解析小数并在右边过滤多余的0?

我不认为有任何标准的数字格式字符串会总是忽略后面不重要的零,我担心。

您可以尝试编写自己的十进制规范化方法,但这可能非常棘手。与 BigInteger 从.NET 4中类是合理可行的,但如果没有它(或类似的东西),这确实是非常困难的。

编辑:好吧,我想这就是你想要的:

using System;
using System.Numerics;

public static class DecimalExtensions
{
    // Avoiding implicit conversions just for clarity
    private static readonly BigInteger Ten = new BigInteger(10);
    private static readonly BigInteger UInt32Mask = new BigInteger(0xffffffffU);

    public static decimal Normalize(this decimal input)
    {
        unchecked
        {
            int[] bits = decimal.GetBits(input);
            BigInteger mantissa = 
                new BigInteger((uint) bits[0]) +
                (new BigInteger((uint) bits[1]) << 32) +
                (new BigInteger((uint) bits[2]) << 64);

            int sign = bits[3] & int.MinValue;            
            int exponent = (bits[3] & 0xff0000) >> 16;

            // The loop condition here is ugly, because we want
            // to do both the DivRem part and the exponent check :(
            while (exponent > 0)
            {
                BigInteger remainder;
                BigInteger divided = BigInteger.DivRem(mantissa, Ten, out remainder);
                if (remainder != BigInteger.Zero)
                {
                    break;
                }
                exponent--;
                mantissa = divided;
            }
            // Okay, now put it all back together again...
            bits[3] = (exponent << 16) | sign;
            // For each 32 bits, convert the bottom 32 bits into a uint (which won't
            // overflow) and then cast to int (which will respect the bits, which
            // is what we want)
            bits[0] = (int) (uint) (mantissa & UInt32Mask);
            mantissa >>= 32;
            bits[1] = (int) (uint) (mantissa & UInt32Mask);
            mantissa >>= 32;
            bits[2] = (int) (uint) (mantissa & UInt32Mask);

            return new decimal(bits);
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            Check(6.000m);
            Check(6000m);
            Check(6m);
            Check(60.00m);
            Check(12345.00100m);
            Check(-100.00m);
        }

        static void Check(decimal d)
        {
            Console.WriteLine("Before: {0}  -  after: {1}", d, d.Normalize());
        }
    }
}

这将从小数中删除所有尾随的零,然后您只需使用 ToString() .

public static class DecimalExtensions
{
    public static Decimal Normalize(this Decimal value)
    {
        return value / 1.000000000000000000000000000000000m;
    }
}

或者,如果需要精确的精度,比如说5个小数位,首先是normalize(),然后乘以1.00000m。

这符合给出的示例,但效果不佳。(我认为)

myDecimal.ToString("#.######") 

还有什么其他要求?您要操作这些值并将被操作的值显示为此小数位数吗?

另一个答案涉及递归性,比如:

  //use like so:
  myTextbox.Text = RemoveTrailingZeroes( myDecimal.ToString() );

private string RemoveTrailingZeroes(string input) {
  if ( input.Contains( "." ) && input.Substring( input.Length - 1 ) == "0" ) { //test the last character == "0"
    return RemoveTrailingZeroes( input.Substring( 0, input.Length - 2 ) ) 
    //drop the last character and recurse again
  }
  return input; //else return the original string
}

如果你想要一个扩展方法,那么这是一个选项

  //use like so:
  myTextbox.Text = myDecimal.ToString().RemoveTrailingZeroes();

private string RemoveTrailingZeroes(this string input) {
  if ( input.Contains( "." ) &&  input.Substring( input.Length - 1 ) == "0" ) { //test the last character == "0"
    return RemoveTrailingZeroes( input.Substring( 0, input.Length - 2 ) ) 
    //drop the last character and recurse again
  }
  return input; //else return the original string
}

补充 input.Contains( "." ) && 根据乔恩·斯基特的评论,但请记住,这会让事情变得异常缓慢。如果你知道你总是有一个小数点,没有像这样的大小写 myDecimal = 6000; 然后您可以删除该测试,或者您可以将其创建为一个类,并根据输入是否包含小数点等,拥有几个私有方法。我选择的是最简单的“它工作”,而不是 Enterprise FizzBuzz

如果您只需要对显示进行此操作,那么使用自定义格式字符串如何?

// decimal has 28/29 significant digits so 30 "#" symbols should be plenty
public static readonly string TrimmedDecimal = "0." + new string('#', 30);

// ...

decimal x = 1.132000m;
Console.WriteLine(x.ToString() + " - " + x.ToString(TrimmedDecimal));  // 1.132

decimal y = 6.000000m;
Console.WriteLine(y.ToString() + " - " + y.ToString(TrimmedDecimal));  // 6

用什么呢 ToScript(“G29”) 在你的小数点上? 这正是你想要达到的目标!