NumberFormatException

[Flo Powers]

Hi, I am doing input from the keyboard and want to safeguard against accidental pressing of the <enter> key without any value having been entered first. I have some simple code here (for demo purposes), and tried to check when it crashes, but don't see a pattern yet (or rather, I see it, but I don't understand it).

This, for example, does not crash, even if I enter nothing before pressing <enter>:

String strLine = SimpleIO.readLine();
System.out.println( strLine );

On the other hand, this will give a NumberFormatException:

String strLine = SimpleIO.readLine();
if( strLine != null )
{
int intLine = Integer.parseInt( strLine );
System.out.println( strLine );
}
else
System.out.println( "You did not input a number!" );

In other words (and as indicated by the name of the exception), the problem occurs when I try to convert to a number. An empty line is not null then? I doesn't seem to work with using double quotation marks either (empty string).I'm not sure how I should be checking for empty input in the conditional statement...

Thanks for any helpful comments!
Flo

[ January 25, 2005: Message edited by: Flo Powers ]
[ January 25, 2005: Message edited by: Flo Powers ]

[Flo Powers]

Hmm, I guess this might be the answer...

String strLine = SimpleIO.readLine();
 
try
{
int intLine = Integer.parseInt( strLine );
}
 
catch( NumberFormatException e)
{
strLine = "You did not input a number!";
}
 
System.out.println( strLine );

If there are better ways, please let me know.

Thanks,
Flo

[ramprasad madathil]

yes, that would be the way to go.
a small thought - if u would wish to accept numbers with decimal places, consider using

   Double.parseDouble(strLine) //in place of Integer.parseInt(strLine)

else if user enters 10.01, it would report "You did not enter a number" :-)

TX,
ram.

[Flo Powers]

Thank you, that makes a lot of sense, of course.
Flo