我想做的是让用户输入一个分子和一个分母，并显示商。我正在使用一种方法，使用Integer.parseInt和try catch来检查数字是否有效。问题是，如果try catch捕获异常并显示错误，它将继续执行代码的下一部分，而不是跳回循环的开头。
public static void main(String[] args) {
String numerator = "";
String denominator = "";
String message = "";
int num = 0;
int den = 0;
int quo = 0;
int yesNo = 0;
do
{
// Displays a JOptionPane asking the user to input a numerator.
numerator = JOptionPane.showInputDialog(null, "Please enter a numerator.", "Enter a Numerator",JOptionPane.QUESTION_MESSAGE);
if(numerator == null)// This breaks the loop if the user clicks the "cancel" button.
break;
num = IsValid(numerator); // Calls the IsValid method, passing it the String entered by the user to validate if it is a proper integer.
denominator = JOptionPane.showInputDialog(null, "Please enter a denominator.", "Enter a Denominator",JOptionPane.QUESTION_MESSAGE);
if(denominator == null) // same as above but for the denominator
break;
den = IsValid(denominator);
/*if(den != 0 && num<den)
{ // This section is unrelated to my problem
quo = num/den;
message = "Numerator: " + num + "\n" +
"Denominator: " + den + "\n" +
"Quotient: ";
JOptionPane.showMessageDialog(null, message,"Calculation",JOptionPane.INFORMATION_MESSAGE);
}*/
yesNo = JOptionPane.showConfirmDialog(null,"Would you like to make another calculation?","Continue?",JOptionPane.YES_NO_OPTION);
}while(yesNo == JOptionPane.YES_OPTION);
}
public static int IsValid(String input)
{
int num = 0; // This method is passed the String that the JOptionPane receives and using Integer.parseInt to make sure it's an integer, and assigns it to int "num".
try{num = Integer.parseInt(input);}
catch(NumberFormatException ex)
{
JOptionPane.showMessageDialog(null, "Error: Invalid number. Please enter a valid number.","Invalid Input",JOptionPane.WARNING_MESSAGE);
}
return num;
// My main issue is that it continues with the code instead of jumping back to the JOptionPane to ask the user for input a second time.
}
基本上，我想调用IsValid（numerator）并将其分配给int“ num”，如果执行了IsValid的catch语句，请跳回到do循环的开头，以便再次显示JOptionPane。如果对分母执行catch语句，则跳回到该语句而不是最开始。我无法将JOptionPane放入方法中，因为该方法将无法确定我是将分子还是分母发送给它。我对Java还是很陌生，并且可以肯定有更简单的方法可以做到这一点，因此可以提供任何帮助。谢谢！
答案 0 :(得分：0)
如果输入无效，对isValid的调用应中断您的程序。最自然的方法是让它引发异常。
然后，您的程序将捕获异常并退出。
package so20190423;
import java.awt.HeadlessException;
import javax.swing.JOptionPane;
public class Snippet {
public static void main(String[] args) {
String numerator = "";
String denominator = "";
String message = "";
int num = 0;
int den = 0;
int quo = 0;
int yesNo = 0;
try {
do
{
// Displays a JOptionPane asking the user to input a numerator.
numerator = JOptionPane.showInputDialog(null, "Please enter a numerator.", "Enter a Numerator",JOptionPane.QUESTION_MESSAGE);
if(numerator == null)// This breaks the loop if the user clicks the "cancel" button.
break;
num = IsValid(numerator); // Calls the IsValid method, passing it the String entered by the user to validate if it is a proper integer.
denominator = JOptionPane.showInputDialog(null, "Please enter a denominator.", "Enter a Denominator",JOptionPane.QUESTION_MESSAGE);
if(denominator == null) // same as above but for the denominator
break;
den = IsValid(denominator);
/*if(den != 0 && num<den)
{ // This section is unrelated to my problem
quo = num/den;
message = "Numerator: " + num + "\n" +
"Denominator: " + den + "\n" +
"Quotient: ";
JOptionPane.showMessageDialog(null, message,"Calculation",JOptionPane.INFORMATION_MESSAGE);
}*/
yesNo = JOptionPane.showConfirmDialog(null,"Would you like to make another calculation?","Continue?",JOptionPane.YES_NO_OPTION);
}while(yesNo == JOptionPane.YES_OPTION);
}
catch(NumberFormatException ex)
{
JOptionPane.showMessageDialog(null, "Error: Invalid number. Please enter a valid number.","Invalid Input",JOptionPane.WARNING_MESSAGE);
}
}
public static int IsValid(String input) throws NumberFormatException
{
int num = 0; // This method is passed the String that the JOptionPane receives and using Integer.parseInt to make sure it's an integer, and assigns it to int "num".
num = Integer.parseInt(input);
return num;
// My main issue is that it continues with the code instead of jumping back to the JOptionPane to ask the user for input a second time.
}
}
您甚至可以将try / catch限制在while循环中，以在输入失败后恢复并再次尝试！
package so20190423;
import java.awt.HeadlessException;
import javax.swing.JOptionPane;
public class Snippet {
public static void main(String[] args) {
String numerator = "";
String denominator = "";
String message = "";
int num = 0;
int den = 0;
int quo = 0;
int yesNo = 0;
do
{
try {
// Displays a JOptionPane asking the user to input a numerator.
numerator = JOptionPane.showInputDialog(null, "Please enter a numerator.", "Enter a Numerator",JOptionPane.QUESTION_MESSAGE);
if(numerator == null)// This breaks the loop if the user clicks the "cancel" button.
break;
num = IsValid(numerator); // Calls the IsValid method, passing it the String entered by the user to validate if it is a proper integer.
denominator = JOptionPane.showInputDialog(null, "Please enter a denominator.", "Enter a Denominator",JOptionPane.QUESTION_MESSAGE);
if(denominator == null) // same as above but for the denominator
break;
den = IsValid(denominator);
/*if(den != 0 && num<den)
{ // This section is unrelated to my problem
quo = num/den;
message = "Numerator: " + num + "\n" +
"Denominator: " + den + "\n" +
"Quotient: ";
JOptionPane.showMessageDialog(null, message,"Calculation",JOptionPane.INFORMATION_MESSAGE);
}*/
yesNo = JOptionPane.showConfirmDialog(null,"Would you like to make another calculation?","Continue?",JOptionPane.YES_NO_OPTION);
}
catch(NumberFormatException ex)
{
JOptionPane.showMessageDialog(null, "Error: Invalid number. Please enter a valid number.","Invalid Input",JOptionPane.WARNING_MESSAGE);
}
}while(yesNo == JOptionPane.YES_OPTION);
}
public static int IsValid(String input) throws NumberFormatException
{
int num = 0; // This method is passed the String that the JOptionPane receives and using Integer.parseInt to make sure it's an integer, and assigns it to int "num".
num = Integer.parseInt(input);
return num;
// My main issue is that it continues with the code instead of jumping back to the JOptionPane to ask the user for input a second time.
}
}
最后一个变体，如果您想重试每个输入，直到输入有效（或被用户取消）：
package so20190423;
import javax.swing.JOptionPane;
public class Snippet {
public static void main(String[] args) {
String numerator = "";
String denominator = "";
String message = "";
int num = 0;
int den = 0;
int quo = 0;
int yesNo = 0;
do {
boolean userCancelled = false;
boolean numeratorIsOk = false;
do {
try {
numerator = JOptionPane.showInputDialog(null, "Please enter a numerator.", "Enter a Numerator",
JOptionPane.QUESTION_MESSAGE);
if (numerator == null) {
userCancelled = true;
} else {
num = IsValid(numerator);
numeratorIsOk = true;
}
} catch (NumberFormatException ex) {
JOptionPane.showMessageDialog(null, "Error: Invalid number. Please enter a valid number.",
"Invalid Input", JOptionPane.WARNING_MESSAGE);
}
} while (!numeratorIsOk && !userCancelled);
if (userCancelled) {
break;
}
boolean denominatorIsOk = false;
do {
try {
denominator = JOptionPane.showInputDialog(null, "Please enter a denominator.",
"Enter a Denominator", JOptionPane.QUESTION_MESSAGE);
if (denominator == null) {
userCancelled = true;
} else {
den = IsValid(denominator);
denominatorIsOk = true;
}
/*
* if(den != 0 && num<den) { // This section is unrelated to my problem quo =
* num/den; message = "Numerator: " + num + "\n" + "Denominator: " + den + "\n"
* + "Quotient: "; JOptionPane.showMessageDialog(null,
* message,"Calculation",JOptionPane.INFORMATION_MESSAGE); }
*/
yesNo = JOptionPane.showConfirmDialog(null, "Would you like to make another calculation?",
"Continue?", JOptionPane.YES_NO_OPTION);
} catch (NumberFormatException ex) {
JOptionPane.showMessageDialog(null, "Error: Invalid number. Please enter a valid number.",
"Invalid Input", JOptionPane.WARNING_MESSAGE);
}
} while (!denominatorIsOk && !userCancelled);
if (userCancelled) {
break;
}
} while (yesNo == JOptionPane.YES_OPTION);
}
public static int IsValid(String input) throws NumberFormatException {
int num = 0; // This method is passed the String that the JOptionPane receives and using
// Integer.parseInt to make sure it's an integer, and assigns it to int "num".
num = Integer.parseInt(input);
return num;
// My main issue is that it continues with the code instead of jumping back to
// the JOptionPane to ask the user for input a second time.
}
}
HTH！