你可以设置为:

decimal decNumber = 23.45600000m;
Console.WriteLine(decNumber.ToString("0.##"));

我从http://dobrzanski.net/2009/05/14/c-decimaltostring-and-how-to-get-rid-of-trailing-zeros/上找到了一个优雅的解决方案

基本上

decimal v=2.4200M;

v.ToString("#.######"); // Will return 2.42. The number of # is how many decimal digits you support.

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;
    }

这很简单。

decimal decNumber = Convert.ToDecimal(value);
        return decNumber.ToString("0.####");

测试。

欢呼:)

非常简单的答案是使用TrimEnd()。这是结果，

double value = 1.00;
string output = value.ToString().TrimEnd('0');

输出为1 如果我的值是1.01，那么输出将是1.01