showInputDialog("What is the height of the triangle?");
if(Double.parseDouble(unknownHeight) > 0){
double knownHeight = Double.parseDouble(unknownHeight);
double triangleArea = (knownBase / 2) * knownHeight;
JOptionPane.showMessageDialog(null, "The area of the triangle is " + triangleArea);
valid = true;
}
else { JOptionPane.showMessageDialog(null, "Please enter a positive number");
JOptionPane.showInputDialog("What is the base length of the triangle?");
}
}
else { JOptionPane.showMessageDialog(null, "Please enter a positive number");
JOptionPane.showInputDialog("What is the height of the triangle?");
}
}else if(menuChoice == 4){
System.exit(0);
} else
JOptionPane.showMessageDialog(null, "Please select from the options given (1-4)!");
}
while(!valid || menuChoice != 4);
}
}
答案 0 :(得分：2)
在第48行，你的陈述if(Double.parseDouble(unknownLength) < 0){的匹配右括号位于第76行，就在} else if(menuChoice == 3){之前。
因此，从逻辑上讲，当输入负数时，您的代码仅运行menuChoice == 2部分。您应该在第一个do-while循环完成后关闭if，因为此时该数字将被（更正为）为正。
您还应该尝试格式化代码。这将使其更具可读性，您可以轻松地看到大括号在使用美化工具（例如Tutorial Point Online Java Formatter）之后不应该排在哪里。
答案 1 :(得分：1)
根据第一个长度是负还是正，你有两个不同的代码块让你自己太复杂了。基本上，您的代码如下所示：
If the length is negative then {
Ask for a new length until it's positive
Now input the width, reject negative values, do the computation,
and output it
} else { // the length is positive
Input the width, reject negative values, do the computation,
and output it
}
关于处理宽度的整个部分在代码中出现两次，这使代码变得不必要复杂，并且使它更容易出错，例如将花括号放在错误的位置（我认为这就是为什么代码不是''表现）。您可以通过重新组织代码来使自己的生活变得更加简单：
If the length is negative then {
Ask for a new length until it's positive
}
// When we get here, the length will be positive. It doesn't matter
// whether we got here because the original length was positive, or whether
// it was negative and the user entered a new value. We're going to
// continue in the same way, either way.
Input the width, reject negative values, do the computation,
and output it
（顺便说一句，如果用户输入0，我不知道你想做什么。你真的没有处理这种情况。）
答案 2 :(得分：0)
据我所知，如果我必须在问题中实现你所要求的，我可以简单地做。
（我只在 JAVA 中提供替代代码，您可以相应地修改 Japplet ）
int a=-1,b=-1;
a=sc.nextInt();
b=sc.nextInt();
if(a>=0 && b>=0){
...
//do the task
}
else{
while(a<0 || b<0){
System.out.println("negative value not allowed, enter +ve value ");
a=sc.nextInt();
b=sc.nextInt();
}
// while exited only if value both a and b are positive
//+ve value
//perform task
}