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C# 大小写转换(金额)实例代码

2019年07月18日  | 移动技术网IT编程  | 我要评论

class：ecanrmb.cs

复制代码 代码如下:

 using system;
 using system.collections.generic;
 using system.linq;
 using system.text;

 namespace jxc.winclient.bll
 {
     public class ecanrmb
     {
         /// <summary>
         /// 转换人民币大小金额
         /// </summary>
         /// <param name="num">金额</param>
         /// <returns>返回大写形式</returns>
         public string cmycurd(decimal num)
         {
             string str1 = "零壹贰叁肆伍陆柒捌玖";            //0-9所对应的汉字
             string str2 = "万仟佰拾亿仟佰拾万仟佰拾元角分"; //数字位所对应的汉字
             string str3 = "";    //从原num值中取出的值
             string str4 = "";    //数字的字符串形式
             string str5 = "";  //人民币大写金额形式
             int i;    //循环变量
             int j;    //num的值乘以100的字符串长度
             string ch1 = "";    //数字的汉语读法
             string ch2 = "";    //数字位的汉字读法
             int nzero = 0;  //用来计算连续的零值是几个
             int temp;            //从原num值中取出的值

             num = math.round(math.abs(num), 2);    //将num取绝对值并四舍五入取2位小数
             str4 = ((long)(num * 100)).tostring();        //将num乘100并转换成字符串形式
             j = str4.length;      //找出最高位
             if (j > 15) { return "溢出"; }
             str2 = str2.substring(15 - j);   //取出对应位数的str2的值。如：200.55,j为5所以str2=佰拾元角分

             //循环取出每一位需要转换的值
             for (i = 0; i < j; i++)
             {
                 str3 = str4.substring(i, 1);          //取出需转换的某一位的值
                 temp = convert.toint32(str3);      //转换为数字
                 if (i != (j - 3) && i != (j - 7) && i != (j - 11) && i != (j - 15))
                 {
                     //当所取位数不为元、万、亿、万亿上的数字时
                     if (str3 == "0")
                     {
                         ch1 = "";
                         ch2 = "";
                         nzero = nzero + 1;
                     }
                     else
                     {
                         if (str3 != "0" && nzero != 0)
                         {
                             ch1 = "零" + str1.substring(temp * 1, 1);
                             ch2 = str2.substring(i, 1);
                             nzero = 0;
                         }
                         else
                         {
                             ch1 = str1.substring(temp * 1, 1);
                             ch2 = str2.substring(i, 1);
                             nzero = 0;
                         }
                     }
                 }
                 else
                 {
                     //该位是万亿，亿，万，元位等关键位
                     if (str3 != "0" && nzero != 0)
                     {
                         ch1 = "零" + str1.substring(temp * 1, 1);
                         ch2 = str2.substring(i, 1);
                         nzero = 0;
                     }
                     else
                     {
                         if (str3 != "0" && nzero == 0)
                         {
                             ch1 = str1.substring(temp * 1, 1);
                             ch2 = str2.substring(i, 1);
                             nzero = 0;
                         }
                         else
                         {
                             if (str3 == "0" && nzero >= 3)
                             {
                                 ch1 = "";
                                 ch2 = "";
                                 nzero = nzero + 1;
                             }
                             else
                             {
                                 if (j >= 11)
                                 {
                                     ch1 = "";
                                     nzero = nzero + 1;
                                 }
                                 else
                                 {
                                     ch1 = "";
                                     ch2 = str2.substring(i, 1);
                                     nzero = nzero + 1;
                                 }
                             }
                         }
                     }
                 }
                 if (i == (j - 11) || i == (j - 3))
                 {
                     //如果该位是亿位或元位，则必须写上
                     ch2 = str2.substring(i, 1);
                 }
                 str5 = str5 + ch1 + ch2;

                 if (i == j - 1 && str3 == "0")
                 {
                     //最后一位（分）为0时，加上“整”
                     str5 = str5 + '整';
                 }
             }
             if (num == 0)
             {
                 str5 = "零元整";
             }
             return str5;
         }

         /**/
         /// <summary>
         /// 一个重载，将字符串先转换成数字在调用cmycurd(decimal num)
         /// </summary>
         /// <param name="num">用户输入的金额，字符串形式未转成decimal</param>
         /// <returns></returns>
         public string cmycurd(string numstr)
         {
             try
             {
                 decimal num = convert.todecimal(numstr);
                 return cmycurd(num);
             }
             catch
             {
                 return "非数字形式！";
             }
         }
     }
 }

调用：

复制代码 代码如下:

  ecanrmb rmb = new ecanrmb();
  string rmb = txtamount.text.replace(",", "").replace("-","");
  label9.text = rmb.cmycurd(rmb);
  label11.text = convert.todecimal(txtamount.text).tostring("###,###,###.00");

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