将BigDecimal舍入到最接近的5美分

Question

我想弄清楚如何将货币金额向上舍入到最接近的5美分.以下显示了我的预期结果

1.03     => 1.05
1.051    => 1.10
1.05     => 1.05
1.900001 => 1.10

我需要结果的精度为2(如上所示).

更新

按照下面的建议,我能做的最好的就是这个

    BigDecimal amount = new BigDecimal(990.49)

    // To round to the nearest .05, multiply by 20, round to the nearest integer, then divide by 20
   def result =  new BigDecimal(Math.ceil(amount.doubleValue() * 20) / 20)
   result.setScale(2, RoundingMode.HALF_UP)

我不相信这是100%犹太人 - 我担心转换到双打时精度可能会丢失.然而,这是迄今为止我提出的最好的,似乎有效.

Answer 1

使用BigDecimal没有任何双打(改进了marcolopes的答案):

public static BigDecimal round(BigDecimal value, BigDecimal increment,
                               RoundingMode roundingMode) {
    if (increment.signum() == 0) {
        // 0 increment does not make much sense, but prevent division by 0
        return value;
    } else {
        BigDecimal divided = value.divide(increment, 0, roundingMode);
        BigDecimal result = divided.multiply(increment);
        return result;
    }
}

舍入模式例如是RoundingMode.HALF_UP.对于你的例子,你真的想要RoundingMode.UP(bd是一个只返回的助手new BigDecimal(input)):

assertEquals(bd("1.05"), round(bd("1.03"), bd("0.05"), RoundingMode.UP));
assertEquals(bd("1.10"), round(bd("1.051"), bd("0.05"), RoundingMode.UP));
assertEquals(bd("1.05"), round(bd("1.05"), bd("0.05"), RoundingMode.UP));
assertEquals(bd("1.95"), round(bd("1.900001"), bd("0.05"), RoundingMode.UP));

另请注意,您的上一个示例中存在错误(将1.900001舍入到1.10).

Answer 2

我会尝试乘以20,舍入到最接近的整数,然后除以20.这是一个黑客,但应该得到正确的答案.

Answer 3

我几年前用Java写过这个:https://github.com/marcolopes/dma/blob/master/org.dma.java/src/org/dma/java/math/BusinessRules.java

/**
 * Rounds the number to the nearest<br>
 * Numbers can be with or without decimals<br>
 */
public static BigDecimal round(BigDecimal value, BigDecimal rounding, RoundingMode roundingMode){

    return rounding.signum()==0 ? value :
        (value.divide(rounding,0,roundingMode)).multiply(rounding);

}


/**
 * Rounds the number to the nearest<br>
 * Numbers can be with or without decimals<br>
 * Example: 5, 10 = 10
 *<p>
 * HALF_UP<br>
 * Rounding mode to round towards "nearest neighbor" unless
 * both neighbors are equidistant, in which case round up.
 * Behaves as for RoundingMode.UP if the discarded fraction is >= 0.5;
 * otherwise, behaves as for RoundingMode.DOWN.
 * Note that this is the rounding mode commonly taught at school.
 */
public static BigDecimal roundUp(BigDecimal value, BigDecimal rounding){

    return round(value, rounding, RoundingMode.HALF_UP);

}


/**
 * Rounds the number to the nearest<br>
 * Numbers can be with or without decimals<br>
 * Example: 5, 10 = 0
 *<p>
 * HALF_DOWN<br>
 * Rounding mode to round towards "nearest neighbor" unless
 * both neighbors are equidistant, in which case round down.
 * Behaves as for RoundingMode.UP if the discarded fraction is > 0.5;
 * otherwise, behaves as for RoundingMode.DOWN.
 */
public static BigDecimal roundDown(BigDecimal value, BigDecimal rounding){

    return round(value, rounding, RoundingMode.HALF_DOWN);

}

Answer 4

您可以使用普通 double 来执行此操作。

double amount = 990.49;
double rounded = ((double) (long) (amount * 20 + 0.5)) / 20;

编辑：对于负数，您需要减去 0.5