[SOORENA]

Ok I'm new to java and I learned how to do try and catch statements and a lot of other stuff. I'm trying to catch the user from enter a negative number or a number above 100. How do I do that? Here is a part of my program where I'm putting that in:

public void findbuttonActionPerformed (ActionEvent event) throws IOException
	{
		try
		{
			value1 = Double.parseDouble (Test1_Text.getText ());
			value2 = Double.parseDouble (Test2_Text.getText ());
			value3 = Double.parseDouble (Test3_Text.getText ());
			value4 = Double.parseDouble (Exam_Text.getText ());

		}
		catch (NumberFormatException e)
		{
			JOptionPane.showMessageDialog (null, "Please Enter Right Value!!!");
			value1 = Double.parseDouble (Test1_Text.getText ());
			value2 = Double.parseDouble (Test2_Text.getText ());
			value3 = Double.parseDouble (Test3_Text.getText ());
			value4 = Double.parseDouble (Exam_Text.getText ());
		}

	}

I thought maybe it was the NumberFormatException but apparently not.

Thanks in advance.

Soorena

Edited by SOORENA, 22 May 2007 - 07:08 PM.

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i dont know that an exception is really needed to do this.

Here's of an example of how I would do this using the command line:

import java.io.*;

public class number {

   public static void main (String[] args) {

	  System.out.print("Enter a number between 0 and 100: ");

	  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

	  String inputString = null;
	  int inputInteger = 0;

	  try {
		 inputString = br.readLine();
		 inputInteger = Integer.parseInt(inputString);
	  } catch (IOException ioe) {
		 System.out.println("Error!");
		 System.exit(1);
	  }

	  // The above is NOT the exception you are looking for, it just makes sure the input # was properly recieved.

	  if (inputInteger < 0 || inputInteger > 100) {
		 System.out.println("Please enter a valid number!");
	  } else {
		 System.out.println(inputInteger);
	  }
   }
}

The part you want to focus on is:

if (inputInteger < 0 || inputInteger > 100) {
		 System.out.println("Please enter a valid number!");
	  } else {
		 System.out.println(inputInteger);
	  }

Any questions, just ask.

[stettybet0]

i dont know that an exception is really needed to do this.

Here's of an example of how I would do this using the command line:

import java.io.*;

public class number {

   public static void main (String[] args) {

	  System.out.print("Enter a number between 0 and 100: ");

	  BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

	  String inputString = null;
	  int inputInteger = 0;

	  try {
		 inputString = br.readLine();
		 inputInteger = Integer.parseInt(inputString);
	  } catch (IOException ioe) {
		 System.out.println("Error!");
		 System.exit(1);
	  }

	  // The above is NOT the exception you are looking for, it just makes sure the input # was properly recieved.

	  if (inputInteger < 0 || inputInteger > 100) {
		 System.out.println("Please enter a valid number!");
	  } else {
		 System.out.println(inputInteger);
	  }
   }
}

The part you want to focus on is:

if (inputInteger < 0 || inputInteger > 100) {
		 System.out.println("Please enter a valid number!");
	  } else {
		 System.out.println(inputInteger);
	  }

Any questions, just ask.

[coyne20]

Another alternative is to use assert statements. This may be a little too advanced for you right now. So i strongly suggest that you stick with the previous users suggestion of an if-then-else validation code block.