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

我想在我使用的静态类中加入，用于在col index和col Label之间进行交互。我对ColumnLabel方法使用了修改后的可接受答案

public static class Extensions
{
    public static string ColumnLabel(this int col)
    {
        var dividend = col;
        var columnLabel = string.Empty;
        int modulo;

        while (dividend > 0)
        {
            modulo = (dividend - 1) % 26;
            columnLabel = Convert.ToChar(65 + modulo).ToString() + columnLabel;
            dividend = (int)((dividend - modulo) / 26);
        } 

        return columnLabel;
    }
    public static int ColumnIndex(this string colLabel)
    {
        // "AD" (1 * 26^1) + (4 * 26^0) ...
        var colIndex = 0;
        for(int ind = 0, pow = colLabel.Count()-1; ind < colLabel.Count(); ++ind, --pow)
        {
            var cVal = Convert.ToInt32(colLabel[ind]) - 64; //col A is index 1
            colIndex += cVal * ((int)Math.Pow(26, pow));
        }
        return colIndex;
    }
}

用这个…

30.ColumnLabel(); // "AD"
"AD".ColumnIndex(); // 30

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

(我知道这个问题与c#有关，但是，如果读者需要用Java做同样的事情，那么下面的内容可能会有用)

事实证明，使用Jakarta POI中的“CellReference”类可以很容易地做到这一点。此外，转换可以以两种方式进行。

// Convert row and column numbers (0-based) to an Excel cell reference
CellReference numbers = new CellReference(3, 28);
System.out.println(numbers.formatAsString());

// Convert an Excel cell reference back into digits
CellReference reference = new CellReference("AC4");
System.out.println(reference.getRow() + ", " + reference.getCol());

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

打印稿

function lengthToExcelColumn(len: number): string {

    let dividend: number = len;
    let columnName: string = '';
    let modulo: number = 0;

    while (dividend > 0) {
        modulo = (dividend - 1) % 26;
        columnName = String.fromCharCode(65 + modulo).toString() + columnName;
        dividend = Math.floor((dividend - modulo) / 26);
    }
    return columnName;
}