java精确到少数的后两位

本文介绍了一种处理价格字符串的方法,该方法能够去除前导零,并调整小数点后的位数,确保数值的有效性和正确性。此外,还展示了如何使用BigDecimal进行数值处理以保持精度。

package com.wm.mad.myself; import java.math.BigDecimal; import junit.framework.TestCase; public class Accurate extends TestCase { public void testAccurate1() { String price = "000000154.54"; //".589"; //"999."; //".999"; //"99.9"; //"99.99"; //"99.999"; //"999.999999999"; //"000005585" if(price == null) { return ; } int i=0; while(price.charAt(i)=='0') { price = price.substring(i+1); i++; } try { Float.parseFloat(price); } catch(Exception e) { return ; } int index = price.indexOf('.'); int len = price.length(); if(index == 0) { index = 1; price = 0 + price; } else if( (index+1) == len) { price = price.substring(0, index); index = -1; } len = price.length(); String priceParse = null; long totalNumber = 0; long headNumber = 0; long endNumber = 0; if(index >= 0) { headNumber = Long.parseLong(price.substring(0, index)); if((index+4) <= len) { endNumber = Long.parseLong(price.substring(index+1, index+4)); totalNumber = headNumber*1000L + endNumber+5; } else if((index+3) <= len){ endNumber = Long.parseLong(price.substring(index+1,len)); totalNumber = headNumber*1000L + endNumber*10; } else if((index+2) <= len) { endNumber = Long.parseLong(price.substring(index+1,len)); totalNumber = headNumber*1000L + endNumber*100; } headNumber = totalNumber/1000; endNumber = totalNumber%1000/10; if(endNumber < 10) { priceParse = headNumber + ".0" + endNumber; } else { priceParse = headNumber + "." + endNumber; } } else { priceParse = price; } System.out.println(priceParse); } public void testAccurate2() { String price = ".589"; //"999."; //".999"; //"99.9"; //"99.99"; //"99.999"; //"999.999999999"; // 保留两位小数并四舍五入 try { price = (new BigDecimal(price).setScale(2, BigDecimal.ROUND_HALF_UP)).toString(); } catch(Exception e) { price = null; } } public static void main(String[] args) { //junit.textui.TestRunner.run(Accurate2.class); Accurate accurate = new Accurate(); accurate.testAccurate1(); accurate.testAccurate2(); } }

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