如何在JAVA的差异原因上为相同的NumberFormatException打印差异消息？

Question

How to print diff Msg for same NumberFormatException on diff cause in JAVA?

try {
 int a=Integer.parseInt(aStr);
int b= Integer.parseInt(bStr);
}catch (NumberFormatException ex) { 
 if ex's cause is from int a;//ex.getCause()=a?
System.out.println("a is not a integer"); 
 if ex's cause is from int b
System.out.println("b is not a integer");
}

Answer 1

      try {
         final int a = Integer.parseInt(aStr);
      } catch (final NumberFormatException ex) {
         System.out.println("a is not a integer");
      }
      try {
         final int b = Integer.parseInt(bStr);
      } catch (final Exception e) {
         System.out.println("b is not a integer");
      }

Answer 2

You can declare the variables in two different try catch...

try {
 int a=Integer.parseInt(aStr);
}catch (NumberFormatException ex) { 

System.out.println("a is not a integer"); 
}
try{
int b= Integer.parseInt(bStr);
}catch (NumberFormatException ex) { 

System.out.println("b is not a integer");
}

Instead of doing that you can keep you try block unchanged and in the catch block print the stack trace by doing this

ex.printStackTrace();

This will give you the line number where the exception occurred which will either be at variable a or b

Answer 3

好吧，它仍然提供了不错的消息，如果您想拥有两个捕获块

Answer 4

Another possibility.

Introduce a string variable before the try block:

String msg = "a is not an integer";
try {
    // parse a
    msg = "b is not an integer";
    // parse b
} catch (...) { println(msg); }

Answer 5

The only alternative to two try catch block would be setting a marker

 boolean aSet = false;
 try{
   int a = Integer.parseInt(aStr);
   aSet = true;
int b = Integer.parseInt(bStr);
 } catch (NumberFormatException ex) {
   if (aset) {....