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

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))

其他回答

如果你只是想要一个没有代码的单元格公式,这里有一个公式:

IF(COLUMN()>=26,CHAR(ROUND(COLUMN()/26,1)+64)&CHAR(MOD(COLUMN(),26)+64),CHAR(COLUMN()+64))

这些我的代码转换特定的数字(索引从1开始)到Excel列。

    public static string NumberToExcelColumn(uint number)
    {
        uint originalNumber = number;

        uint numChars = 1;
        while (Math.Pow(26, numChars) < number)
        {
            numChars++;

            if (Math.Pow(26, numChars) + 26 >= number)
            {
                break;
            }               
        }

        string toRet = "";
        uint lastValue = 0;

        do
        {
            number -= lastValue;

            double powerVal = Math.Pow(26, numChars - 1);
            byte thisCharIdx = (byte)Math.Truncate((columnNumber - 1) / powerVal);
            lastValue = (int)powerVal * thisCharIdx;

            if (numChars - 2 >= 0)
            {
                double powerVal_next = Math.Pow(26, numChars - 2);
                byte thisCharIdx_next = (byte)Math.Truncate((columnNumber - lastValue - 1) / powerVal_next);
                int lastValue_next = (int)Math.Pow(26, numChars - 2) * thisCharIdx_next;

                if (thisCharIdx_next == 0 && lastValue_next == 0 && powerVal_next == 26)
                {
                    thisCharIdx--;
                    lastValue = (int)powerVal * thisCharIdx;
                }
            }

            toRet += (char)((byte)'A' + thisCharIdx + ((numChars > 1) ? -1 : 0));

            numChars--;
        } while (numChars > 0);

        return toRet;
    }

我的单元测试:

    [TestMethod]
    public void Test()
    {
        Assert.AreEqual("A", NumberToExcelColumn(1));
        Assert.AreEqual("Z", NumberToExcelColumn(26));
        Assert.AreEqual("AA", NumberToExcelColumn(27));
        Assert.AreEqual("AO", NumberToExcelColumn(41));
        Assert.AreEqual("AZ", NumberToExcelColumn(52));
        Assert.AreEqual("BA", NumberToExcelColumn(53));
        Assert.AreEqual("ZZ", NumberToExcelColumn(702));
        Assert.AreEqual("AAA", NumberToExcelColumn(703));
        Assert.AreEqual("ABC", NumberToExcelColumn(731));
        Assert.AreEqual("ACQ", NumberToExcelColumn(771));
        Assert.AreEqual("AYZ", NumberToExcelColumn(1352));
        Assert.AreEqual("AZA", NumberToExcelColumn(1353));
        Assert.AreEqual("AZB", NumberToExcelColumn(1354));
        Assert.AreEqual("BAA", NumberToExcelColumn(1379));
        Assert.AreEqual("CNU", NumberToExcelColumn(2413));
        Assert.AreEqual("GCM", NumberToExcelColumn(4823));
        Assert.AreEqual("MSR", NumberToExcelColumn(9300));
        Assert.AreEqual("OMB", NumberToExcelColumn(10480));
        Assert.AreEqual("ULV", NumberToExcelColumn(14530));
        Assert.AreEqual("XFD", NumberToExcelColumn(16384));
    }

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

谢谢你的回答!!帮助我想出了这些帮助函数,与我正在Elixir/Phoenix中工作的谷歌Sheets API进行一些交互

以下是我想到的(可能需要一些额外的验证和错误处理)

长生不老药:

def number_to_column(number) do
  cond do
    (number > 0 && number <= 26) ->
      to_string([(number + 64)])
    (number > 26) ->
      div_col = number_to_column(div(number - 1, 26))
      remainder = rem(number, 26)
      rem_col = cond do
        (remainder == 0) ->
          number_to_column(26)
        true ->
          number_to_column(remainder)
      end
      div_col <> rem_col
    true ->
      ""
  end
end

逆函数是:

def column_to_number(column) do
  column
    |> to_charlist
    |> Enum.reverse
    |> Enum.with_index
    |> Enum.reduce(0, fn({char, idx}, acc) ->
      ((char - 64) * :math.pow(26,idx)) + acc
    end)
    |> round
end

还有一些测试:

describe "test excel functions" do
  @excelTestData [{"A", 1}, {"Z",26}, {"AA", 27}, {"AB", 28}, {"AZ", 52},{"BA", 53}, {"AAA", 703}]

  test "column to number" do
    Enum.each(@excelTestData, fn({input, expected_result}) ->
      actual_result = BulkOnboardingController.column_to_number(input)
      assert actual_result == expected_result
    end)
  end

  test "number to column" do
    Enum.each(@excelTestData, fn({expected_result, input}) ->
      actual_result = BulkOnboardingController.number_to_column(input)
      assert actual_result == expected_result
    end)
  end
end

(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());