No, there is no better way.

Actually you have an error in your pattern. What you want is:

DecimalFormat df = new DecimalFormat("#.00"); 

Note the "00", meaning exactly two decimal places.

If you use "#.##" (# means "optional" digit), it will drop trailing zeroes - ie new DecimalFormat("#.##").format(3.0d); prints just "3", not "3.00".

You need to take the floor of the double value first - then format it.

Math.floor(double)

Returns the largest (closest to positive infinity) double value that is less than or equal to the argument and is equal to a mathematical integer.

So use something like:

double v = Math.floor(res * 100) / 100.0;

Other alternatives include using BigDecimal.

public void test() {
    double d = 0.29;
    System.out.println("d=" + d);
    System.out.println("floor(d*100)/100=" + Math.floor(d * 100) / 100);
    System.out.println("BigDecimal d=" + BigDecimal.valueOf(d).movePointRight(2).round(MathContext.UNLIMITED).movePointLeft(2));
}

prints

d=0.29
floor(d*100)/100=0.28
BigDecimal d=0.29

If you want that for display purposes, use java.text.DecimalFormat:

 new DecimalFormat("#.##").format(dblVar);

If you need it for calculations, use java.lang.Math:

 Math.floor(value * 100) / 100;

Call setRoundingMode to set the RoundingMode appropriately:

String s1 = "10.126";
Double f1 = Double.parseDouble(s1);
DecimalFormat df = new DecimalFormat(".00");
df.setRoundingMode(RoundingMode.DOWN); // Note this extra step
System.out.println(df.format(f1));

Output

10.12