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BigDecimal Java. How to append zeros in front

My question is basically the following:
When I use a value with BigDecimal, how do I append zeros in front of a random number?
Say I want to have a number <10 following an entirely random pattern. Now i want to add zeros in front of the number, so the actual amount adds up to 10 numbers.
Here's an example:
BigDecimal num = new BigDecimal(2353);
Now I want to have that ouput:
0000002353
Is there a function that appends numbers to a BigDecimal type?
I couldn't find any.
I tried using a while loop that checks whether the number is less than ten. But I don't understand the Big Decimal well enough to actually compare integral values to the BigDecimal types.
Thanks for any help in advance!

If you use a BigInteger instead (or any integer type, such as int or long) you can format the value with
String.format("%010d", BigInteger.valueOf(2353))
The leading 0 in the format strings means pad with 0, the following 10 is the desired length...

BigDecimal is meant to be used for storing large floating point numbers. Since in a floating-point number there isn't any difference between 0000002353 and 2353, there is no reasonable way to append leading 0's to a BigDecimal just as there is no reasonable way to append leading 0's to a normal float. According to the behavior you're looking for, I would suggest using a String to store your number, and then convert to and from BigDecimal when you want to perform any operations.
To compare an integral type to a BigDecimal, first convert the variable to a BigDecimal and then call BigDecimal's compareTo method. More info is in this question.

Since you're interested in formatting the number, you might want to look at DecimalFormat class, which allows to format floating point and integer numbers according to the specified pattern.
BigDecimal num = new BigDecimal(2353);
DecimalFormat f1 = new DecimalFormat("0000000000");
DecimalFormat f2 = new DecimalFormat("0,000,000,000");
System.out.println(f1.format(num));
System.out.println(f2.format(num));
Output:
0000002353
0,000,002,353

If the maximum number of digits is 10 and only whole numbers are allowed you don't need anything more than to use long with standard formatting:
long myNumber = 123456;
System.out.printf("%010d%n", myNumber);

Formatting Double to String? [duplicate]

This question already has answers here:
Why does floating-point arithmetic not give exact results when adding decimal fractions?
(31 answers)
Closed 3 years ago.
String.format("%,.0f", 200000000000000000000000.0)
-> 199,999,999,999,999,980,000,000
why?

Understand the Double Data type - it is an approximation of amount and scale.
The Following assignment:
double d = 2.00000000000f;
will generate a value of 1.9999999 at times when printed. What you are seeing here is magnification of that. Double Data types also have a maximum (implementation-dependant) of how many places of significance they can support (upto 15 generally) - which is why the last 6 digits are all zeros (0)
For your particular solution, if you don't require Floating-point Data, stick to Integer.

It is because current processors(and most VMs) work like that if use default data types. Here it is explained in details
If you want precision use BigDecimal. This class is specifically intended for situations like this - to be used in currency related stuff and scientific calculations.
To format decimals in proper way Java has DecimalFormat
String pattern = "###,###.###";
DecimalFormat decimalFormat = new DecimalFormat(pattern);
String format = decimalFormat.format(123456789.123);
System.out.println(format); // -> 123.456.789,123
Here is nice tutorial about it
Hope it helps.

My problem has been solved
String.format("%,.0f", BigDecimal( 200000000000000000000000.0, MathContext.DECIMAL64))

How to display decimal numbers in Java? [duplicate]

This question already has answers here:
How to round a number to n decimal places in Java
(39 answers)
Closed 8 years ago.
I'm using float to display decimal numbers, but sometimes it doesn't display correct result.
For example, for 6.2/1000 the result is 0.0061999997.
I know why is this happening, but I wonder is there a way to display correct result, in this case, 0.0062?
EDIT:
How to round a number to n decimal places in Java does not answer to my question, so why did you marked my question as already been answered in other place?
Numbers I wrote are only example. In the app user can enter any number and divide / multiply number with any other number, so the result maybe won't have any decimal points, maybe it will have 4 decimals, maybe it will have 7 decimals,...

First, you need to understand that this isn't just a display issue - if you want to avoid displaying incorrect values, it helps to have the right values to start with.
You should use BigDecimal instead of float. That stores the value as an integer scaled by a factor of 10exp rather than the 2exp used by double and float.
If the BigDecimal.toString doesn't format the result the way you want, use DecimalFormat to perform the formatting instead.

Format the decimal places use this:
DecimalFormat df = new DecimalFormat();
df.setMaximumFractionDigits(decimalPlace);
String formatedValue = df.format(value);
Hope this will resolve your query.

Getting numbers after decimal Java [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Strange floating-point behaviour in a Java program
Why does JSP/JSTL division by 1000 sometimes give remainder?
I am trying to get the numbers after the decimal.
ex: 60.4 -> 0.4
Yet, when do
double a = 60.4 % 1;
it comes out to be 0.3999999999999986.
Why is this? And how could it be fixed?

Use fixed-point types
BigDecimal src = new BigDecimal("60.4");
BigDecimal a = src.remainder(BigDecimal.ONE);

You can use DecimalFormat to do your desired task.

OK here is how you can do: How to get the numbers after the decimal point? (java)
I think this is exactly what you are looking for. So essentially you can use:
double x = d - Math.floor(d);
OR
BigDecimal class for exact digits after decimal.

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);