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

I need to know if a variable contains a number or not.
I have some conditional formatiing that is taking place, and the variable could either be set to nothing, null, or a number.  Is there a JSP function like this?

String myvar = null
myvar = request.getParameter("myvar");
if(myvar.IsNumber()){
     //Do Something
}else{
     //Do Something else...
}

Thanks in advance,
-MD
mderbinAsked:
Who is Participating?
 
rrzConnect With a Mentor Commented:
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
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rrzCommented:
Did you try that code ? won't it throw a null pointer ?  

String myvar = null
myvar = request.getParameter("myvar");
if(myvar != null && myvar.IsNumber()){
     //Do Something
}else{
     //Do Something else...
}  

Maybe  you  should look at JSTL if you are looking for functions.   rrz
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rrzCommented:
Here is old article ( we are using JSTL 1.0 ).
http://java.sun.com/developer/technicalArticles/javaserverpages/faster/
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rrzCommented:
>we are using JSTL 1.0 ).
No, we are using  1.1  now  !
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rrzCommented:
Or we could use
Float.parseFloat(myvar);
in same way as the above code.
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TimYatesConnect With a Mentor Commented:
I think rrz's way of catching the parse exception is the only way...
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boonlengCommented:
You can use regex to check whther the value is a number:
JDK 1.3 needs to use org.apache.regexp.RE, JDK 1.4 can do regex in String object.

RE re = new RE("^[0-9\\.]+$");
if (re.match(value)) {
}
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TimYatesCommented:
That doesn't work, as

     System.out.println( java.util.regex.Pattern.matches( "^[0-9\\.]+$", "0.3.2") ) ;

prints "true"
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boonlengConnect With a Mentor Commented:
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;
      }
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TimYatesCommented:
> any better suggestion?

yes, the way rrz@871311 does it ;-)
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boonlengCommented:
hmm... true also :)
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pat5starCommented:
How about this?

    public static boolean isNumber(String in) {
        if (null == in || in.trim().length() == 0) return false;
        return in.matches("\\d+");
    }

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TimYatesCommented:
>  How about this?

Will that work for 5.3?
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pat5starCommented:
No, but this will:

    public static boolean isNumber(String in) {
        if (null == in || in.trim().length() == 0) return false;
        return in.matches("\\d*(?:\\.?\\d+)?");
    }

This allows any number. If the variable contains a decimal it must have at least 1 number following it. Here are some example results:

.321 = true
2.45 = true
23.1 = true
23.   = false
230  = true
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TimYatesCommented:
what about

9.0E-3

?

;-)

My point here, is that the best way of doing it is rrz's way that he posted first ;-)
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pat5starConnect With a Mentor Commented:
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
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pat5starCommented:
Those results were bugging me today so I thought about it more and realized what made the exception check method so much faster. The test was with a valid float so the isNumber2() method never once threw an exception. If you use a value for the test that isn't a valid float, the benchmarks show a whole different picture:

Method using regex completed in 5.796 seconds.
Method using exception check completed in 9.125 seconds.
Method using regex with regex pattern precompiled completed in 1.891 seconds.

That makes isNumber3() look a whole lot better :)

-Pat

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rrzCommented:
>Is there a JSP function like this?
There probably should be one.  
Javascript has isNaN() .
Why not validate before  submission ?
Here is another way.

    public static boolean isNumber(String var){
                          int pointCount = 0;
                          for(int i=0;i<var.length();i++){
                                   char c = var.charAt(i);
                                   if(var.charAt(i) == '.')pointCount++;
                                   if(!Character.isDigit(c)|| pointCount > 1) return false;
                          }
                          return true;
    }
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