| | List |
| Subject: | Re: check if a variable is number |
| Poster: | Lewlew@nospam.lewscanon.com |
| Date: | Fri, 23 Mar 2007 17:01:19 -0500 |
| Related Postings: | 1 2 3 |
CodeForTea@gmail.com wrote:
> On Mar 23, 10:14 am, "Andrew Thompson" wrote:
>> On Mar 24, 12:54 am, "Lara" wrote:
>> ...
>>
>>> String firstnum =JOptionPane.showInputDialog("Enter first number");
>>> How can I check if (firstnum) is number or not befor converting the
>>> String to Integer??
>> You do not have to. You might try it,
>> and catch the result of failure. If it
>> fails, it means the string does not
>> represent an integer.
>>
>> Andrew T.
>
> private static Pattern p = Pattern.compile("\\d+");
>
>
> public static void main(String[] args) {
>
> String s = "3456";
> Matcher m = p.matcher(s);
> int number = 0;
> if (m.matches()) {
> number = Integer.parseInt(s);
> System.out.println("The number entered is = " + number);
> } else {
> System.out.println("Not a number = " + s);
> }
> int number2 = 0;
> String s2 = "x3456";
> m = p.matcher(s2);
> if (m.matches()) {
> number = Integer.parseInt(s);
> System.out.println("The numner entered is = " + number2);
> } else {
> System.out.println("Not a number = " + s2);
> }
>
> }
String s = obtainSomeValueThatMightBeNumeric();
int n;
try
{
n = Integer.parseInt( s );
System.err.println( "Yes a number: \""+ s +"\" = "+ n );
}
catch ( NumberFormatException e )
{
System.err.println( "not a number: \""+ s +"\"" );
}
-- Lew
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