下面是一个基于零的列索引的更简单的解决方案

 public static string GetColumnIndexNumberToExcelColumn(int columnIndex)
        {
            int offset = columnIndex % 26;
            int multiple = columnIndex / 26;

            int initialSeed = 65;//Represents column "A"
            if (multiple == 0)
            {
                return Convert.ToChar(initialSeed + offset).ToString();
            }

            return $"{Convert.ToChar(initialSeed + multiple - 1)}{Convert.ToChar(initialSeed + offset)}";
        }

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

我是这样做的:

private string GetExcelColumnName(int columnNumber)
{
    string columnName = "";

    while (columnNumber > 0)
    {
        int modulo = (columnNumber - 1) % 26;
        columnName = Convert.ToChar('A' + modulo) + columnName;
        columnNumber = (columnNumber - modulo) / 26;
    } 

    return columnName;
}

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));            
    }

这是我在PHP中的超级后期实现。这个是递归的。我是在发现这篇文章之前写的。我想看看其他人是否已经解决了这个问题……

public function GetColumn($intNumber, $strCol = null) {

    if ($intNumber > 0) {
        $intRem = ($intNumber - 1) % 26;
        $strCol = $this->GetColumn(intval(($intNumber - $intRem) / 26), sprintf('%s%s', chr(65 + $intRem), $strCol));
    }

    return $strCol;
}

    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));
    }