Develop Reference

Java formatting decimals to 2 digits precision

Right now I am doing this
DecimalFormat df = new DecimalFormat("#.##");
//usage: df.format(someDouble);
But I don't know if this is the correct way to "round off" the display of a double to 2 decimal points. Will this work even if the number of digits to the left of the decimal point > 1?

You should use:
DecimalFormat df = new DecimalFormat("#.00");
DecimalFormat df = new DecimalFormat("#.##"); means, that those decimal places are optional. You can see the difference when applying on 1.2 or 1.0.
--Documentation

To answer the question about the rounding, the documentation tell it :
Rounding
DecimalFormat provides rounding modes defined in RoundingMode for formatting. By default, it uses RoundingMode.HALF_EVEN.
So this is up to you to set the RoundingMode that you want with the setter.
public void setRoundingMode(RoundingMode roundingMode)

Will this work even if the number of digits to the left of the decimal point > 1?
Yes it will.

Java BigDecimal from String without decimal places but asume last two digits as decimal

A have a String: "1234567"
I want to transfer this String to a BigDecimal with this format:
12345.67
--> The last two digets shall be treated as decimal places.
Is there any method for that, or do I have to use a resource consuming way like:
String numberUnformated = 1234567;
BigDecimal number = new BigDecimal(amount.substring(0,amount.length()-2)+"."+amount.substring(amount.length()-2))
Thank you

You could use
BigDecimal number = new BigDecimal("1234567") .movePointLeft(2):
Don't use doubles as suggested because you would certainly lose precision and with BigDecimal you won't. It's important when working with currencies.

You could divide it by 100 like this:
BigDecimal number = new BigDecimal("1234567")
.divide(BigDecimal.valueOf(100L));

Why Float return different value and how can i solve it, Java

I am using Float to saving value. But when i save it seven and more then value in Float its return me different format value for example..
Float f=1234567.98
then float return me in string
1.23457e+06
What means of this and how can i solve it. Kindly please help me about this
Thanks in Advance

It means one of two things. Either your preferred printing method only shows a certain number of digits by default, or you are requesting more precision than a Float offers. The wikipedia page for single precision floats says that you can expect 6-9 decimal digits will be represented accurately. So that's probably your problem. You should be able to solve it by using a Double to store your value and DecimalFormat to customize the output.

To format a decimal use:
http://docs.oracle.com/javase/tutorial/i18n/format/decimalFormat.html
In this case the format you want is ###.##
An example of using this would be
float f = 123456.78f;
DecimalFormat format = new DecimalFormat("###.##');
String formatted = format.format(f);
Note that if you are using this for monetary calculations you should use BigDecimal instead. Floats are an efficient way of storing real numbers, but they can only represent so many. Here is an example of using BigDecimal:
BigDecimal bd = new BigDecimal(123456.78f);
bd = bd.add(3.4f);
bd = bd.divide(4.3f);//this division with floats may have returned an inexact answer
bd.setScale(2);//Makes sure we only care about 2 decimal places
String strValue = bd.toString();

Java DecimalFormat to double

I'm trying to prevent a java double from printing in scientific notation. I know I can use DecimalFormat for this, but this produces a String. How can I produce an actual double?
I've tried:
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(8);
double n = Double.parseDouble(df.format(z));
However, this still produces a double in scientific notation.

A format is not a property of a double, the data type. This defines just a set of values.
Printing always requires a conversion to a String, so DecimalFormat is fine.
As for using an entirely different set of decimal fractions as values, see java.math.BigDecimal. This is the set of "all" decimal fractions, with a dynamic number of fractional digits, operations and control of round-off.

Setting precision in Java

If I want a double to have 9 decimal places, do I have to convert it to a string and then back to a double to do this (string methods are the only methods I'm seeing for setting the precision)?. In any case, what is the conventional way for setting the precision for a double if, for example, I want my method to return a double with 9 decimal places.

You're getting confused between a double's internal representation and it's display. Internally double numbers are always stored the same way but you can of course use a formater like DecimalFormat to return 9 decimal points from your double number.

Use Big Decimal to get 9 decimal places
or
double d = 1.2345672626346;
DecimalFormat df = new DecimalFormat("#.#########");
System.out.print(df.format(d));

You can't control precision with primitives in Java. You need to use BigDecimal. Read this excellent tutorial..

You could also use:
DecimalFormat df = DecimalFormat.getInstance(Locale.getDefault());
df.setMinimumFractionDigits(9); //sets 9 digits after the '.'

String formatedString = String.format("%.2f");
it set the presition to 5 digit o any.
if you only want to print it use this way.
System.out.printf("%.5f",123.2342622467);