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;
    }
}

这些我的代码转换特定的数字(索引从1开始)到Excel列。

    public static string NumberToExcelColumn(uint number)
    {
        uint originalNumber = number;

        uint numChars = 1;
        while (Math.Pow(26, numChars) < number)
        {
            numChars++;

            if (Math.Pow(26, numChars) + 26 >= number)
            {
                break;
            }               
        }

        string toRet = "";
        uint lastValue = 0;

        do
        {
            number -= lastValue;

            double powerVal = Math.Pow(26, numChars - 1);
            byte thisCharIdx = (byte)Math.Truncate((columnNumber - 1) / powerVal);
            lastValue = (int)powerVal * thisCharIdx;

            if (numChars - 2 >= 0)
            {
                double powerVal_next = Math.Pow(26, numChars - 2);
                byte thisCharIdx_next = (byte)Math.Truncate((columnNumber - lastValue - 1) / powerVal_next);
                int lastValue_next = (int)Math.Pow(26, numChars - 2) * thisCharIdx_next;

                if (thisCharIdx_next == 0 && lastValue_next == 0 && powerVal_next == 26)
                {
                    thisCharIdx--;
                    lastValue = (int)powerVal * thisCharIdx;
                }
            }

            toRet += (char)((byte)'A' + thisCharIdx + ((numChars > 1) ? -1 : 0));

            numChars--;
        } while (numChars > 0);

        return toRet;
    }

我的单元测试:

    [TestMethod]
    public void Test()
    {
        Assert.AreEqual("A", NumberToExcelColumn(1));
        Assert.AreEqual("Z", NumberToExcelColumn(26));
        Assert.AreEqual("AA", NumberToExcelColumn(27));
        Assert.AreEqual("AO", NumberToExcelColumn(41));
        Assert.AreEqual("AZ", NumberToExcelColumn(52));
        Assert.AreEqual("BA", NumberToExcelColumn(53));
        Assert.AreEqual("ZZ", NumberToExcelColumn(702));
        Assert.AreEqual("AAA", NumberToExcelColumn(703));
        Assert.AreEqual("ABC", NumberToExcelColumn(731));
        Assert.AreEqual("ACQ", NumberToExcelColumn(771));
        Assert.AreEqual("AYZ", NumberToExcelColumn(1352));
        Assert.AreEqual("AZA", NumberToExcelColumn(1353));
        Assert.AreEqual("AZB", NumberToExcelColumn(1354));
        Assert.AreEqual("BAA", NumberToExcelColumn(1379));
        Assert.AreEqual("CNU", NumberToExcelColumn(2413));
        Assert.AreEqual("GCM", NumberToExcelColumn(4823));
        Assert.AreEqual("MSR", NumberToExcelColumn(9300));
        Assert.AreEqual("OMB", NumberToExcelColumn(10480));
        Assert.AreEqual("ULV", NumberToExcelColumn(14530));
        Assert.AreEqual("XFD", NumberToExcelColumn(16384));
    }

巧合和优雅的Ruby版本:

def col_name(col_idx)
    name = ""
    while col_idx>0
        mod     = (col_idx-1)%26
        name    = (65+mod).chr + name
        col_idx = ((col_idx-mod)/26).to_i
    end
    name
end

这是我在PHP中的超级后期实现。这个是递归的。我是在发现这篇文章之前写的。我想看看其他人是否已经解决了这个问题……

public function GetColumn($intNumber, $strCol = null) {

    if ($intNumber > 0) {
        $intRem = ($intNumber - 1) % 26;
        $strCol = $this->GetColumn(intval(($intNumber - $intRem) / 26), sprintf('%s%s', chr(65 + $intRem), $strCol));
    }

    return $strCol;
}

我今天必须做这个工作，我的实现使用递归:

private static string GetColumnLetter(string colNumber)
{
    if (string.IsNullOrEmpty(colNumber))
    {
        throw new ArgumentNullException(colNumber);
    }

    string colName = String.Empty;

    try
    {
        var colNum = Convert.ToInt32(colNumber);
        var mod = colNum % 26;
        var div = Math.Floor((double)(colNum)/26);
        colName = ((div > 0) ? GetColumnLetter((div - 1).ToString()) : String.Empty) + Convert.ToChar(mod + 65);
    }
    finally
    {
        colName = colName == String.Empty ? "A" : colName;
    }

    return colName;
}

该方法将数字视为字符串，而以“0”开头的数字(A = 0)

我想在我使用的静态类中加入，用于在col index和col Label之间进行交互。我对ColumnLabel方法使用了修改后的可接受答案

public static class Extensions
{
    public static string ColumnLabel(this int col)
    {
        var dividend = col;
        var columnLabel = string.Empty;
        int modulo;

        while (dividend > 0)
        {
            modulo = (dividend - 1) % 26;
            columnLabel = Convert.ToChar(65 + modulo).ToString() + columnLabel;
            dividend = (int)((dividend - modulo) / 26);
        } 

        return columnLabel;
    }
    public static int ColumnIndex(this string colLabel)
    {
        // "AD" (1 * 26^1) + (4 * 26^0) ...
        var colIndex = 0;
        for(int ind = 0, pow = colLabel.Count()-1; ind < colLabel.Count(); ++ind, --pow)
        {
            var cVal = Convert.ToInt32(colLabel[ind]) - 64; //col A is index 1
            colIndex += cVal * ((int)Math.Pow(26, pow));
        }
        return colIndex;
    }
}

用这个…

30.ColumnLabel(); // "AD"
"AD".ColumnIndex(); // 30