十进制限制整体和小数部分

Question

How do you restrict the whole part to 15 and decimal part to 8 like 15.8 using BigDecimal

For Example:

String quant= "1000";
String price = "123456789012345.12345678";
final int contant = 100;
BigDecimal bd1;
BigDecimal bd2;
String value = "";
bd1 = new BigDecimal(price).multiply(new BigDecimal(quant));
bd2 = bd1.divide(new BigDecimal(contant));
value = bd2.toPlainString();

Output value coming as 1234567890123451.2345678 . So can I restrict the whole part up to 15 and decimal part up to 8.

Answer 1

You can validate as follows

String price = "123456789012345.12345678";
    String[] str=price.split("\\.");
    if(!((str[0].length()==15)&&(str[1].length()==8))){
        System.out.println("not a valid format");
    }

Answer 2

If you restrict the 'whole' part to 15, then you will lose precision, why would you do that? You can restrict (round) the decimal part by using BigDecimal.setScale . setScale can take an integer (how many decimal parts should be used) and a rounding strategy, eg BigDecimal.ROUND_UP .

Check the JavaDoc from BigDecimal for further information.

Answer 3

Not sure why you would do that, but here is a way-

int contant = 100;
String price = "123456789012345.12345678224423";
String quant = "1000";                
BigDecimal bd1 = new BigDecimal(price).multiply(new BigDecimal(quant));
BigDecimal bd2 = bd1.divide(new BigDecimal(contant));    

final int WHOLE_MAX = 15;
final int SCALE = 8;

String value = bd2.setScale(SCALE, RoundingMode.HALF_UP).toPlainString();

String tempWhole = value.substring(0, value.indexOf('.')); // Digits before the decimal point

String wholePart =   tempWhole.substring(0, Math.min(tempWhole.length(), WHOLE_MAX)); // Gets the first 15 digits
String decimalPart = value.substring(value.indexOf('.'));

value = wholePart.concat(decimalPart);
System.out.println(value);

Output:

123456789012345.23456782

Answer 4

The BigDecimal is described with precision. You can set it using MathContext .

MathContext context = new MathContext(15 + 8, RoundingMode.HALF_UP);

When you apply to your need

BigDecimal value = new BigDecimal("1234567890123456789.12345678",context);

The out put will be

1234567890123456789.1235

This is the expected way that what is erased from the number is the smallest thing. If you wan to set upper limit for the number you will need to declare your own type and check that result is grater then limit, then you have an overflow error.

You can implement your type that controls it by extending BigDecimal class

private class BigNumber extends BigDecimal {

        private final int exponent;
        private final int mantissa;
        private BigDecimal limit;

        public BigNumber(String val, int exponent, int mantissa) {
            super(val);
            this.exponent = exponent;
            this.mantissa = mantissa;

            char[] limit = new char[exponent];
            Arrays.fill(limit, '9');

            this.limit = new BigDecimal(limit).add(BigDecimal.ONE);
        }

        @Override
        public BigDecimal add(BigDecimal augend) {

            BigDecimal result = super.add(augend);

            if(result.compareTo(limit) < 0) {
                return result;
            }

            throw new RuntimeException("Nuber to large");
        }
    }