NodeJS实现:

/**
* getColumnFromIndex
* Helper that returns a column value (A-XFD) for an index value (integer).
* The column follows the Common Spreadsheet Format e.g., A, AA, AAA.
* See https://stackoverflow.com/questions/181596/how-to-convert-a-column-number-eg-127-into-an-excel-column-eg-aa/3444285#3444285
* @param numVal: Integer
* @return String
*/
getColumnFromIndex: function(numVal){
   var dividend = parseInt(numVal);
   var columnName = '';
   var modulo;
   while (dividend > 0) {
      modulo = (dividend - 1) % 26;
      columnName = String.fromCharCode(65 + modulo) + columnName;
      dividend = parseInt((dividend - modulo) / 26);
   }
   return columnName;
},

将excel列字母(如AA)转换为数字(如25)。反过来说:

/**
* getIndexFromColumn
* Helper that returns an index value (integer) for a column value (A-XFD).
* The column follows the Common Spreadsheet Format e.g., A, AA, AAA.
* See https://stackoverflow.com/questions/9905533/convert-excel-column-alphabet-e-g-aa-to-number-e-g-25
* @param strVal: String
* @return Integer
*/
getIndexFromColumn: function(val){
   var base = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ', i, j, result = 0;
   for (i = 0, j = val.length - 1; i < val.length; i += 1, j -= 1) {
      result += Math.pow(base.length, j) * (base.indexOf(val[i]) + 1);
   }
   return result;
}

虽然我在这方面姗姗来迟，但格雷厄姆的答案远非最佳。特别是，你不需要使用模数，调用ToString()和apply (int)强制转换。考虑到在c#世界中的大多数情况下，您将从0开始编号，以下是我的修订:

public static string GetColumnName(int index) // zero-based
{
    const byte BASE = 'Z' - 'A' + 1;
    string name = String.Empty;

    do
    {
        name = Convert.ToChar('A' + index % BASE) + name;
        index = index / BASE - 1;
    }
    while (index >= 0);

    return name;
}

我的解决方案基于Graham, Herman Kan和desseim的回答，使用StringBuilder:

internal class Program
{
    #region get_excel_col_name
    /// <summary>
    /// Returns the name of the column by its number
    /// </summary>
    /// <param name="col_num">Column number</param>
    /// <returns>Column name</returns>
    /// <remarks>Numbering columns from zero</remarks>
    private static string get_excel_col_name(int col_num)
    {
        StringBuilder sb = new StringBuilder(2);
        if (col_num >= 0)
        {
            do
            {
                sb.Insert(0, (char)(col_num % 26 + 65));
                col_num /= 26;
            }
            while (--col_num >= 0);
        }
        return sb.ToString();
    }
    #endregion

    private static void Main(string[] args)
    {
        Console.WriteLine(get_excel_col_name(34));//outputs AI
        Console.ReadKey(true);
    }
}

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;
}

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));            
    }

只是抛出一个简单的使用递归的两行c#实现，因为这里所有的答案似乎都比必要的复杂得多。

/// <summary>
/// Gets the column letter(s) corresponding to the given column number.
/// </summary>
/// <param name="column">The one-based column index. Must be greater than zero.</param>
/// <returns>The desired column letter, or an empty string if the column number was invalid.</returns>
public static string GetColumnLetter(int column) {
    if (column < 1) return String.Empty;
    return GetColumnLetter((column - 1) / 26) + (char)('A' + (column - 1) % 26);
}