?
Solved

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

Posted on 2009-05-05
7
Medium Priority
?
6,498 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
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 4
  • 2
7 Comments
 
LVL 86

Assisted Solution

by:CEHJ
CEHJ earned 400 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 1600 total points
ID: 24302850
String s = t3.getText().trim();
n_id = (s.length()==0 ? 0 : Integer.parseInt(s));
0
Optimize your web performance

What's in the eBook?
- Full list of reasons for poor performance
- Ultimate measures to speed things up
- Primary web monitoring types
- KPIs you should be monitoring in order to increase your ROI

 
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 1600 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 1600 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.

0

Featured Post

Enroll in August's Course of the Month

August's CompTIA IT Fundamentals course includes 19 hours of basic computer principle modules and prepares you for the certification exam. It's free for Premium Members, Team Accounts, and Qualified Experts!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

INTRODUCTION Working with files is a moderately common task in Java.  For most projects hard coding the file names, using parameters in configuration files, or using command-line arguments is sufficient.   However, when your application has vi…
Are you developing a Java application and want to create Excel Spreadsheets? You have come to the right place, this article will describe how you can create Excel Spreadsheets from a Java Application. For the purposes of this article, I will be u…
Viewers will learn about if statements in Java and their use The if statement: The condition required to create an if statement: Variations of if statements: An example using if statements:
This video teaches viewers about errors in exception handling.
Suggested Courses
Course of the Month15 days, 7 hours left to enroll

743 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question