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Java Discussion Thread
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Oct 22nd, 2007

Help with GUI

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Hi everybody, I'm new to GUI and was wondering if someone could help me with my fraction calculator if they input text in the fields or if they leave something blank. I'm not quite sure how to do that. Here's my code, and if someone could guide me to how to do it, I'd be very thankful.

(Toggle Plain Text)

fractionNumeratorOne = (int)(Integer.parseInt(firstNumerator.getText()));
      fractionDenominatorOne = (int)(Integer.parseInt(firstDenominator.getText()));
      fractionNumeratorTwo = (int)(Integer.parseInt(secondNumerator.getText()));
      fractionDenominatorTwo = (int)(Integer.parseInt(secondDenominator.getText()));
      if (fractionDenominatorOne==0 || fractionDenominatorTwo==0)
      {
        JOptionPane.showMessageDialog(null, "Zero isn't the right number, it will equal a positive infinity","Error", JOptionPane.ERROR_MESSAGE);
       
      }
      if(fractionNumeratorOne==)      {
      	JOptionPane.showMessageDialog(null, "Zero isn't the right number, it will equal a positive infinity","Error", JOptionPane.ERROR_MESSAGE);
      }

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PL.P27

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Oct 23rd, 2007

Re: Help with GUI

basically you need to check if they have a value. if the enter nothing then the string will be null or "". If you try to set a integer equal to null, you will get an exception, so you need to check the value of the fraction inputs before you turn them into integers.
do so this simly use an if statement

java Syntax (Toggle Plain Text)
if(firstDenominator.getText().equals(""){
//handle null value
}
if(firstDenominator.getText().equals(""){
//handle null value
}

hope that helps

nschessnerd

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Posting Whiz in Training

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Oct 23rd, 2007

Re: Help with GUI

Thanks a lot, what if they enter letters, how would I change the if statement?

PL.P27

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Oct 23rd, 2007

Re: Help with GUI

hmm,
There is no clean way i can think of, but you can do

java Syntax (Toggle Plain Text)
try{
int num=Integer.parseInt(firstDenominator.getText());
//if it gets to here its valid
}catch(NumberFormatException e){
//invalid number
}
try{
int num=Integer.parseInt(firstDenominator.getText());
//if it gets to here its valid
}catch(NumberFormatException e){
//invalid number
}

Last edited by nschessnerd; Oct 23rd, 2007 at 12:53 am.

nschessnerd

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Oct 23rd, 2007

Re: Help with GUI

also how about if they enter a decimal

PL.P27

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Oct 23rd, 2007

Re: Help with GUI

You can do Double.parseDouble instead of Integer.parseInt

nschessnerd

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Oct 23rd, 2007

Re: Help with GUI

Thanks a lot everything really helped. I appreciate the time that you have given to help me. May you have a good day!

PL.P27

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Oct 23rd, 2007

Re: Help with GUI

you too :-)

nschessnerd

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