CODEWITHSUNDEEP
java
Pawan
Pawan

Reputation: 32321

How to eliminate the exponential result in this case

if the Value is something like this (0.0007 ) ends with 3 zeros after decimal , i am getting the result as 4.0E-4 .

Please tell me how to fix this

This is my program .

package com;
import java.text.DecimalFormat;
public class Test {
    public static void main(String args[]) {
        try {
            String result = "";
            Test test = new Test();
            double value = 0.0004;
            if (value < 1) {
                result = test.numberFormat(value, 4);
            } else {
                result = test.numberFormat(value, 2);
            }
            System.out.println(result);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    public String numberFormat(double d, int decimals) {
        if (2 == decimals)
            return new DecimalFormat("#,###,###,##0.00").format(d);
        else if (0 == decimals)
            return new DecimalFormat("#,###,###,##0").format(d);
        else if (3 == decimals)
            return new DecimalFormat("#,###,###,##0.000").format(d);
        else if (4 == decimals)
            new DecimalFormat("#,###,###,##0.00##").format(d);
        return String.valueOf(d);
    }

}

Upvotes: 1

Views: 280

Answers (2)

arshajii
arshajii

Reputation: 129507

How about this instead:

public static String numberFormat(double d, int decimals) {
    return String.format("%." + decimals + "f", d);
}

Seems cleaner than what you're currently doing.

Upvotes: 2

tobias_k
tobias_k

Reputation: 82899

You just forgot the return statement.

else if (4 == decimals)
    return new DecimalFormat("#,###,###,##0.00##").format(d);

So in the 4 case, you use DecimalFormat to format your number, but then return just the normal string representation for the double, after the final else.

Upvotes: 6

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