这个答案是用javaScript写的:

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

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

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))

    public string ToBase26(int number)
    {
        if (number < 0) return String.Empty;

        int remainder = number % 26;
        int value = number / 26;

        return value == 0 ?
            String.Format("{0}", Convert.ToChar(65 + remainder)) :
            String.Format("{0}{1}", ToBase26(value - 1), Convert.ToChar(65 + remainder));
    }

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));
    }

int nCol = 127;
string sChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol >= 26)
{
    int nChar = nCol % 26;
    nCol = (nCol - nChar) / 26;
    // You could do some trick with using nChar as offset from 'A', but I am lazy to do it right now.
    sCol = sChars[nChar] + sCol;
}
sCol = sChars[nCol] + sCol;

更新:Peter的评论是正确的。这就是我在浏览器中编写代码的结果。:-)我的解决方案是不编译，它是最左边的字母，它是在反向顺序构建字符串-现在都固定了。

除了bug之外，该算法基本上是将一个数字从10进制转换为26进制。

更新2:Joel Coehoorn是对的-上面的代码将返回AB为27。如果它是一个以26为底的实数，AA就等于A, Z之后的下一个数字就是BA。

int nCol = 127;
string sChars = "0ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string sCol = "";
while (nCol > 26)
{
    int nChar = nCol % 26;
    if (nChar == 0)
        nChar = 26;
    nCol = (nCol - nChar) / 26;
    sCol = sChars[nChar] + sCol;
}
if (nCol != 0)
    sCol = sChars[nCol] + sCol;

在perl中，对于1 (A)， 27 (AA)等输入。

sub excel_colname {
  my ($idx) = @_;       # one-based column number
  --$idx;               # zero-based column index
  my $name = "";
  while ($idx >= 0) {
    $name .= chr(ord("A") + ($idx % 26));
    $idx   = int($idx / 26) - 1;
  }
  return scalar reverse $name;
}