There is a similar question on SO which suggests using NumberFormat which is what I have done.
I am using the parse() method of NumberFormat.
public static void main(String[] args) throws ParseException{
DecToTime dtt = new DecToTime();
dtt.decToTime("1.930000000000E+02");
}
public void decToTime(String angle) throws ParseException{
DecimalFormat dform = new DecimalFormat();
//ParsePosition pp = new ParsePosition(13);
Number angleAsNumber = dform.parse(angle);
System.out.println(angleAsNumber);
}
The result I get is
1.93
I didn't really expect this to work because 1.930000000000E+02 is a pretty unusual looking number, do I have to do some string parsing first to remove the zeros? Or is there a quick and elegant way?
Memorize the String.format syntax so you can convert your doubles and BigDecimals to strings of whatever precision without e notation:
This java code:
double dennis = 0.00000008880000d;
System.out.println(dennis);
System.out.println(String.format("%.7f", dennis));
System.out.println(String.format("%.9f", new BigDecimal(dennis)));
System.out.println(String.format("%.19f", new BigDecimal(dennis)));
Prints:
8.88E-8
0.0000001
0.000000089
0.0000000888000000000
When you use DecimalFormat with an expression in scientific notation, you need to specify a pattern. Try something like
DecimalFormat dform = new DecimalFormat("0.###E0");
See the javadocs for DecimalFormat -- there's a section marked "Scientific Notation".
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