in Java, how to delete all 0s in float? - java

I'd like to change float like this way:
10.5000 -> 10.5
10.0000 -> 10
How can I delete all zeros after the decimal point, and change it either float (if there's non-zeros) or int (if there were only zeros)?
Thanks in advance.


Why not try regexp?
new Float(10.25000f).toString().replaceAll("\\.?0*$", "")


Well the trick is that floats and doubles themselves don't really have trailing zeros per se; it's just the way they are printed (or initialized as literals) that might show them. Consider these examples:
Float.toString(10.5000); // => "10.5"
Float.toString(10.0000); // => "10.0"
You can use a DecimalFormat to fix the example of "10.0":
new java.text.DecimalFormat("#").format(10.0); // => "10"


java.math.BigDecimal has a stripTrailingZeros() method, which will achieve what you're looking for.
BigDecimal myDecimal = new BigDecimal(myValue);
myDecimal.stripTrailingZeros();
myValue = myDecimal.floatValue();


This handles it with two different formatters:
double d = 10.5F;
DecimalFormat formatter = new DecimalFormat("0");
DecimalFormat decimalFormatter = new DecimalFormat("0.0");
String s;
if (d % 1L > 0L) s = decimalFormatter.format(d);
else s = formatter.format(d);
System.out.println("s: " + s);


You just need to use format class like following:
new java.text.DecimalFormat("#.#").format(10.50000);


Format your numbers for your output as required. You cannot delete the internal "0" values.


Try using System.out.format
Heres a link which allows c style formatting
http://docs.oracle.com/javase/tutorial/java/data/numberformat.html


I had the same issue and find a workaround in the following link:
StackOverFlow - How to nicely format floating numbers to string without unnecessary decimal 0
The answer from JasonD was the one I followed. It's not locale-dependent which was good for my issue and didn't have any problem with long values.
Hope this help.
ADDING CONTENT FROM LINK ABOVE:
public static String fmt(double d) {
if(d == (long) d)
return String.format("%d",(long)d);
else
return String.format("%s",d);
}
Produces:
232
0.18
1237875192
4.58
0
1.2345

Related

Is Java assigning incorrect value to double variable? [duplicate]

I want to print a double value in Java without exponential form.
double dexp = 12345678;
System.out.println("dexp: "+dexp);
It shows this E notation: 1.2345678E7.
I want it to print it like this: 12345678
What is the best way to prevent this?

Java prevent E notation in a double:
Five different ways to convert a double to a normal number:
import java.math.BigDecimal;
import java.text.DecimalFormat;
public class Runner {
public static void main(String[] args) {
double myvalue = 0.00000021d;
//Option 1 Print bare double.
System.out.println(myvalue);
//Option2, use decimalFormat.
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(8);
System.out.println(df.format(myvalue));
//Option 3, use printf.
System.out.printf("%.9f", myvalue);
System.out.println();
//Option 4, convert toBigDecimal and ask for toPlainString().
System.out.print(new BigDecimal(myvalue).toPlainString());
System.out.println();
//Option 5, String.format
System.out.println(String.format("%.12f", myvalue));
}
}
This program prints:
2.1E-7
.00000021
0.000000210
0.000000210000000000000001085015324114868562332958390470594167709350585
0.000000210000
Which are all the same value.
Protip: If you are confused as to why those random digits appear beyond a certain threshold in the double value, this video explains: computerphile why does 0.1+0.2 equal 0.30000000000001?
http://youtube.com/watch?v=PZRI1IfStY0

You could use printf() with %f:
double dexp = 12345678;
System.out.printf("dexp: %f\n", dexp);
This will print dexp: 12345678.000000. If you don't want the fractional part, use
System.out.printf("dexp: %.0f\n", dexp);
0 in %.0f means 0 places in fractional part i.e no fractional part. If you want to print fractional part with desired number of decimal places then instead of 0 just provide the number like this %.8f. By default fractional part is printed up to 6 decimal places.
This uses the format specifier language explained in the documentation.
The default toString() format used in your original code is spelled out here.

In short:
If you want to get rid of trailing zeros and Locale problems, then you should use:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); // 340 = DecimalFormat.DOUBLE_FRACTION_DIGITS
System.out.println(df.format(myValue)); // Output: 0.00000021
Explanation:
Why other answers did not suit me:
Double.toString() or System.out.println or FloatingDecimal.toJavaFormatString uses scientific notations if double is less than 10^-3 or greater than or equal to 10^7
By using %f, the default decimal precision is 6, otherwise you can hardcode it, but it results in extra zeros added if you have fewer decimals. Example:
double myValue = 0.00000021d;
String.format("%.12f", myvalue); // Output: 0.000000210000
By using setMaximumFractionDigits(0); or %.0f you remove any decimal precision, which is fine for integers/longs, but not for double:
double myValue = 0.00000021d;
System.out.println(String.format("%.0f", myvalue)); // Output: 0
DecimalFormat df = new DecimalFormat("0");
System.out.println(df.format(myValue)); // Output: 0
By using DecimalFormat, you are local dependent. In French locale, the decimal separator is a comma, not a point:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0");
df.setMaximumFractionDigits(340);
System.out.println(df.format(myvalue)); // Output: 0,00000021
Using the ENGLISH locale makes sure you get a point for decimal separator, wherever your program will run.
Why using 340 then for setMaximumFractionDigits?
Two reasons:
setMaximumFractionDigits accepts an integer, but its implementation has a maximum digits allowed of DecimalFormat.DOUBLE_FRACTION_DIGITS which equals 340
Double.MIN_VALUE = 4.9E-324 so with 340 digits you are sure not to round your double and lose precision.

You can try it with DecimalFormat. With this class you are very flexible in parsing your numbers.
You can exactly set the pattern you want to use.
In your case for example:
double test = 12345678;
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(0);
System.out.println(df.format(test)); //12345678

I've got another solution involving BigDecimal's toPlainString(), but this time using the String-constructor, which is recommended in the javadoc:
this constructor is compatible with the values returned by Float.toString and Double.toString. This is generally the preferred way to convert a float or double into a BigDecimal, as it doesn't suffer from the unpredictability of the BigDecimal(double) constructor.
It looks like this in its shortest form:
return new BigDecimal(myDouble.toString()).stripTrailingZeros().toPlainString();
NaN and infinite values have to be checked extra, so looks like this in its complete form:
public static String doubleToString(Double d) {
if (d == null)
return null;
if (d.isNaN() || d.isInfinite())
return d.toString();
return new BigDecimal(d.toString()).stripTrailingZeros().toPlainString();
}
This can also be copied/pasted to work nicely with Float.
For Java 7 and below, this results in "0.0" for any zero-valued Doubles, so you would need to add:
if (d.doubleValue() == 0)
return "0";

Java/Kotlin compiler converts any value greater than 9999999 (greater than or equal to 10 million) to scientific notation ie. Epsilion notation.
Ex: 12345678 is converted to 1.2345678E7
Use this code to avoid automatic conversion to scientific notation:
fun setTotalSalesValue(String total) {
var valueWithoutEpsilon = total.toBigDecimal()
/* Set the converted value to your android text view using setText() function */
salesTextView.setText( valueWithoutEpsilon.toPlainString() )
}

This will work as long as your number is a whole number:
double dnexp = 12345678;
System.out.println("dexp: " + (long)dexp);
If the double variable has precision after the decimal point it will truncate it.

I needed to convert some double to currency values and found that most of the solutions were OK, but not for me.
The DecimalFormat was eventually the way for me, so here is what I've done:
public String foo(double value) //Got here 6.743240136E7 or something..
{
DecimalFormat formatter;
if(value - (int)value > 0.0)
formatter = new DecimalFormat("0.00"); // Here you can also deal with rounding if you wish..
else
formatter = new DecimalFormat("0");
return formatter.format(value);
}
As you can see, if the number is natural I get - say - 20000000 instead of 2E7 (etc.) - without any decimal point.
And if it's decimal, I get only two decimal digits.

I think everyone had the right idea, but all answers were not straightforward.
I can see this being a very useful piece of code. Here is a snippet of what will work:
System.out.println(String.format("%.8f", EnterYourDoubleVariableHere));
the ".8" is where you set the number of decimal places you would like to show.
I am using Eclipse and it worked no problem.
Hope this was helpful. I would appreciate any feedback!

The following code detects if the provided number is presented in scientific notation. If so it is represented in normal presentation with a maximum of '25' digits.
static String convertFromScientificNotation(double number) {
// Check if in scientific notation
if (String.valueOf(number).toLowerCase().contains("e")) {
System.out.println("The scientific notation number'"
+ number
+ "' detected, it will be converted to normal representation with 25 maximum fraction digits.");
NumberFormat formatter = new DecimalFormat();
formatter.setMaximumFractionDigits(25);
return formatter.format(number);
} else
return String.valueOf(number);
}

This may be a tangent.... but if you need to put a numerical value as an integer (that is too big to be an integer) into a serializer (JSON, etc.) then you probably want "BigInterger"
Example:
value is a string - 7515904334
We need to represent it as a numerical in a Json message:
{
"contact_phone":"800220-3333",
"servicer_id":7515904334,
"servicer_name":"SOME CORPORATION"
}
We can't print it or we'll get this:
{
"contact_phone":"800220-3333",
"servicer_id":"7515904334",
"servicer_name":"SOME CORPORATION"
}
Adding the value to the node like this produces the desired outcome:
BigInteger.valueOf(Long.parseLong(value, 10))
I'm not sure this is really on-topic, but since this question was my top hit when I searched for my solution, I thought I would share here for the benefit of others, lie me, who search poorly. :D

use String.format ("%.0f", number)
%.0f for zero decimal
String numSring = String.format ("%.0f", firstNumber);
System.out.println(numString);

I had this same problem in my production code when I was using it as a string input to a math.Eval() function which takes a string like "x + 20 / 50"
I looked at hundreds of articles... In the end I went with this because of the speed. And because the Eval function was going to convert it back into its own number format eventually and math.Eval() didn't support the trailing E-07 that other methods returned, and anything over 5 dp was too much detail for my application anyway.
This is now used in production code for an application that has 1,000+ users...
double value = 0.0002111d;
String s = Double.toString(((int)(value * 100000.0d))/100000.0d); // Round to 5 dp
s display as: 0.00021

This will work not only for a whole numbers:
double dexp = 12345678.12345678;
BigDecimal bigDecimal = new BigDecimal(Double.toString(dexp));
System.out.println("dexp: "+ bigDecimal.toPlainString());

My solution:
String str = String.format ("%.0f", yourDouble);

For integer values represented by a double, you can use this code, which is much faster than the other solutions.
public static String doubleToString(final double d) {
// check for integer, also see https://stackoverflow.com/a/9898613/868941 and
// https://github.com/google/guava/blob/master/guava/src/com/google/common/math/DoubleMath.java
if (isMathematicalInteger(d)) {
return Long.toString((long)d);
} else {
// or use any of the solutions provided by others, this is the best
DecimalFormat df =
new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); // 340 = DecimalFormat.DOUBLE_FRACTION_DIGITS
return df.format(d);
}
}
// Java 8+
public static boolean isMathematicalInteger(final double d) {
return StrictMath.rint(d) == d && Double.isFinite(d);
}

This works for me. The output will be a String.
String.format("%.12f", myvalue);

Good way to convert scientific e notation
String.valueOf(YourDoubleValue.longValue())

tinylog formatting for double values [duplicate]

I want to print a double value in Java without exponential form.
double dexp = 12345678;
System.out.println("dexp: "+dexp);
It shows this E notation: 1.2345678E7.
I want it to print it like this: 12345678
What is the best way to prevent this?

Java prevent E notation in a double:
Five different ways to convert a double to a normal number:
import java.math.BigDecimal;
import java.text.DecimalFormat;
public class Runner {
public static void main(String[] args) {
double myvalue = 0.00000021d;
//Option 1 Print bare double.
System.out.println(myvalue);
//Option2, use decimalFormat.
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(8);
System.out.println(df.format(myvalue));
//Option 3, use printf.
System.out.printf("%.9f", myvalue);
System.out.println();
//Option 4, convert toBigDecimal and ask for toPlainString().
System.out.print(new BigDecimal(myvalue).toPlainString());
System.out.println();
//Option 5, String.format
System.out.println(String.format("%.12f", myvalue));
}
}
This program prints:
2.1E-7
.00000021
0.000000210
0.000000210000000000000001085015324114868562332958390470594167709350585
0.000000210000
Which are all the same value.
Protip: If you are confused as to why those random digits appear beyond a certain threshold in the double value, this video explains: computerphile why does 0.1+0.2 equal 0.30000000000001?
http://youtube.com/watch?v=PZRI1IfStY0

You could use printf() with %f:
double dexp = 12345678;
System.out.printf("dexp: %f\n", dexp);
This will print dexp: 12345678.000000. If you don't want the fractional part, use
System.out.printf("dexp: %.0f\n", dexp);
0 in %.0f means 0 places in fractional part i.e no fractional part. If you want to print fractional part with desired number of decimal places then instead of 0 just provide the number like this %.8f. By default fractional part is printed up to 6 decimal places.
This uses the format specifier language explained in the documentation.
The default toString() format used in your original code is spelled out here.

In short:
If you want to get rid of trailing zeros and Locale problems, then you should use:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); // 340 = DecimalFormat.DOUBLE_FRACTION_DIGITS
System.out.println(df.format(myValue)); // Output: 0.00000021
Explanation:
Why other answers did not suit me:
Double.toString() or System.out.println or FloatingDecimal.toJavaFormatString uses scientific notations if double is less than 10^-3 or greater than or equal to 10^7
By using %f, the default decimal precision is 6, otherwise you can hardcode it, but it results in extra zeros added if you have fewer decimals. Example:
double myValue = 0.00000021d;
String.format("%.12f", myvalue); // Output: 0.000000210000
By using setMaximumFractionDigits(0); or %.0f you remove any decimal precision, which is fine for integers/longs, but not for double:
double myValue = 0.00000021d;
System.out.println(String.format("%.0f", myvalue)); // Output: 0
DecimalFormat df = new DecimalFormat("0");
System.out.println(df.format(myValue)); // Output: 0
By using DecimalFormat, you are local dependent. In French locale, the decimal separator is a comma, not a point:
double myValue = 0.00000021d;
DecimalFormat df = new DecimalFormat("0");
df.setMaximumFractionDigits(340);
System.out.println(df.format(myvalue)); // Output: 0,00000021
Using the ENGLISH locale makes sure you get a point for decimal separator, wherever your program will run.
Why using 340 then for setMaximumFractionDigits?
Two reasons:
setMaximumFractionDigits accepts an integer, but its implementation has a maximum digits allowed of DecimalFormat.DOUBLE_FRACTION_DIGITS which equals 340
Double.MIN_VALUE = 4.9E-324 so with 340 digits you are sure not to round your double and lose precision.

You can try it with DecimalFormat. With this class you are very flexible in parsing your numbers.
You can exactly set the pattern you want to use.
In your case for example:
double test = 12345678;
DecimalFormat df = new DecimalFormat("#");
df.setMaximumFractionDigits(0);
System.out.println(df.format(test)); //12345678

I've got another solution involving BigDecimal's toPlainString(), but this time using the String-constructor, which is recommended in the javadoc:
this constructor is compatible with the values returned by Float.toString and Double.toString. This is generally the preferred way to convert a float or double into a BigDecimal, as it doesn't suffer from the unpredictability of the BigDecimal(double) constructor.
It looks like this in its shortest form:
return new BigDecimal(myDouble.toString()).stripTrailingZeros().toPlainString();
NaN and infinite values have to be checked extra, so looks like this in its complete form:
public static String doubleToString(Double d) {
if (d == null)
return null;
if (d.isNaN() || d.isInfinite())
return d.toString();
return new BigDecimal(d.toString()).stripTrailingZeros().toPlainString();
}
This can also be copied/pasted to work nicely with Float.
For Java 7 and below, this results in "0.0" for any zero-valued Doubles, so you would need to add:
if (d.doubleValue() == 0)
return "0";

Java/Kotlin compiler converts any value greater than 9999999 (greater than or equal to 10 million) to scientific notation ie. Epsilion notation.
Ex: 12345678 is converted to 1.2345678E7
Use this code to avoid automatic conversion to scientific notation:
fun setTotalSalesValue(String total) {
var valueWithoutEpsilon = total.toBigDecimal()
/* Set the converted value to your android text view using setText() function */
salesTextView.setText( valueWithoutEpsilon.toPlainString() )
}

This will work as long as your number is a whole number:
double dnexp = 12345678;
System.out.println("dexp: " + (long)dexp);
If the double variable has precision after the decimal point it will truncate it.

I needed to convert some double to currency values and found that most of the solutions were OK, but not for me.
The DecimalFormat was eventually the way for me, so here is what I've done:
public String foo(double value) //Got here 6.743240136E7 or something..
{
DecimalFormat formatter;
if(value - (int)value > 0.0)
formatter = new DecimalFormat("0.00"); // Here you can also deal with rounding if you wish..
else
formatter = new DecimalFormat("0");
return formatter.format(value);
}
As you can see, if the number is natural I get - say - 20000000 instead of 2E7 (etc.) - without any decimal point.
And if it's decimal, I get only two decimal digits.

I think everyone had the right idea, but all answers were not straightforward.
I can see this being a very useful piece of code. Here is a snippet of what will work:
System.out.println(String.format("%.8f", EnterYourDoubleVariableHere));
the ".8" is where you set the number of decimal places you would like to show.
I am using Eclipse and it worked no problem.
Hope this was helpful. I would appreciate any feedback!

The following code detects if the provided number is presented in scientific notation. If so it is represented in normal presentation with a maximum of '25' digits.
static String convertFromScientificNotation(double number) {
// Check if in scientific notation
if (String.valueOf(number).toLowerCase().contains("e")) {
System.out.println("The scientific notation number'"
+ number
+ "' detected, it will be converted to normal representation with 25 maximum fraction digits.");
NumberFormat formatter = new DecimalFormat();
formatter.setMaximumFractionDigits(25);
return formatter.format(number);
} else
return String.valueOf(number);
}

This may be a tangent.... but if you need to put a numerical value as an integer (that is too big to be an integer) into a serializer (JSON, etc.) then you probably want "BigInterger"
Example:
value is a string - 7515904334
We need to represent it as a numerical in a Json message:
{
"contact_phone":"800220-3333",
"servicer_id":7515904334,
"servicer_name":"SOME CORPORATION"
}
We can't print it or we'll get this:
{
"contact_phone":"800220-3333",
"servicer_id":"7515904334",
"servicer_name":"SOME CORPORATION"
}
Adding the value to the node like this produces the desired outcome:
BigInteger.valueOf(Long.parseLong(value, 10))
I'm not sure this is really on-topic, but since this question was my top hit when I searched for my solution, I thought I would share here for the benefit of others, lie me, who search poorly. :D

use String.format ("%.0f", number)
%.0f for zero decimal
String numSring = String.format ("%.0f", firstNumber);
System.out.println(numString);

I had this same problem in my production code when I was using it as a string input to a math.Eval() function which takes a string like "x + 20 / 50"
I looked at hundreds of articles... In the end I went with this because of the speed. And because the Eval function was going to convert it back into its own number format eventually and math.Eval() didn't support the trailing E-07 that other methods returned, and anything over 5 dp was too much detail for my application anyway.
This is now used in production code for an application that has 1,000+ users...
double value = 0.0002111d;
String s = Double.toString(((int)(value * 100000.0d))/100000.0d); // Round to 5 dp
s display as: 0.00021

This will work not only for a whole numbers:
double dexp = 12345678.12345678;
BigDecimal bigDecimal = new BigDecimal(Double.toString(dexp));
System.out.println("dexp: "+ bigDecimal.toPlainString());

My solution:
String str = String.format ("%.0f", yourDouble);

For integer values represented by a double, you can use this code, which is much faster than the other solutions.
public static String doubleToString(final double d) {
// check for integer, also see https://stackoverflow.com/a/9898613/868941 and
// https://github.com/google/guava/blob/master/guava/src/com/google/common/math/DoubleMath.java
if (isMathematicalInteger(d)) {
return Long.toString((long)d);
} else {
// or use any of the solutions provided by others, this is the best
DecimalFormat df =
new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); // 340 = DecimalFormat.DOUBLE_FRACTION_DIGITS
return df.format(d);
}
}
// Java 8+
public static boolean isMathematicalInteger(final double d) {
return StrictMath.rint(d) == d && Double.isFinite(d);
}

This works for me. The output will be a String.
String.format("%.12f", myvalue);

Good way to convert scientific e notation
String.valueOf(YourDoubleValue.longValue())

How to parse Float to String exactly?

As the title said:
I tried:
Float.toString(float);
String.valueOf(float);
Float.toHexString(float);
float.toString();
But I found if the Float value = 100.00;
Covert to String value will be 100.0.
How to avoid it? How to be exactly?
Thanks in advance.
Edits-------------
The answers are point to that those which specific the decimal places.

To be exact, you'd better try to format your Float to String using the NumberFormat class hierarchy : http://docs.oracle.com/javase/6/docs/api/java/text/NumberFormat.html

Its "stupid" to keep 2 zeros at the end. All you've to do is add as many zeros as needed at the moment you're printing it, but internally, it's going to be saved as x.0
Example:
printf ("%.2f", 3.14159);
Prints:
3.14

if you are bent upon doing this..
when you get str = "100.0";
String[] strNew = str.split(".");
if(strNew[1].length==1)
{
str=str+"0";
}
BTW A VERY BAD WAY ...

float f = 100.00000f;
String expected = String.format("%.6f", f);
The output of this will be :
100.000000
The length of the numbers after the floating point is done by you.
String.format("%.6f", f) == 100.000000
String.format("%.2f", f) == 100.00

Converting exponential value in java to a number format

I am trying to read the values from excel sheet using java. When i type more than 10 letters in a cell in excel it is displaying in exponential form like "9.78313E+2". but this is not the real number what i given.
Can any body help me out in this. How can i convert the above exponential form to original number using java language.
Thanks in advance

You can convert as follows,
for example:
new BigDecimal("406770000244E+12").toBigInteger();

Double.parseDouble("9.78313E+2");
gives me
978.313
For more info see the doc.
Following your further queries below, if you've entered 4256411411 and Excel is presenting this as 4.26E+09, putting that value into parseDouble() will only give you 4260000000. If you want the original, perhaps you need to output the Excel file in a fuller format for your Java program, and/or query it using a Java/Excel API (e.g. POI)

Sorry, but none of the answers above Double.parseDouble() and Double.valueOf()... solved my problem, and I continued to get the exponential 'E' value...
This link has a much better approach for the problem, and as I've written there - there is a very good solution:
I needed to convert some double to currency values, and fount that most to the solution are OK but not for me.
The DecimalFormat was eventually the way for me, so here is what I've done:
public String foo(double value) //Got here 6.743240136E7 or something..
{
DecimalFormat formatter;
if(value - (int)value > 0.0)
formatter = new DecimalFormat("0.00"); //Here you can also deal with rounding if you wish..
else
formatter = new DecimalFormat("0");
return formatter.format(value);
}
As you can see, if the number is natural I get - say - 20000000 instead of 2E7 (etc) - without any decimal point.
and if it's decimal, I get only 2 decimal digits.
Hope this will help.

You can use BigDecimal, if you want the exact value that you have in Excel Sheet: -
BigDecimal bd = new BigDecimal("4256411411");
System.out.println(bd.doubleValue());
// If you are sure that's not a floating point number, then use
System.out.println(bd.longValue());
Prints: -
4.256411411E9
4256411411

Try this definitely gona work
double value = 2.06E //real 205809104.13
BigDecimal.valueOf(value)
work for me

Before you read the value from excel sheet format your Column to number.
This may be helps to you
UPDATED
HSSFCell cellE1 = row1.getCell((short) 4);
cellE1.setCellType(HSSFCell.CELL_TYPE_NUMERIC);
Double e1Val = cellE1.getNumericCellValue();
BigDecimal bd = new BigDecimal(e1Val.toString());
long lonVal = bd.longValue();
System.out.println(lonVal);

You can convert easily with the following methods:
Double.valueOf("9.78313E+2").longValue() or
BigDecimal bd = new BigDecimal("9.78313E+2");
long val = bd.longValue();
Assuming that the given number is in a String form.

You can also use wrapper classes :
Double bd=new Double(4445566622);
System.out.println(bd.longValue());
Outputs -4445566622

i had same problem when i only needed String Data that is "1744949451" but it give "1.744949451E9"
so this worked for me
XSSFCell cell = cells.getCell(j);
String value = cell.toString();
if (cell.getCellType() == XSSFCell.CELL_TYPE_NUMERIC) {
//cell.setCellType(XSSFCell.CELL_TYPE_STRING);
value = cell.getRawValue();
}
Log.i("LOG", value + " " + cell.getCellType());

This answer worked for me:
Double bd = new Double(4445566622);
System.out.println(bd.longValue());
// Outputs -4445566622

`Double value = double value ;
Long longValue = value.longValue(); String strCellValue1 = new String(longValue.toString().format("%f",value).replaceAll("\\,?0*$", ""));`
declare a double value and convert to long
convert to string and formated to float the double value finally replace all the value like 123456789,0000 to 123456789

Have to convert the cell into number format before reading the cell value. Below is the code snippet that is used to get the actual value that is in exponential format:
nextCell.setCellType(XSSFCell.CELL_TYPE_NUMERIC);
Double doubleValue = nextCell.getNumericCellValue();
BigDecimal bd = new BigDecimal(doubleValue.toString());
long lonVal = bd.longValue();
String phoneNumber = Long.toString(lonVal).trim();
System.out.print("PhoneNumber " + phoneNumber);
Blog has been wirtten to showcase the actual result.
Regards,
Ankur

Try the following pattern:
Double dblValue = Double.parseDouble("1.99E+07");
String str = String.format("%.2f", dblValue);
System.out.println(str);
Output:
run:
19900000,00
BUILD SUCCESSFUL (total time: 0 seconds)

How to use Java's DecimalFormat for "smart" currency formatting?

I'd like to use Java's DecimalFormat to format doubles like so:
#1 - 100 -> $100
#2 - 100.5 -> $100.50
#3 - 100.41 -> $100.41
The best I can come up with so far is:
new DecimalFormat("'$'0.##");
But this doesn't work for case #2, and instead outputs "$100.5"
Edit:
A lot of these answers are only considering cases #2 and #3 and not realizing that their solution will cause #1 to format 100 as "$100.00" instead of just "$100".

Does it have to use DecimalFormat?
If not, it looks like the following should work:
String currencyString = NumberFormat.getCurrencyInstance().format(currencyNumber);
//Handle the weird exception of formatting whole dollar amounts with no decimal
currencyString = currencyString.replaceAll("\\.00", "");

Use NumberFormat:
NumberFormat n = NumberFormat.getCurrencyInstance(Locale.US);
double doublePayment = 100.13;
String s = n.format(doublePayment);
System.out.println(s);
Also, don't use doubles to represent exact values. If you're using currency values in something like a Monte Carlo method (where the values aren't exact anyways), double is preferred.
See also: Write Java programs to calculate and format currency

Try
new DecimalFormat("'$'0.00");
Edit:
I Tried
DecimalFormat d = new DecimalFormat("'$'0.00");
System.out.println(d.format(100));
System.out.println(d.format(100.5));
System.out.println(d.format(100.41));
and got
$100.00
$100.50
$100.41

Try using
DecimalFormat.setMinimumFractionDigits(2);
DecimalFormat.setMaximumFractionDigits(2);

You can check "is number whole or not" and choose needed number format.
public class test {
public static void main(String[] args){
System.out.println(function(100d));
System.out.println(function(100.5d));
System.out.println(function(100.42d));
}
public static String function(Double doubleValue){
boolean isWholeNumber=(doubleValue == Math.round(doubleValue));
DecimalFormatSymbols formatSymbols = new DecimalFormatSymbols(Locale.GERMAN);
formatSymbols.setDecimalSeparator('.');
String pattern= isWholeNumber ? "#.##" : "#.00";
DecimalFormat df = new DecimalFormat(pattern, formatSymbols);
return df.format(doubleValue);
}
}
will give exactly what you want:
100
100.50
100.42

You can use the following format:
DecimalFormat dformat = new DecimalFormat("$#.##");

I know its too late. However following worked for me :
DecimalFormatSymbols otherSymbols = new DecimalFormatSymbols(Locale.UK);
new DecimalFormat("\u00A4#######0.00",otherSymbols).format(totalSale);
\u00A4 : acts as a placeholder for currency symbol
#######0.00 : acts as a placeholder pattern for actual number with 2 decimal
places precision.
Hope this helps whoever reads this in future :)

You can try by using two different DecimalFormat objects based on the condition as follows:
double d=100;
double d2=100.5;
double d3=100.41;
DecimalFormat df=new DecimalFormat("'$'0.00");
if(d%1==0){ // this is to check a whole number
DecimalFormat df2=new DecimalFormat("'$'");
System.out.println(df2.format(d));
}
System.out.println(df.format(d2));
System.out.println(df.format(d3));
Output:-
$100
$100.50
$100.41

You could use the Java Money API to achieve this. (although this is not using DecialFormat)
long amountInCents = ...;
double amountInEuro = amountInCents / 100.00;
String customPattern;
if (minimumOrderValueInCents % 100 == 0) {
customPattern = "# ¤";
} else {
customPattern = "#.## ¤";
}
Money minDeliveryAmount = Money.of(amountInEuro, "EUR");
MonetaryAmountFormat formatter = MonetaryFormats.getAmountFormat(AmountFormatQueryBuilder.of(Locale.GERMANY)
.set(CurrencyStyle.SYMBOL)
.set("pattern", customPattern)
.build());
System.out.println(minDeliveryAmount);

printf also works.
Example:
double anyNumber = 100;
printf("The value is %4.2f ", anyNumber);
Output:
The value is 100.00
4.2 means force the number to have two digits after the decimal. The 4 controls how many digits to the right of the decimal.

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