在Delphi (Pascal)中:

function GetExcelColumnName(columnNumber: integer): string;
var
  dividend, modulo: integer;
begin
  Result := '';
  dividend := columnNumber;
  while dividend > 0 do begin
    modulo := (dividend - 1) mod 26;
    Result := Chr(65 + modulo) + Result;
    dividend := (dividend - modulo) div 26;
  end;
end;

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;
}

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;
    }

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;

前面的答案大部分是正确的。下面是将列号转换为excel列的另一种方法。 如果我们把它看作一个基转换，解就很简单了。简单地，将列号转换为以26为基数，因为只有26个字母。 你可以这样做:

步骤:

将列设置为商 从商变量中减去1(从上一步)，因为我们需要以97为a的ASCII表结束。 除以26，得到余数。 在余数上加97并转换为字符(因为97在ASCII表中是a) 商变成了新的商/ 26(因为我们可能会越过26列) 继续这样做，直到商大于0，然后返回结果

下面是这样做的代码:)

def convert_num_to_column(column_num):
    result = ""
    quotient = column_num
    remainder = 0
    while (quotient >0):
        quotient = quotient -1
        remainder = quotient%26
        result = chr(int(remainder)+97)+result
        quotient = int(quotient/26)
    return result

print("--",convert_num_to_column(1).upper())

在查看了这里提供的所有版本后，我决定自己使用递归来做一个。

这是我的vb.net版本:

Function CL(ByVal x As Integer) As String
    If x >= 1 And x <= 26 Then
        CL = Chr(x + 64)
    Else
        CL = CL((x - x Mod 26) / 26) & Chr((x Mod 26) + 1 + 64)
    End If
End Function