static string[] ExcelColumnAlphabetIdentifiers = new string[] { "", "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" };
 public static string ExcelColumnAlphabetIdentifier( int ColumnNumber)
    {
        StringBuilder sb = new StringBuilder();
        int remainder = ColumnNumber;
        do
        {
            sb.Append(ExcelColumnAlphabetIdentifiers[remainder % 26]);
            remainder = remainder / 26;
        }
        while (remainder > 0);
       return sb.ToString();
    }

如果有人需要在没有VBA的Excel中做到这一点，这里有一种方法:

=SUBSTITUTE(ADDRESS(1;colNum;4);"1";"")

其中colNum是列号

在VBA中:

Function GetColumnName(colNum As Integer) As String
    Dim d As Integer
    Dim m As Integer
    Dim name As String
    d = colNum
    name = ""
    Do While (d > 0)
        m = (d - 1) Mod 26
        name = Chr(65 + m) + name
        d = Int((d - m) / 26)
    Loop
    GetColumnName = name
End Function

看到了另一个VBA答案-这可以在excel-vba中用1行UDF完成:

Function GetColLetter(ByVal colID As Integer) As String
    If colID > Columns.Count Then
        Err.Raise 9, , "Column index out of bounds"
    Else
        GetColLetter = Split(Cells(1, colID).Address, "$")(1)
    End If
End Function

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

private String getColumn(int c) {
    String s = "";
    do {
        s = (char)('A' + (c % 26)) + s;
        c /= 26;
    } while (c-- > 0);
    return s;
}

它不是以26为底，系统中没有0。如果有的话，'Z'后面应该是'BA'而不是'AA'。

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)