一个非常低级的方法，但我相信这将是最高效的方法，只使用快速整数计算(没有缓慢的字符串解析和区域性敏感的方法):

public static decimal Normalize(this decimal d)
{
    int[] bits = decimal.GetBits(d);

    int sign = bits[3] & (1 << 31);
    int exp = (bits[3] >> 16) & 0x1f;

    uint a = (uint)bits[2]; // Top bits
    uint b = (uint)bits[1]; // Middle bits
    uint c = (uint)bits[0]; // Bottom bits

    while (exp > 0 && ((a % 5) * 6 + (b % 5) * 6 + c) % 10 == 0)
    {
        uint r;
        a = DivideBy10((uint)0, a, out r);
        b = DivideBy10(r, b, out r);
        c = DivideBy10(r, c, out r);
        exp--;
    }

    bits[0] = (int)c;
    bits[1] = (int)b;
    bits[2] = (int)a;
    bits[3] = (exp << 16) | sign;
    return new decimal(bits);
}

private static uint DivideBy10(uint highBits, uint lowBits, out uint remainder)
{
    ulong total = highBits;
    total <<= 32;
    total = total | (ulong)lowBits;

    remainder = (uint)(total % 10L);
    return (uint)(total / 10L);
}

这个怎么样:

public static string TrimEnd(this decimal d)
    {
        string str = d.ToString();
        if (str.IndexOf(".") > 0)
        {
            str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "0+?$", " ");
            str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "[.]$", " ");
        }
        return str;
    }

string.Format("{0:G29}", decimal.Parse("2.00"))

string.Format("{0:G29}", decimal.Parse(Your_Variable))

这是可行的:

decimal source = 2.4200m;
string output = ((double)source).ToString();

或者如果你的初始值是string:

string source = "2.4200";
string output = double.Parse(source).ToString();

请注意这条评论。

下面的代码将能够删除后面的0。我知道这很难，但很有效。

private static string RemoveTrailingZeros(string input) 
{
    for (int i = input.Length - 1; i > 0; i-- )
    {
        if (!input.Contains(".")) break;
        if (input[i].Equals('0'))
        {
            input= input.Remove(i);
        }
        else break;
    }
    return input;
}

一个非常低级的方法，但我相信这将是最高效的方法，只使用快速整数计算(没有缓慢的字符串解析和区域性敏感的方法):

public static decimal Normalize(this decimal d)
{
    int[] bits = decimal.GetBits(d);

    int sign = bits[3] & (1 << 31);
    int exp = (bits[3] >> 16) & 0x1f;

    uint a = (uint)bits[2]; // Top bits
    uint b = (uint)bits[1]; // Middle bits
    uint c = (uint)bits[0]; // Bottom bits

    while (exp > 0 && ((a % 5) * 6 + (b % 5) * 6 + c) % 10 == 0)
    {
        uint r;
        a = DivideBy10((uint)0, a, out r);
        b = DivideBy10(r, b, out r);
        c = DivideBy10(r, c, out r);
        exp--;
    }

    bits[0] = (int)c;
    bits[1] = (int)b;
    bits[2] = (int)a;
    bits[3] = (exp << 16) | sign;
    return new decimal(bits);
}

private static uint DivideBy10(uint highBits, uint lowBits, out uint remainder)
{
    ulong total = highBits;
    total <<= 32;
    total = total | (ulong)lowBits;

    remainder = (uint)(total % 10L);
    return (uint)(total / 10L);
}