public static string ConvertToAlphaColumnReferenceFromInteger(int columnReference)
    {
        int baseValue = ((int)('A')) - 1 ;
        string lsReturn = String.Empty; 

        if (columnReference > 26) 
        {
            lsReturn = ConvertToAlphaColumnReferenceFromInteger(Convert.ToInt32(Convert.ToDouble(columnReference / 26).ToString().Split('.')[0]));
        } 

        return lsReturn + Convert.ToChar(baseValue + (columnReference % 26));            
    }

到目前为止，所有的解决方案都包含迭代或递归，这让我感到惊讶。

这是我的解，在常数时间内运行(没有循环)。此解决方案适用于所有可能的Excel列，并检查输入是否可以转换为Excel列。可能的列在[A, XFD]或[1,16384]范围内。(这取决于你的Excel版本)

private static string Turn(uint col)
{
    if (col < 1 || col > 16384) //Excel columns are one-based (one = 'A')
        throw new ArgumentException("col must be >= 1 and <= 16384");

    if (col <= 26) //one character
        return ((char)(col + 'A' - 1)).ToString();

    else if (col <= 702) //two characters
    {
        char firstChar = (char)((int)((col - 1) / 26) + 'A' - 1);
        char secondChar = (char)(col % 26 + 'A' - 1);

        if (secondChar == '@') //Excel is one-based, but modulo operations are zero-based
            secondChar = 'Z'; //convert one-based to zero-based

        return string.Format("{0}{1}", firstChar, secondChar);
    }

    else //three characters
    {
        char firstChar = (char)((int)((col - 1) / 702) + 'A' - 1);
        char secondChar = (char)((col - 1) / 26 % 26 + 'A' - 1);
        char thirdChar = (char)(col % 26 + 'A' - 1);

        if (thirdChar == '@') //Excel is one-based, but modulo operations are zero-based
            thirdChar = 'Z'; //convert one-based to zero-based

        return string.Format("{0}{1}{2}", firstChar, secondChar, thirdChar);
    }
}

这是一个javascript版本，根据格雷厄姆的代码

function (columnNumber) {
    var dividend = columnNumber;
    var columnName = "";
    var modulo;

    while (dividend > 0) {
        modulo = (dividend - 1) % 26;
        columnName = String.fromCharCode(65 + modulo) + columnName;
        dividend = parseInt((dividend - modulo) / 26);
    }

    return columnName;
};

(我知道这个问题与c#有关，但是，如果读者需要用Java做同样的事情，那么下面的内容可能会有用)

事实证明，使用Jakarta POI中的“CellReference”类可以很容易地做到这一点。此外，转换可以以两种方式进行。

// Convert row and column numbers (0-based) to an Excel cell reference
CellReference numbers = new CellReference(3, 28);
System.out.println(numbers.formatAsString());

// Convert an Excel cell reference back into digits
CellReference reference = new CellReference("AC4");
System.out.println(reference.getRow() + ", " + reference.getCol());

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;

}

精炼原始的解决方案(在c#中):

public static class ExcelHelper
{
    private static Dictionary<UInt16, String> l_DictionaryOfColumns;

    public static ExcelHelper() {
        l_DictionaryOfColumns = new Dictionary<ushort, string>(256);
    }

    public static String GetExcelColumnName(UInt16 l_Column)
    {
        UInt16 l_ColumnCopy = l_Column;
        String l_Chars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        String l_rVal = "";
        UInt16 l_Char;


        if (l_DictionaryOfColumns.ContainsKey(l_Column) == true)
        {
            l_rVal = l_DictionaryOfColumns[l_Column];
        }
        else
        {
            while (l_ColumnCopy > 26)
            {
                l_Char = l_ColumnCopy % 26;
                if (l_Char == 0)
                    l_Char = 26;

                l_ColumnCopy = (l_ColumnCopy - l_Char) / 26;
                l_rVal = l_Chars[l_Char] + l_rVal;
            }
            if (l_ColumnCopy != 0)
                l_rVal = l_Chars[l_ColumnCopy] + l_rVal;

            l_DictionaryOfColumns.ContainsKey(l_Column) = l_rVal;
        }

        return l_rVal;
    }
}