如何在c#中将数值转换为Excel列名,而不使用直接从Excel中获取值的自动化。

Excel 2007的范围可能是1到16384,这是它支持的列数。结果值应以excel列名的形式出现,例如A、AA、AAA等。


当前回答

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);
    }
}

其他回答

您可能需要两种方式转换,例如从Excel列地址(如AAZ)到整数,以及从任何整数到Excel。下面的两个方法就可以做到这一点。假设基于1的索引,“数组”中的第一个元素是元素1。 这里没有大小限制,所以你可以使用ERROR这样的地址,这将是列号2613824…

public static string ColumnAdress(int col)
{
  if (col <= 26) { 
    return Convert.ToChar(col + 64).ToString();
  }
  int div = col / 26;
  int mod = col % 26;
  if (mod == 0) {mod = 26;div--;}
  return ColumnAdress(div) + ColumnAdress(mod);
}

public static int ColumnNumber(string colAdress)
{
  int[] digits = new int[colAdress.Length];
  for (int i = 0; i < colAdress.Length; ++i)
  {
    digits[i] = Convert.ToInt32(colAdress[i]) - 64;
  }
  int mul=1;int res=0;
  for (int pos = digits.Length - 1; pos >= 0; --pos)
  {
    res += digits[pos] * mul;
    mul *= 26;
  }
  return res;
}

抱歉,这是Python而不是c#,但至少结果是正确的:

def ColIdxToXlName(idx):
    if idx < 1:
        raise ValueError("Index is too small")
    result = ""
    while True:
        if idx > 26:
            idx, r = divmod(idx - 1, 26)
            result = chr(r + ord('A')) + result
        else:
            return chr(idx + ord('A') - 1) + result


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

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);
    }
}

我今天必须做这个工作,我的实现使用递归:

private static string GetColumnLetter(string colNumber)
{
    if (string.IsNullOrEmpty(colNumber))
    {
        throw new ArgumentNullException(colNumber);
    }

    string colName = String.Empty;

    try
    {
        var colNum = Convert.ToInt32(colNumber);
        var mod = colNum % 26;
        var div = Math.Floor((double)(colNum)/26);
        colName = ((div > 0) ? GetColumnLetter((div - 1).ToString()) : String.Empty) + Convert.ToChar(mod + 65);
    }
    finally
    {
        colName = colName == String.Empty ? "A" : colName;
    }

    return colName;
}

该方法将数字视为字符串,而以“0”开头的数字(A = 0)

(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());