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;

下面是一个Actionscript版本:

private var columnNumbers:Array = ['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'];

    private function getExcelColumnName(columnNumber:int) : String{
        var dividend:int = columnNumber;
        var columnName:String = "";
        var modulo:int;

        while (dividend > 0)
        {
            modulo = (dividend - 1) % 26;
            columnName = columnNumbers[modulo] + columnName;
            dividend = int((dividend - modulo) / 26);
        } 

        return columnName;
    }

似乎很多答案都比必要的要复杂得多。下面是一个基于上面描述的递归的通用Ruby答案:

这个答案的一个好处是，它不局限于26个英文字母。你可以在COLUMNS常量中定义任何你喜欢的范围，它会做正确的事情。

  # vim: ft=ruby
  class Numeric
    COLUMNS = ('A'..'Z').to_a

    def to_excel_column(n = self)
      n < 1 ?  '' : begin
        base = COLUMNS.size
        to_excel_column((n - 1) / base) + COLUMNS[(n - 1) % base]
      end
    end
  end

  # verify:
  (1..52).each { |i| printf "%4d => %4s\n", i, i.to_excel_column }

这将打印以下内容，例如:

   1 =>    A
   2 =>    B
   3 =>    C
  ....
  33 =>   AG
  34 =>   AH
  35 =>   AI
  36 =>   AJ
  37 =>   AK
  38 =>   AL
  39 =>   AM
  40 =>   AN
  41 =>   AO
  42 =>   AP
  43 =>   AQ
  44 =>   AR
  45 =>   AS
  46 =>   AT
  47 =>   AU
  48 =>   AV
  49 =>   AW
  50 =>   AX
  51 =>   AY
  52 =>   AZ

T-sql (sql server 18)

第一页的解决方案副本

CREATE FUNCTION dbo.getExcelColumnNameByOrdinal(@RowNum int)  
RETURNS varchar(5)   
AS   
BEGIN  
    DECLARE @dividend int = @RowNum;
    DECLARE @columnName varchar(max) = '';
    DECLARE @modulo int;

    WHILE (@dividend > 0)
    BEGIN  
        SELECT @modulo = ((@dividend - 1) % 26);
        SELECT @columnName = CHAR((65 + @modulo)) + @columnName;
        SELECT @dividend = CAST(((@dividend - @modulo) / 26) as int);
    END
    RETURN 
       @columnName;

END;

已经有30多个解决方案了，但这里是我的一行c#解决方案……

public string IntToExcelColumn(int i)
{
    return ((i<16926? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<2730? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + (i<26? "" : ((char)((((i/26)-1)%26)+65)).ToString()) + ((char)((i%26)+65)));
}

递归很简单。

public static string GetStandardExcelColumnName(int columnNumberOneBased)
{
  int baseValue = Convert.ToInt32('A');
  int columnNumberZeroBased = columnNumberOneBased - 1;

  string ret = "";

  if (columnNumberOneBased > 26)
  {
    ret = GetStandardExcelColumnName(columnNumberZeroBased / 26) ;
  }

  return ret + Convert.ToChar(baseValue + (columnNumberZeroBased % 26) );
}