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java.lang.NumberFormatException: For input string: "" problem...

Posted on 2009-05-05
7
6,478 Views
Last Modified: 2013-12-22
Hi there;

What is the problem with the following code? i am getting number format exception but the code seems true...How to fix it without haing drastic changes?

As snippet:

Best regards...
Student.txt
StudentEx.txt
0
Comment
Question by:jazzIIIlove
  • 4
  • 2
7 Comments
 
LVL 86

Assisted Solution

by:CEHJ
CEHJ earned 100 total points
ID: 24302842
You need to check that the contents of 't3' is a valid number before attempting to parse it or catch the exception (the former is better)
0
 
LVL 92

Expert Comment

by:objects
ID: 24302843
>        n_id=Integer.parseInt(t3.getText().trim());

need to check the string isn't empty

0
 
LVL 92

Accepted Solution

by:
objects earned 400 total points
ID: 24302850
String s = t3.getText().trim();
n_id = (s.length()==0 ? 0 : Integer.parseInt(s));
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LVL 12

Author Comment

by:jazzIIIlove
ID: 24302963
sorry:
objects:
>>String s = t3.getText().trim();
>>n_id = (s.length()==0 ? 0 : Integer.parseInt(s));

Below is same as above? (I am not used to that ? format)...Could you clarify?

if(s.length() == 0)
            n_id = Integer.parseInt(s);
        else n_id = 0;
0
 
LVL 92

Assisted Solution

by:objects
objects earned 400 total points
ID: 24302980
almost :)

if(s.length() == 0)
  n_id = 0;
else
   n_id = Integer.parseInt(s);

0
 
LVL 86

Expert Comment

by:CEHJ
ID: 24303013
The easiest way to avoid exceptions of any kind would probably be
if ("" + t3.getText().matches("\\d+")) {
    // valid
}
else {
   // don't bother trying
}

Open in new window

0
 
LVL 92

Assisted Solution

by:objects
objects earned 400 total points
ID: 24303045
If you really want to check if its an int then use the following

http://helpdesk.objects.com.au/java/how-to-check-if-a-string-is-an-integer-using-a-regular-expression

expensive operation though and not necessary in your case.

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