Formatting Double into a String in iReport - java

I'm doing a report, that I need to join 4 variables in one.
If I treat the variables separately I can format them with no problem.
But when I merge them into a String, the double value comes as 0.0 instead of 0.00
How can I make it comes as the original, 0.00?
The code right now looks like this:
$F{someDoubleField} + "a string" + $F{anotherDoubleField} + "another string"
It prints:
0.0 a string 0.0 another string
instead of:
0.00 a string 0.00 another string
Remember that iReport uses Java, so maybe, Java code can help me out.


Use like below:
new DecimalFormat("0.00").format(doubleField) + "str"


Do the next:
Open Properties window for your report (right-mouse click on top
node in tree view -> Properties)
Set the Language option to Java
Use such code in your text field expression:
String.format("%.2f",$V{Sum}) + " " + $F{unit}
Here
2 - number of digits after dot
$V{Sum} - Double or Floaf variable (or field - $F{...})


use DecimalFormat
here is howto:
double d = 0.00;
NumberFormat nf = new DecimalFormat("0.00");
String format = nf.format(d);
System.out.println(d); //0.00


Perhaps something like:
new DecimalFormat("0.00").format(new java.math.Double(($F{amount} == null) ? 0 : $F{amount}))

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 format a number to include a "+/-" only when it is Not 0?

I want to display double number with +/- sign and I am using this decimal format for that
double d = 0;
DecimalFormat nf = new DecimalFormat("+#0.0;-#0.0");
System.out.println(nf.format(d));
d = -1;
System.out.println(nf.format(d));
d = 1;
System.out.println(nf.format(d));
and I am getting out-put like this
+0.0
-1.0
+1.0
but I want 0.0 without + sign like
0.0
-1.0
+1.0
Thanks

Maybe better and more simple code:
String style = d > 0 ? "+%.2f" : "% .2f";
System.out.println(String.format(style, d));

As with this, I'm not entirely sure there is a way to solve this with only decimal format, however a slight modification of the regex given in that answer (Again, Credit: #Bohemian) seems to solve the problem.
System.out.println(nf.format(d).replaceAll("^[-+](?=0(.0*)?$)", ""));
This simply looks for numbers in the format of the number 0 (i.e 0 followed by some number of zeros after a decimal point), removing the sign preceding it.

If you search on this forum there already exists an answer to your question. The answer is available at https://stackoverflow.com/a/11929642/4284989
You cannot satisfy your requirement using format symbols alone. You will have to replace the sign symbol with "" (empty) after applying the format.
For clarity, the below snippet will give you your required answer.
public class FormatTester {
public static void main(String[] args){
DecimalFormat nf = new DecimalFormat("#,##0.0");
String formattedValue = nf.format(Double.parseDouble(args[0]));
formattedValue = formattedValue.replaceAll( "^-(?=0(.0*)?$)", "");
System.out.println(">>>> "+formattedValue+" <<<<");
}
}

How to format a printed line that is looped in Java

I'm pretty new to Java programming and couldn't find an answer to my problem anywhere. Basically, I have successfully created a program that builds a chart of Celsius to Fahrenheit conversions and Fahrenheit to Celsius conversions, however my looped print statements are not lined up correctly after the number 9 similar this:
9.0 48.2 40.0 4.44
10.0 50.0 41.0 5.0
I was required to use two separate methods to calculate the conversions and then call them within the main method. Here is the main method with the println statement that I am reffering to:
public static void main(String[]args){
double celsius = 1;
double fahrenheit = 32;
while(celsius <= 50 && fahrenheit <= 120){
double toFarhenheit = celsiusToFahrenheit(celsius);
double toCelsius = fahrenheitToCelsius(fahrenheit);
DecimalFormat fardec = new DecimalFormat("#.##");
toFarhenheit = Double.valueOf(fardec.format(toFarhenheit));
DecimalFormat celsdec = new DecimalFormat("#.##");
toCelsius = Double.valueOf(celsdec.format(toCelsius));
System.out.println(celsius + " " + toFarhenheit + " " + fahrenheit +
" " +toCelsius);
celsius++;
fahrenheit++;
}
}
To make a long story short, is there anyway to use a printf with this kind of long print statement so that the numbers will line up with one another?
In the past I have used printf %3d and %5d and the like to line integers up, however, I couldn't get this to work at all with this particular print statement.
Any ideas and/or help would be much appreciated.

Use System.out.printf(...) and a format String to output your data in regular columns. Avoid using \t as it is unreliable. For example please look here.
Eventually your code would look like:
System.out.printf(formatString, celsius, toFarhenheit, fahrenheit, toCelsius);
Where the formatString is a String that uses printf format specifiers and width constants that would allow for pretty output. I'll let you experiment with format Strings. It would also end with "%n" so that it becomes in effect a println with formatting.

Adding to what Hovercraft Full Of Eels said, using System.out.printf without "\t" is a better solution.
For example, you should be able to do something like this:
String myformat = "%0$10s";
Explanation of the format:
%0s identifies your output as a string
$10 tells it to ensure that a minimum of 10 characters are written to the output. Hence, you'll have a fixed width.
See http://docs.oracle.com/javase/6/docs/api/java/util/Formatter.html#syntax for some more details
[haven't used java in a while so someone do correct me if I'm off]

Use \t to format them as this spaces them out evenly.
System.out.println(celsius + "\t" + toFarhenheit + "\t" + fahrenheit +
"\t" +toCelsius);

Using "\t" or using printf is probably not going to help you out as what ever space is being added takes into consideration the 2 string literals. For Eg. 9.0 is 3 chars long and 10.0 is 4 chars long.. so in this case spaces applied are correct but your string literals itself are of different length.
Try changing the code like below, use one more hash.
DecimalFormat fardec = new DecimalFormat("##.##");

Conversion from exponential form to decimal in Java

I want to convert exponential to decimal. e.g. 1.234E3 to 1234.

It is not really a conversion, but about how you display the number. You can use NumberFormat to specify how the number should be displayed.
Check the difference:
double number = 100550000.75;
NumberFormat formatter = new DecimalFormat("#0.00");
System.out.println(number);
System.out.println(formatter.format(number));

How about BigDecimal.valueOf(doubleToFormat).toPlainString()

While working with Doubles and Long numbers in Java you will see that most of the value are displayed in Exponential form.
For Example: In following we are multiplying 2.35 with 10000 and the result is printed.
//Division example
Double a = 2.85d / 10000;
System.out.println("1. " + a.doubleValue());
//Multiplication example
a = 2.85d * 100000000;
System.out.println("2. " + a.doubleValue());
Result:
2.85E-4
2.85E8
Thus you can see the result is printed in exponential format. Now you may want to display the result in pure decimal format like: 0.000285 or 285000000. You can do this simply by using class java.math.BigDecimal. In following example we are using BigDecimal.valueOf() to convert the Double value to BigDecimal and than .toPlainString() to convert it into plain decimal string.
import java.math.BigDecimal;
//..
//..
//Division example
Double a = 2.85d / 10000;
System.out.println("1. " + BigDecimal.valueOf(a).toPlainString());
//Multiplication example
a = 2.85d * 100000000;
System.out.println("2. " + BigDecimal.valueOf(a).toPlainString());
Result:
0.000285
285000000
The only disadvantage of the above method is that it generates long strings of number. You may want to restrict the value and round off the number to 5 or 6 decimal point. For this you can use java.text.DecimalFormat class. In following example we are rounding off the number to 4 decimal point and printing the output.
import java.text.DecimalFormat;
//..
//..
Double a = 2.85d / 10000;
DecimalFormat formatter = new DecimalFormat("0.0000");
System.out.println(formatter .format(a));
Result:
0.0003
I have just tried to compress this code with one line, it will print value of 'a' with two decimal places:
new DecimalFormat("0.00").format(BigDecimal.valueOf(a).toPlainString());
Happy Converting :)

you can turn it into a String using
DecimalFormat

the answer by #b.roth is correct only if it is country specific. I used that method and got i18n issue , because the new DecimalFormat("#0.00) takes the decimal seperator of the particular country. For ex if a country uses decimal seperation as "," , then the formatted value will be in 0,00 ( ex.. 1.2e2 will be 120.00 in some places and 120,00 ) in some places due to i18n issue as said here..
the method that i prefer is `(new BigDecimal("1.2e2").toPlainString() )

just add following tag to jspx:-
<f:convertNumber maxFractionDigits="4" minFractionDigits="2" groupingUsed="false"/>

String data = Long.toString((long) 3.42E8);
System.out.println("**************"+data);

try the following
long l;
double d; //It holds the double value.such as 1.234E3
l=Double.valueOf(time_d).longValue();
you get the decimal value in the variable l.

You can do:
BigDecimal
.valueOf(value)
.setScale(decimalLimit, RoundingMode.HALF_UP)
.toPlainString()

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