Java big decimal number format exception

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傲寒
傲寒 2021-02-19 11:03

Why does the code below throw a java number format exception?

BigDecimal d = new BigDecimal(\"10934,375\");
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  • 2021-02-19 11:29

    You can use NumberFormat to choose the Locale, see the example:

            String numberToFormat = "1.900,35";
            NumberFormat formatter = NumberFormat.getNumberInstance(Locale.GERMAN);
            Number number = formatter.parse(numberToFormat);
            BigDecimal decimal = BigDecimal.valueOf(number.doubleValue());
    
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  • 2021-02-19 11:34

    The problem is that constructor of BigDecimal requires decimal number format where decimals come right after decimal dot . instead of decimal comma , so the right format for this specific case would be:

    BigDecimal d = new BigDecimal("10934.375");
    
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  • 2021-02-19 11:41

    Yes, the BigDecimal class does not take any Locale into account in its constructor that takes a String, as can be read in the Javadoc of this constructor:

    the fraction consists of a decimal point followed by zero or more decimal digits.

    If you want to parse according to a different Locale, one that uses the comma as decimals separator, you need to use java.text.DecimalFormat with a specific Locale.

    Example:

    DecimalFormat fmt = new DecimalFormat("0.0", new DecimalFormatSymbols(Locale.GERMAN));
    fmt.setParseBigDecimal(true);
    BigDecimal n = (BigDecimal) fmt.parse("10934,375");
    

    Note: you need to get an instance of DecimalFormat (a subclass of NumberFormat) to be able to call the method setParseBigDecimal. Otherwise it returns a Double instead, which is a binary floating point number, and binary floating point numbers cannot accurately represent many decimal fractions. So that would cause a loss of accuracy in many cases.

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