How to get the double value that is only two digit after decimal point.
For example if the a = 190253.80846153846 then the result value should be like a = 190253.80
Try: I have try with this:
public static DecimalFormat twoDForm = new DecimalFormat("#0.00");
in code
a = Double.parseDouble(twoDForm.format(((a))));
But i got the value like 190253.81 instead of that i want 190253.80
So what should i have to change for it ??
Because Math.round() Returns the closest int to the argument. The result is rounded to an integer by adding 1/2, taking the floor of the result, and casting the result to type int.
Use Math.floor()
Example
public static double roundMyData(double Rval, int numberOfDigitsAfterDecimal) {
double p = (float)Math.pow(10,numberOfDigitsAfterDecimal);
Rval = Rval * p;
double tmp = Math.floor(Rval);
System.out.println("~~~~~~tmp~~~~~"+tmp);
return (double)tmp/p;
}
Complete Source code
class ZiggyTest2{
public static void main(String[] args) {
double num = 190253.80846153846;
double round = roundMyData(num,2);
System.out.println("Rounded data: " + round);
}
public static double roundMyData(double Rval, int numberOfDigitsAfterDecimal) {
double p = (float)Math.pow(10,numberOfDigitsAfterDecimal);
Rval = Rval * p;
double tmp = Math.floor(Rval);
System.out.println("~~~~~~tmp~~~~~"+tmp);
return (double)tmp/p;
}
}
Try this,
make a object of BigDecimal
double a = 190253.80846153846;
BigDecimal bd = new BigDecimal(a);
BigDecimal res = bd.setScale(2, RoundingMode.DOWN);
System.out.println("" + res.toPlainString());
With no libraries:
a = (float) (((int)(a * 100)) / 100.0f);
or, for double:
a = (double) (((int)(a * 100)) / 100.0);
i think that is going to round of the value, check this
(float)Math.round(value * 100) / 100
from this link round of decimal number
or this example
Following code works for me.
public static double round(double value, int places) {
//here 2 means 2 places after decimal
long factor = (long) Math.pow(10, 2);
value = value * factor;
long tmp = Math.round(value);
return (double) tmp / factor;
}
Here is another way to write it that is similar to Shabbir's ^ but I think is easier to read. I actually wanted to round it; if it showed .349, I wanted to see .35 - but if you don't want that then you can use Math.floor in place of my Math.round:
public static double round(double time){
time = Math.round(100*time);
return time /= 100;
}
Here is my complete code: I am working on a program to find 3 random 1 min time intervals in a given time segment, for a research project im working on.
import java.lang.*;
import java.util.*;
class time{
public static void main (String[] args){
int time = Integer.parseInt(args[0]);
System.out.println("time is: "+time);
//randomly select one eligible start time among many
//the time must not start at the end, a full min is needed from times given
time = time - 1;
double start1 = Math.random()*time;
System.out.println(start1);
start1 = round(start1);
System.out.println(start1);
}
public static double round(double time){
time = Math.round(100*time);
return time /= 100;
}
}
to run it, with current output:
[bharthur@unix2 edu]$ java time 12
time is: 12
10.757832858914
10.76
[bharthur@unix2 edu]$ java time 12
time is: 12
0.043720864837211715
0.04
double dValue = 10.12345;
try{
String str = Double.toString(dValue*100);`
str = str.split("[.]")[0];
dValue = Double.parseDouble(str)/100;
} catch (Exception e) {
e.printStackTrace();
}
System.out.println(dValue);
Use this code. You can get the desired output you want.
