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;
}

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)

 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();
    }

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())

我在VB中使用这个。NET 2003和它的工作良好…

Private Function GetExcelColumnName(ByVal aiColNumber As Integer) As String
    Dim BaseValue As Integer = Convert.ToInt32(("A").Chars(0)) - 1
    Dim lsReturn As String = String.Empty

    If (aiColNumber > 26) Then
        lsReturn = GetExcelColumnName(Convert.ToInt32((Format(aiColNumber / 26, "0.0").Split("."))(0)))
    End If

    GetExcelColumnName = lsReturn + Convert.ToChar(BaseValue + (aiColNumber Mod 26))
End Function

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) );
}