How do I fix the error of NumberFormatException everytime I run the program?

Question

I have to create a love calculator object for my computer science class. However, every time I compile and run the program I keep ending up with the error:

java.lang.NumberFormatException:

For input string: "70.78%" (in sun.misc.FloatingDecimal)

My code for the method:

public double calculate()
{
    double value1;
    double value2;
    double sqrt1;
    double sqrt2;
    double relationship;
    sqrt1 = Math.sqrt(name1.length());
    sqrt2 = Math.sqrt(name2.length());
    value1 = (Math.pow(name1.length(), 3)/(Math.random()+0.1))* sqrt1;
    value2 = (Math.pow(name2.length(), 3)/(Math.random()+0.1))* sqrt2;

    if(value1 > value2)
    {
        relationship = value2 / value1;
    }
    else
    {
        relationship = value1 / value2;
    }
    NumberFormat nf = NumberFormat.getPercentInstance();
    nf.setMinimumFractionDigits(2);
    return Double.parseDouble(nf.format(relationship));
}

I attempted to convert it to a float. I tried to separate it by declaring and initializing another double variable and returning that instead but they didn't work. I looked up solutions and most said to use a try and catch but I don't understand how that would work (since I just began the class and am a beginner).

How would I use a try and catch for this situation?

score 2 · Accepted Answer · answered Sep 30 '15 at 03:02

2

NumberFormat is meant to create a human readable string. 70.78% isn't a number. 70.78 is, but with the percent sign, it's a string. It seems like what you're trying to do is use the number formatting functionality to round the number. This question has some suggestions for how to properly round a number and keep it as a number.

To answer your other question, the proper way to use a try/catch would be like this:

 double result;
 try{
   result = Double.parseDouble(nf.format(relationship));
 }catch(NumberFormatException e){
   e.printStackTrace();
   result = 0.0;
 }
 return result;

But the only thing that will do is cause your program to not crash and you'll always get 0.0 returned from the calculate() method. Instead you need to fix the source of the exception.

answered Sep 30 '15 at 03:02

Chris Thompson

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Yeah I'm trying to round the number and keep it as a number, but also include the % sign. Is it possible to do this? – immmjimmy Sep 30 '15 at 03:05
@immmjimmy No. A `double` cannot store a percent sign. – Andreas Sep 30 '15 at 03:06
You'd need to return a `String` instead. A `double` is _only_ the number – Chris Thompson Sep 30 '15 at 03:10
So would it be possible to return result + "%"? EDIT: I just saw your answer Chris – immmjimmy Sep 30 '15 at 03:12

James Jithin · Answer 2 · 2015-09-30T03:30:41.013

0

As you see the exception stacktrace, the exception is caused by string 70.78%. This is because it has a symbol where Double.parseDouble() is expecting a string in format of double value representation like 70.78. If you are looking for an output like 70.78% and if the value is correct, you may return nf.format(relationship) and change your return type of calculate to String.

Update

The following values are valid to be parsed as double: "1.2", "1", ".2", "0.2", "1.2D", "1.", to say a valid number representation holdable by the datatype.

edited Sep 30 '15 at 03:30

answered Sep 30 '15 at 03:11

James Jithin

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Changing it to String did work. When would a double/int/float not work in String? – immmjimmy Sep 30 '15 at 03:15

How do I fix the error of NumberFormatException everytime I run the program?

2 Answers2

Update