Function for finding out if a varible holds a number i.e., IsNumber()?

You could use something like the following.  
<%
      try{ Integer.parseInt(myvar);
                                 out.print("myvar is a number");
      }
         catch(NumberFormatException nfe){
                                out.print("myvar is not a number");
         }
%>

But hopefully someone  else has something prettier.                 rrz

I think rrz's way of catching the parse exception is the only way...

opps...
modify a bit, try to use pattern {n} but cant seem to get it work, so check base on index.
any better suggestion?

      RE re = new RE("^[0-9\\.]+$");
      if (re.match(value)) {
            int idx = value.indexOf(".");
            if (idx > -1) {
                  return !(value.indexOf(".", idx + 1) > -1);
            }
            return true;
      }

If the OP specifies his exact requirements (like whether he has to handle cases like your last example) then the proper regex can be built to handle that. I disagree that rrz's method is the best way to do it. Checking whether something is true or not by determining whether an exception is thrown is a bad design pattern. I was also going to back this argument up by mentioning that it would be more expensive to test this way but I was suprised by the results.

Running this code:

import java.util.regex.Pattern;
import java.util.regex.Matcher;

public class FindNumber {
   
    private static final Pattern p = Pattern.compile("\\d*(?:\\.?\\d+)?");
   
    public static void main(String[] args) {
        double start, end;
       
        // Use regex, isNumber1()
        start = System.currentTimeMillis();
        for (int i=0; i<1000000; i++) {
            isNumber1("55.55");
        }
        end = System.currentTimeMillis();
        System.out.println("Method using regex completed in " + (end - start)/1000 + " seconds.");
       
        // Use exception check, isNumber2()
        start = System.currentTimeMillis();
        for (int i=0; i<1000000; i++) {
            isNumber2("55.55");
        }
        end = System.currentTimeMillis();
        System.out.println("Method using exception check completed in " + (end - start)/1000 + " seconds.");
       
        // Use regex, with regex pattern precompiled, isNumber3()
        start = System.currentTimeMillis();
        for (int i=0; i<1000000; i++) {
            isNumber3("55.55");
        }
        end = System.currentTimeMillis();
        System.out.println("Method using regex with regex pattern precompiled completed in " + (end - start)/1000 + " seconds.");
    }
   
    public static boolean isNumber1(String in) {
        if (null == in || in.trim().length() == 0) return false;
        return in.matches("\\d*(?:\\.?\\d+)?");
    }
   
    public static boolean isNumber2(String in) {
        try {
            float value = Float.parseFloat(in);
            return true;
        } catch (NumberFormatException ignored) {
            return false;
        }
    }
   
    public static boolean isNumber3(String in) {
        if (null == in || in.trim().length() == 0) return false;
        Matcher m = p.matcher(in);
        return m.matches();
    }
}

Resulted in these benchmarks:

Method using regex completed in 5.579 seconds.
Method using exception check completed in 0.593 seconds.
Method using regex with regex pattern precompiled completed in 1.641 seconds.

So my last argument is out the window :P

Whatever the OP chooses to use is up to him. It seems wrong to me to use isNumber2() but the benchmarks don't lie! I really expected the regex method to be much faster. Either way, I found this topic interesting and learned something new today :)

-Pat