如何在c#中将数值转换为Excel列名,而不使用直接从Excel中获取值的自动化。

Excel 2007的范围可能是1到16384,这是它支持的列数。结果值应以excel列名的形式出现,例如A、AA、AAA等。


当前回答

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);
}

其他回答

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));            
    }

..并转换为php:

function GetExcelColumnName($columnNumber) {
    $columnName = '';
    while ($columnNumber > 0) {
        $modulo = ($columnNumber - 1) % 26;
        $columnName = chr(65 + $modulo) . $columnName;
        $columnNumber = (int)(($columnNumber - $modulo) / 26);
    }
    return $columnName;
}
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;

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

在Java中实现相同

public String getExcelColumnName (int columnNumber) 
    {     
        int dividend = columnNumber;   
        int i;
        String columnName = "";     
        int modulo;     
        while (dividend > 0)     
        {        
            modulo = (dividend - 1) % 26;         
            i = 65 + modulo;
            columnName = new Character((char)i).toString() + columnName;        
            dividend = (int)((dividend - modulo) / 26);    
        }       
        return columnName; 
    }