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;
}

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'。

巧合和优雅的Ruby版本:

def col_name(col_idx)
    name = ""
    while col_idx>0
        mod     = (col_idx-1)%26
        name    = (65+mod).chr + name
        col_idx = ((col_idx-mod)/26).to_i
    end
    name
end

另一种VBA方式

Public Function GetColumnName(TargetCell As Range) As String
    GetColumnName = Split(CStr(TargetCell.Cells(1, 1).Address), "$")(1)
End Function

我正在尝试在Java中做同样的事情… 我写了以下代码:

private String getExcelColumnName(int columnNumber) {

    int dividend = columnNumber;
    String columnName = "";
    int modulo;

    while (dividend > 0)
    {
        modulo = (dividend - 1) % 26;

        char val = Character.valueOf((char)(65 + modulo));

        columnName += val;

        dividend = (int)((dividend - modulo) / 26);
    } 

    return columnName;
}

现在，一旦我用columnNumber = 29运行它，它给我的结果=“CA”(而不是“AC”) 有什么意见吗? 我知道我可以通过StringBuilder....反转它但看着格雷厄姆的回答，我有点困惑....

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));            
    }