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; 
    }  

NodeJS实现:

/**
* getColumnFromIndex
* Helper that returns a column value (A-XFD) for an index value (integer).
* The column follows the Common Spreadsheet Format e.g., A, AA, AAA.
* See https://stackoverflow.com/questions/181596/how-to-convert-a-column-number-eg-127-into-an-excel-column-eg-aa/3444285#3444285
* @param numVal: Integer
* @return String
*/
getColumnFromIndex: function(numVal){
   var dividend = parseInt(numVal);
   var columnName = '';
   var modulo;
   while (dividend > 0) {
      modulo = (dividend - 1) % 26;
      columnName = String.fromCharCode(65 + modulo) + columnName;
      dividend = parseInt((dividend - modulo) / 26);
   }
   return columnName;
},

将excel列字母(如AA)转换为数字(如25)。反过来说:

/**
* getIndexFromColumn
* Helper that returns an index value (integer) for a column value (A-XFD).
* The column follows the Common Spreadsheet Format e.g., A, AA, AAA.
* See https://stackoverflow.com/questions/9905533/convert-excel-column-alphabet-e-g-aa-to-number-e-g-25
* @param strVal: String
* @return Integer
*/
getIndexFromColumn: function(val){
   var base = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ', i, j, result = 0;
   for (i = 0, j = val.length - 1; i < val.length; i += 1, j -= 1) {
      result += Math.pow(base.length, j) * (base.indexOf(val[i]) + 1);
   }
   return result;
}

．．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;
}

下面是一个Actionscript版本:

private var columnNumbers:Array = ['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'];

    private function getExcelColumnName(columnNumber:int) : String{
        var dividend:int = columnNumber;
        var columnName:String = "";
        var modulo:int;

        while (dividend > 0)
        {
            modulo = (dividend - 1) % 26;
            columnName = columnNumbers[modulo] + columnName;
            dividend = int((dividend - modulo) / 26);
        } 

        return columnName;
    }

private String getColumn(int c) {
    String s = "";
    do {
        s = (char)('A' + (c % 26)) + s;
        c /= 26;
    } while (c-- > 0);
    return s;
}

它不是以26为底，系统中没有0。如果有的话，'Z'后面应该是'BA'而不是'AA'。

到目前为止，所有的解决方案都包含迭代或递归，这让我感到惊讶。

这是我的解，在常数时间内运行(没有循环)。此解决方案适用于所有可能的Excel列，并检查输入是否可以转换为Excel列。可能的列在[A, XFD]或[1,16384]范围内。(这取决于你的Excel版本)

private static string Turn(uint col)
{
    if (col < 1 || col > 16384) //Excel columns are one-based (one = 'A')
        throw new ArgumentException("col must be >= 1 and <= 16384");

    if (col <= 26) //one character
        return ((char)(col + 'A' - 1)).ToString();

    else if (col <= 702) //two characters
    {
        char firstChar = (char)((int)((col - 1) / 26) + 'A' - 1);
        char secondChar = (char)(col % 26 + 'A' - 1);

        if (secondChar == '@') //Excel is one-based, but modulo operations are zero-based
            secondChar = 'Z'; //convert one-based to zero-based

        return string.Format("{0}{1}", firstChar, secondChar);
    }

    else //three characters
    {
        char firstChar = (char)((int)((col - 1) / 702) + 'A' - 1);
        char secondChar = (char)((col - 1) / 26 % 26 + 'A' - 1);
        char thirdChar = (char)(col % 26 + 'A' - 1);

        if (thirdChar == '@') //Excel is one-based, but modulo operations are zero-based
            thirdChar = 'Z'; //convert one-based to zero-based

        return string.Format("{0}{1}{2}", firstChar, secondChar, thirdChar);
    }
}