java金额数字转中文工具类详解

本文实例为大家分享了java金额数字转中文工具类的具体代码，供大家参考，具体内容如下

java金额数字转中文工具类ConvertNum.java

package light.mvc.utils;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
* 金额数字转中文工具类
*
* @author ardo
*
*/
public class ConvertNum {

/**
* 把金额阿拉伯数字转换为汉字表示，小数点后四舍五入保留两位
* 还有一种方法可以在转换的过程中不考虑连续0的情况，然后对最终的结果进行一次遍历合并连续的零
*/
public static String[] ChineseNum = new String[] { "零", "壹", "贰", "叁", "肆",
 "伍", "陆", "柒", "捌", "玖" };

public static String NumToChinese(double num) {
if (num > 99999999999999.99 || num < -99999999999999.99)
 throw new IllegalArgumentException(
  "参数值超出允许范围 (-99999999999999.99 ～ 99999999999999.99)！");
boolean negative = false;// 正负标号
if (num < 0) {
 negative = true;
 num = num * (-1);
}
long temp = Math.round(num * 100);
int numFen = (int) (temp ％ 10);// 分
temp = temp / 10;
int numJiao = (int) (temp ％ 10);// 角
temp = temp / 10;
// 此时temp只包含整数部分
int[] parts = new int[20];// 将金额整数部分分为在0-9999之间数的各个部分
int numParts = 0;// 记录把原来金额整数部分分割为几个部分
for (int i = 0;; i++) {
 if (temp == 0)
 break;
 int part = (int) (temp ％ 10000);
 parts[i] = part;
 temp = temp / 10000;
 numParts++;
}
boolean beforeWanIsZero = true;// 标志位，记录万的下一级是否为0
String chineseStr = "";
for (int i = 0; i < numParts; i++) {
 String partChinese = partConvert(parts[i]);
 if (i ％ 2 == 0) {
 if ("".equals(partChinese))
  beforeWanIsZero = true;
 else
  beforeWanIsZero = false;
 }
 if (i != 0) {
 if (i ％ 2 == 0)// 亿的部分
  chineseStr = "亿" + chineseStr;
 else {
  if ("".equals(partChinese) && !beforeWanIsZero)// 如果“万”对应的
        // part 为
        // 0，而“万”下面一级不为
        // 0，则不加“万”，而加“零”
  chineseStr = "零" + chineseStr;
  else {
  if (parts[i - 1] < 1000 && parts[i - 1] > 0)// 如果万的部分不为0，而万前面的部分小于1000大于0，则万后面应该跟零
   chineseStr = "零" + chineseStr;
  chineseStr = "万" + chineseStr;
  }
 }
 }
 chineseStr = partChinese + chineseStr;
}
if ("".equals(chineseStr))// 整数部分为0，则表示为零元
 chineseStr = ChineseNum[0];
else if (negative)// 整数部分部位0，但是为负数
 chineseStr = "负" + chineseStr;
chineseStr = chineseStr + "元";
if (numFen == 0 && numJiao == 0) {
 chineseStr = chineseStr + "整";
} else if (numFen == 0) {// 0分
 chineseStr = chineseStr + ChineseNum[numJiao] + "角";
} else {
 if (numJiao == 0)
 chineseStr = chineseStr + "零" + ChineseNum[numFen] + "分";
 else
 chineseStr = chineseStr + ChineseNum[numJiao] + "角"
  + ChineseNum[numFen] + "分";
}
return chineseStr;
}

// 转换拆分后的每个部分，0-9999之间
public static String partConvert(int partNum) {
if (partNum < 0 || partNum > 10000) {
 throw new IllegalArgumentException("参数必须是大于等于0或小于10000的整数");
}
String[] units = new String[] { "", "拾", "佰", "仟" };
int temp = partNum;
String partResult = new Integer(partNum).toString();
int partResultLength = partResult.length();
boolean lastIsZero = true;// 记录上一位是否为0
String chineseStr = "";
for (int i = 0; i < partResultLength; i++) {
 if (temp == 0)// 高位无数字
 break;
 int digit = temp ％ 10;
 if (digit == 0) {
 if (!lastIsZero)// 如果前一个数字不是0则在当前汉字串前加零
  chineseStr = "零" + chineseStr;
 lastIsZero = true;
 } else {
 chineseStr = ChineseNum[digit] + units[i] + chineseStr;
 lastIsZero = false;
 }
 temp = temp / 10;
}
return chineseStr;
}

public static void main(String args[]) {
double num = 0;
System.out.println("请输入金额数字,-1退出");
try {
 BufferedReader br = new BufferedReader(new InputStreamReader(
  System.in));
 num = Double.parseDouble(br.readLine());
} catch (IOException e) {
 System.out.println(e.toString());
}
while (num != -1) {
 System.out.println(num + NumToChinese(num));
 try {
 BufferedReader br = new BufferedReader(new InputStreamReader(
  System.in));
 num = Double.parseDouble(br.readLine());
 } catch (IOException e) {
 System.out.println(e.toString());
 }
}
System.out.println("其他测试：");
System.out.println("100120: " + NumToChinese(100120));
System.out.println("25000000000005.999: "
 + NumToChinese(25000000000005.999));
System.out.println("45689263.626: " + NumToChinese(45689263.626));
System.out.println("0.69457: " + NumToChinese(0.69457));
System.out.println("253.0: " + NumToChinese(253.0));
System.out.println("0: " + NumToChinese(0));
}
}

来源：https://blog.csdn.net/ardo_pass/article/details/78703030