T-sql (sql server 18)

第一页的解决方案副本

CREATE FUNCTION dbo.getExcelColumnNameByOrdinal(@RowNum int)  
RETURNS varchar(5)   
AS   
BEGIN  
    DECLARE @dividend int = @RowNum;
    DECLARE @columnName varchar(max) = '';
    DECLARE @modulo int;

    WHILE (@dividend > 0)
    BEGIN  
        SELECT @modulo = ((@dividend - 1) % 26);
        SELECT @columnName = CHAR((65 + @modulo)) + @columnName;
        SELECT @dividend = CAST(((@dividend - @modulo) / 26) as int);
    END
    RETURN 
       @columnName;

END;

另一个解决方案:

private void Foo()
{
   l_ExcelApp = new Excel.ApplicationClass();
   l_ExcelApp.ReferenceStyle = Excel.XlReferenceStyle.xlR1C1;
   // ... now reference by R[row]C[column], Ex. A1 <==> R1C1, C6 <==> R3C6, ...
}

在这里查看更多- Excel中的单元格引用!作者:Nitin Paranjape博士

Objective-C实现:

-(NSString*)getColumnName:(int)n {
     NSString *name = @"";
     while (n>0) {
     n--;
     char c = (char)('A' + n%26);
     name = [NSString stringWithFormat:@"%c%@",c,name];
     n = n/26;
  }    
     return name;

}

迅速实现:

func getColumnName(n:Int)->String{
 var columnName = ""
 var index = n
 while index>0 {
     index--
     let char = Character(UnicodeScalar(65 + index%26))
     columnName = "\(char)\(columnName)"
     index = index / 26
 }
 return columnName

}

答案是基于:https://stackoverflow.com/a/4532562/2231118

抱歉，这是Python而不是c#，但至少结果是正确的:

def excel_column_number_to_name(column_number):
    output = ""
    index = column_number-1
    while index >= 0:
        character = chr((index%26)+ord('A'))
        output = output + character
        index = index/26 - 1

    return output[::-1]


for i in xrange(1, 1024):
    print "%4d : %s" % (i, excel_column_number_to_name(i))

通过这些测试用例:

列号:494286 => ABCDZ 列号:27 => 列号:52 => AZ

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;

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;
}