The BigDecimal(double) constructor can have unpredictable behaviors. It is preferable to use BigDecimal(String) or BigDecimal.valueOf(double).
System.out.println(new BigDecimal(135.69)); //135.68999999999999772626324556767940521240234375
System.out.println(new BigDecimal("135.69")); // 135.69
System.out.println(new BigDecimal("13۵.۶9")); // 135.69
System.out.println(BigDecimal.valueOf(135.69)); // 135.69
The documentation for BigDecimal(double) explains in detail:
Answer from Marlon Patrick on Stack Overflow
- The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.
- The String constructor, on the other hand, is perfectly predictable: writing new BigDecimal("0.1") creates a BigDecimal which is exactly equal to 0.1, as one would expect. Therefore, it is generally recommended that the String constructor be used in preference to this one.
- When a double must be used as a source for a BigDecimal, note that this constructor provides an exact conversion; it does not give the same result as converting the double to a String using the Double.toString(double) method and then using the BigDecimal(String) constructor. To get that result, use the static valueOf(double) method.
The BigDecimal(double) constructor can have unpredictable behaviors. It is preferable to use BigDecimal(String) or BigDecimal.valueOf(double).
System.out.println(new BigDecimal(135.69)); //135.68999999999999772626324556767940521240234375
System.out.println(new BigDecimal("135.69")); // 135.69
System.out.println(new BigDecimal("13۵.۶9")); // 135.69
System.out.println(BigDecimal.valueOf(135.69)); // 135.69
The documentation for BigDecimal(double) explains in detail:
- The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.
- The String constructor, on the other hand, is perfectly predictable: writing new BigDecimal("0.1") creates a BigDecimal which is exactly equal to 0.1, as one would expect. Therefore, it is generally recommended that the String constructor be used in preference to this one.
- When a double must be used as a source for a BigDecimal, note that this constructor provides an exact conversion; it does not give the same result as converting the double to a String using the Double.toString(double) method and then using the BigDecimal(String) constructor. To get that result, use the static valueOf(double) method.
String currency = "135.69";
System.out.println(new BigDecimal(currency));
//will print 135.69
Check out setParseBigDecimal in DecimalFormat. With this setter, parse will return a BigDecimal for you.
String value = "1,000,000,000.999999999999999";
BigDecimal money = new BigDecimal(value.replaceAll(",", ""));
System.out.println(money);
Full code to prove that no NumberFormatException is thrown:
import java.math.BigDecimal;
public class Tester {
public static void main(String[] args) {
// TODO Auto-generated method stub
String value = "1,000,000,000.999999999999999";
BigDecimal money = new BigDecimal(value.replaceAll(",", ""));
System.out.println(money);
}
}
Output
1000000000.999999999999999
It seems like there is no other way since java.Lang.Number doesn't have a method which returns a BigDecimal type. Anyway it makes sense because BigDecimal only accepts strings which are properly formatted not like "2.105,88" but like "2105.88".
Let me show your my code:
import java.math.BigDecimal;
import java.text.DecimalFormat;
import java.text.NumberFormat;
import java.text.ParseException;
import java.util.Locale;
public class JavaMain {
public static void main(String[] args) {
String numberString = "2.105,88";
//using casting
try {
DecimalFormat df = (DecimalFormat) NumberFormat.getInstance(Locale.GERMAN);
df.setParseBigDecimal(true);
BigDecimal bd = (BigDecimal) df.parseObject(numberString);
System.out.println(bd.toString());
} catch (ParseException e) {
e.printStackTrace();
}
//your way short version
NumberFormat nf = NumberFormat.getInstance(Locale.GERMAN);
try {
BigDecimal bd1 = new BigDecimal(nf.parse(numberString).toString());
System.out.println(bd1.toString());
} catch (ParseException e) {
e.printStackTrace();
}
String numberStringFixed = "2105.88";
//direct string formatted
System.out.println(new BigDecimal(numberStringFixed));;
//direct but erroneous way if the string is not formatted
System.out.println(new BigDecimal(numberString));;
}
}
I hope this helps!
DecimalFormat has a method called setParseBigDecimal that causes parse() to return a BigDecimal. You just need to cast the returned Number.
String numberString = "2.105,88";
NumberFormat nf = NumberFormat.getInstance(Locale.GERMAN);
if (nf instanceof DecimalFormat) {
DecimalFormat df = (DecimalFormat) nf;
df.setParseBigDecimal(true);
BigDecimal parsed = (BigDecimal) df.parse(numberString);
System.out.println(parsed);
}
Output:
2105.88
setParseBigDecimal was introduced in Java 1.5.