虽然我在这方面姗姗来迟，但格雷厄姆的答案远非最佳。特别是，你不需要使用模数，调用ToString()和apply (int)强制转换。考虑到在c#世界中的大多数情况下，您将从0开始编号，以下是我的修订:

public static string GetColumnName(int index) // zero-based
{
    const byte BASE = 'Z' - 'A' + 1;
    string name = String.Empty;

    do
    {
        name = Convert.ToChar('A' + index % BASE) + name;
        index = index / BASE - 1;
    }
    while (index >= 0);

    return name;
}

以下是Graham在Powershell中的代码:

function ConvertTo-ExcelColumnID {
param (
    [parameter(Position = 0,
        HelpMessage = "A 1-based index to convert to an excel column ID. e.g. 2 => 'B', 29 => 'AC'",
        Mandatory = $true)]
    [int]$index
);

[string]$result = '';
if ($index -le 0 ) {
    return $result;
}

while ($index -gt 0) {
    [int]$modulo = ($index - 1) % 26;
    $character = [char]($modulo + [int][char]'A');
    $result = $character + $result;
    [int]$index = ($index - $modulo) / 26;
}

return $result;

}

 static string[] ExcelColumnAlphabetIdentifiers = new string[] { "", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", 
     "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z" };
 public static string ExcelColumnAlphabetIdentifier( int ColumnNumber)
    {
        StringBuilder sb = new StringBuilder();
        int remainder = ColumnNumber;
        do
        {
            sb.Append(ExcelColumnAlphabetIdentifiers[remainder % 26]);
            remainder = remainder / 26;
        }
        while (remainder > 0);
       return sb.ToString();
    }

抱歉，这是Python而不是c#，但至少结果是正确的:

def ColIdxToXlName(idx):
    if idx < 1:
        raise ValueError("Index is too small")
    result = ""
    while True:
        if idx > 26:
            idx, r = divmod(idx - 1, 26)
            result = chr(r + ord('A')) + result
        else:
            return chr(idx + ord('A') - 1) + result


for i in xrange(1, 1024):
    print "%4d : %s" % (i, ColIdxToXlName(i))

我在VB中使用这个。NET 2003和它的工作良好…

Private Function GetExcelColumnName(ByVal aiColNumber As Integer) As String
    Dim BaseValue As Integer = Convert.ToInt32(("A").Chars(0)) - 1
    Dim lsReturn As String = String.Empty

    If (aiColNumber > 26) Then
        lsReturn = GetExcelColumnName(Convert.ToInt32((Format(aiColNumber / 26, "0.0").Split("."))(0)))
    End If

    GetExcelColumnName = lsReturn + Convert.ToChar(BaseValue + (aiColNumber Mod 26))
End Function

我在我的第一篇文章中发现了一个错误，所以我决定坐下来算算。我发现用来识别Excel列的数字系统不是另一个人说的26进制系统。以10为基数考虑以下情况。你也可以用字母表中的字母来做这件事。

空间 :.........................S1, s2, s3: S1, s2, s3 ....................................0,00, 000:..A aa aaa ....................................1,01, 001:..B ab aab ....................................…，…，…:……，…，… ....................................9,99,999:..Z, zz, ZZZ 空间中的总状态:10,100,1000:26,676,17576 国家总 :............... 1110年 ................ 18278年

Excel在以26为基数的字母空格中对列进行编号。你可以看到，一般来说，状态空间的级数是a, a^2, a^3，…对于以a为底的情况，状态的总数是a + a^2 + a^3 + ... .

Suppose you want to find the total number of states A in the first N spaces. The formula for doing so is A = (a)(a^N - 1 )/(a-1). This is important because we need to find the space N that corresponds to our index K. If I want to find out where K lies in the number system I need to replace A with K and solve for N. The solution is N = log{base a} (A (a-1)/a +1). If I use the example of a = 10 and K = 192, I know that N = 2.23804… . This tells me that K lies at the beginning of the third space since it is a little greater than two.

The next step is to find exactly how far in the current space we are. To find this, subtract from K the A generated using the floor of N. In this example, the floor of N is two. So, A = (10)(10^2 – 1)/(10-1) = 110, as is expected when you combine the states of the first two spaces. This needs to be subtracted from K because these first 110 states would have already been accounted for in the first two spaces. This leaves us with 82 states. So, in this number system, the representation of 192 in base 10 is 082.

使用基本索引为0的c#代码是

    private string ExcelColumnIndexToName(int Index)
    {
        string range = string.Empty;
        if (Index < 0 ) return range;
        int a = 26;
        int x = (int)Math.Floor(Math.Log((Index) * (a - 1) / a + 1, a));
        Index -= (int)(Math.Pow(a, x) - 1) * a / (a - 1);
        for (int i = x+1; Index + i > 0; i--)
        {
            range = ((char)(65 + Index % a)).ToString() + range;
            Index /= a;
        }
        return range;
    }

/ /旧邮政

c#中的零基础解决方案。

    private string ExcelColumnIndexToName(int Index)
    {
        string range = "";
        if (Index < 0 ) return range;
        for(int i=1;Index + i > 0;i=0)
        {
            range = ((char)(65 + Index % 26)).ToString() + range;
            Index /= 26;
        }
        if (range.Length > 1) range = ((char)((int)range[0] - 1)).ToString() + range.Substring(1);
        return range;
    }