使用BigDecimal去掉小数点后无用的0 使用BigDecimal去掉小数点后无用

如题：BigDecimal去掉小数点后无用的0

解决：BigDecimal，有方法解决stripTrailingZeros()
demo
方法二：

Java BigDecimal 小数点处理

保留两位小数

方法一：
方式二：
方式三：

四舍五入

如题：BigDecimal去掉小数点后无用的0 比如：数据库存储的是Decimal(5,2)类型保留两位数。
如果展示数据5.00，5.10等字样感觉很不爽，如何做呢？
只战术5和5.1

解决：BigDecimal，有方法解决stripTrailingZeros()
看源码：

/*** Returns a string representation of this {@code BigDecimal}* without an exponent field.For values with a positive scale,* the number of digits to the right of the decimal point is used* to indicate scale.For values with a zero or negative scale,* the resulting string is generated as if the value were* converted to a numerically equal value with zero scale and as* if all the trailing zeros of the zero scale value were present* in the result.** The entire string is prefixed by a minus sign character '-'* ('\ u002D') if the unscaled value is less than* zero. No sign character is prefixed if the unscaled value is* zero or positive.** Note that if the result of this method is passed to the* {@linkplain #BigDecimal(String) string constructor}, only the* numerical value of this {@code BigDecimal} will necessarily be* recovered; the representation of the new {@code BigDecimal}* may have a different scale.In particular, if this* {@code BigDecimal} has a negative scale, the string resulting* from this method will have a scale of zero when processed by* the string constructor.** (This method behaves analogously to the {@code toString}* method in 1.4 and earlier releases.)** @return a string representation of this {@code BigDecimal}* without an exponent field.* @since 1.5* @see #toString()* @see #toEngineeringString()*/public String toPlainString() {if(scale==0) {if(intCompact!=INFLATED) {return Long.toString(intCompact); } else {return intVal.toString(); }}if(this.scale<0) { // No decimal pointif(signum()==0) {return "0"; }int tailingZeros = checkScaleNonZero((-(long)scale)); StringBuilder buf; if(intCompact!=INFLATED) {buf = new StringBuilder(20+tailingZeros); buf.append(intCompact); } else {String str = intVal.toString(); buf = new StringBuilder(str.length()+tailingZeros); buf.append(str); }for (int i = 0; i < tailingZeros; i++)buf.append('0'); return buf.toString(); }String str ; if(intCompact!=INFLATED) {str = Long.toString(Math.abs(intCompact)); } else {str = intVal.abs().toString(); }return getValueString(signum(), str, scale); } /*** Returns a {@code BigDecimal} which is numerically equal to* this one but with any trailing zeros removed from the* representation.For example, stripping the trailing zeros from* the {@code BigDecimal} value {@code 600.0}, which has* [{@code BigInteger}, {@code scale}] components equals to* [6000, 1], yields {@code 6E2} with [{@code BigInteger},* {@code scale}] components equals to [6, -2].If* this BigDecimal is numerically equal to zero, then* {@code BigDecimal.ZERO} is returned.** @return a numerically equal {@code BigDecimal} with any* trailing zeros removed.* @since 1.5*/public BigDecimal stripTrailingZeros() {if (intCompact == 0 || (intVal != null && intVal.signum() == 0)) {return BigDecimal.ZERO; } else if (intCompact != INFLATED) {return createAndStripZerosToMatchScale(intCompact, scale, Long.MIN_VALUE); } else {return createAndStripZerosToMatchScale(intVal, scale, Long.MIN_VALUE); }}

demo

public class StringUtils {public static void main(String[] args) {System.out.println(BigDecimal.ZERO); System.out.println(new BigDecimal("2.0")); System.out.println(new Double("0")); System.out.println(new BigDecimal("2.00")); String d = new BigDecimal("100.10").stripTrailingZeros().toPlainString(); System.out.println(d); System.out.println(new BigDecimal("100.10").stripTrailingZeros().toPlainString()); }}

结果：

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方法二：

private static final DecimalFormat decimalFormat = new DecimalFormat("###################.###########"); public static void main(String[] args) throws Exception{System.out.print( "格式化结果:"); System.out.println(decimalFormat.format(new BigDecimal("10.10"))); }

【使用BigDecimal去掉小数点后无用的0】结果：

文章图片

Java BigDecimal 小数点处理
保留两位小数

方法一：

{doublec=3.154215; java.text.DecimalFormat myformat=new java.text.DecimalFormat("0.00"); String str = myformat.format(c); }

方式二：

{java.text.DecimalFormatdf=newjava.text.DecimalFormat("#.00"); df.format(你要格式化的数字); 例：new java.text.DecimalFormat("#.00").format(3.1415926)#.00 表示两位小数 #.0000四位小数以此类推...}

方式三：

{double d = 3.1415926; String result = String .format("%.2f"); %.2f %. 表示小数点前任意位数2 表示两位小数格式后的结果为f 表示浮点型}

四舍五入

{doublef=111231.5585; BigDecimalb=newBigDecimal(f); //保留2位小数doublef1=b.setScale(2,BigDecimal.ROUND_HALF_UP).doubleValue(); }

public class PreciseCompute {//默认除法运算精度private static final int DEF_DIV_SCALE = 10; /*** 提供精确的加法运算。* @param v1 被加数* @param v2 加数* @return 两个参数的和*/ public static double add(double v1, double v2) {BigDecimal b1 = new BigDecimal(Double.toString(v1)); BigDecimal b2 = new BigDecimal(Double.toString(v2)); return b1.add(b2).doubleValue(); } /*** 提供精确的减法运算。* @param v1 被减数* @param v2 减数* @return 两个参数的差*/ public static double sub(double v1, double v2) {BigDecimal b1 = new BigDecimal(Double.toString(v1)); BigDecimal b2 = new BigDecimal(Double.toString(v2)); return b1.subtract(b2).doubleValue(); } /*** 提供精确的乘法运算。* @param v1 被乘数* @param v2 乘数* @return 两个参数的积*/public static double mul(double v1, double v2) {BigDecimal b1 = new BigDecimal(Double.toString(v1)); BigDecimal b2 = new BigDecimal(Double.toString(v2)); return b1.multiply(b2).doubleValue(); } /*** 提供（相对）精确的除法运算，当发生除不尽的情况时，精确到* 小数点以后10位，以后的数字四舍五入。* @param v1 被除数* @param v2 除数* @return 两个参数的商*/ public static double div(double v1, double v2) {return div(v1, v2, DEF_DIV_SCALE); } /*** 提供（相对）精确的除法运算。当发生除不尽的情况时，由scale参数指* 定精度，以后的数字四舍五入。* @param v1 被除数* @param v2 除数* @param scale 表示表示需要精确到小数点以后几位。* @return 两个参数的商*/public static double div(double v1, double v2, int scale) {if (scale < 0) {throw new IllegalArgumentException("The scale must be a positive integer or zero"); }BigDecimal b1 = new BigDecimal(Double.toString(v1)); BigDecimal b2 = new BigDecimal(Double.toString(v2)); return b1.divide(b2, scale, BigDecimal.ROUND_HALF_UP).doubleValue(); } /*** 提供精确的小数位四舍五入处理。* @param v 需要四舍五入的数字* @param scale 小数点后保留几位* @return 四舍五入后的结果*/public static double round(double v, int scale) {if (scale < 0) {throw new IllegalArgumentException("The scale must be a positive integer or zero"); }BigDecimal b = new BigDecimal(Double.toString(v)); BigDecimal ne = new BigDecimal("1"); return b.divide(one, scale, BigDecimal.ROUND_HALF_UP).doubleValue(); }}

my code：

private BigDecimal formatComma2BigDecimal(Object obj) {String val = String.valueOf(obj); if (val == null)return new BigDecimal("0.00"); val = val.replaceAll(",", ""); if (!isNumber(val))return new BigDecimal("0.00"); BigDecimal decimal = new BigDecimal(val); decimal = decimal.setScale(2, RoundingMode.HALF_UP); return decimal; } private String formatCommaAnd2Point(Object obj) {BigDecimal decimal = formatComma2BigDecimal(obj); DecimalFormat df = new DecimalFormat("#,###.00"); String decimalStr = df.format(decimal).equals(".00")?"0.00":df.format(decimal); if(decimalStr.startsWith(".")){decimalStr = "0"+decimalStr; }return decimalStr; } private boolean isDouble(String value) {try {Double.parseDouble(value); if (value.contains("."))return true; return false; } catch (NumberFormatException e) {return false; } } private boolean isInteger(String value) {try {Integer.parseInt(value); return true; } catch (NumberFormatException e) {return false; } } private boolean isNumber(String value) {return isInteger(value) || isDouble(value); }

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