将A转换为1 B到2 ... Z转换为26，然后将AA转换为27 AB到28（Excel中的列索引到列引用）

Question

是否有人将算法或逻辑转换为1，B为2，...，Z为26，然后是AA到27，AB到28等。

换句话说，将列索引转换为Excel中的列引用。

Answer 1

这是一个简单的LINQ表达式：

static int TextToNumber(this string text) {
    return text
        .Select(c => c - 'A' + 1)
        .Aggregate((sum, next) => sum*26 + next);
}

此测试

Console.WriteLine(" A -> " + "A".TextToNumber());
Console.WriteLine(" B -> " + "B".TextToNumber());
Console.WriteLine(" Z -> " + "Z".TextToNumber());
Console.WriteLine("AA -> " + "AA".TextToNumber());
Console.WriteLine("AB -> " + "AB".TextToNumber());

将产生此输出：

 A -> 1
 B -> 2
 Z -> 26
AA -> 27
AB -> 28

<强>更新 这是相同的代码，但针对的是.NET 2.0：

static int TextToNumber(string text) {
    int sum = 0;
    foreach (char c in text) {
        sum = sum*26 + c - 'A' + 1;
    }
    return sum;
}

Answer 2

看看这些

/// <summary>
/// 1 -> A<br/>
/// 2 -> B<br/>
/// 3 -> C<br/>
/// ...
/// </summary>
/// <param name="column"></param>
/// <returns></returns>
public static string ExcelColumnFromNumber(int column)
{
    string columnString = "";
    decimal columnNumber = column;
    while (columnNumber > 0)
    {
        decimal currentLetterNumber = (columnNumber - 1) % 26;
        char currentLetter = (char)(currentLetterNumber + 65);
        columnString = currentLetter + columnString;
        columnNumber = (columnNumber - (currentLetterNumber + 1)) / 26;
    }
    return columnString;
}

/// <summary>
/// A -> 1<br/>
/// B -> 2<br/>
/// C -> 3<br/>
/// ...
/// </summary>
/// <param name="column"></param>
/// <returns></returns>
public static int NumberFromExcelColumn(string column)
{
    int retVal = 0;
    string col = column.ToUpper();
    for (int iChar = col.Length - 1; iChar >= 0; iChar--)
    {
        char colPiece = col[iChar];
        int colNum = colPiece - 64;
        retVal = retVal + colNum * (int)Math.Pow(26, col.Length - (iChar + 1));
    }
    return retVal;
}

Answer 3

如果你喜欢在客户端上完成，这是一个JavaScript代码

<script type="text/javascript" lang="javascript">
function Alphabet2Numeric(mystr) {
    mystr = mystr.toUpperCase(); //Hence the ASCII code 64 down there
    var sum = 0;
    for (var i = 0; i < mystr.length; i++) {
        sum = sum * 26 + mystr.charCodeAt(i) - 64; //returns 1 for 'a' and 2 for 'b' so on and so forth.
    }
    return sum;
}
</script>

Answer 4

同样的问题，不同的语言：PHP。

function charToInt($char)
{
   $array = array_flip(range(a, z));
   return $array[$char] + 1;
}

echo charToInt('c');

outputs: 3

Answer 5

从 A 到 ZZ : ((字符数 - 1) * 26 ) + 最后一个字符的 ASCII%65 + 第一个字符的 ASCII %65 * (26^(字符长度 -1)) -1 [仅在正数时添加]