如何将Excel表格列名转换为数字？

Question

我在想将Excel表格中的列名转换成数字的最佳方法是什么。

我正在使用Excel Package，这是一个很好的库，用于处理 .xlsx 文档。但是这个库不幸没有包括此功能。

OBS：在这个库中，第一列 A 对应数字1。

- Victor Rodrigues

Answer 1

该函数应适用于任意长度的列名。

public static int GetColumnNumber(string name)
{
    int number = 0;
    int pow = 1;
    for (int i = name.Length - 1; i >= 0; i--)
    {
        number += (name[i] - 'A' + 1) * pow;
        pow *= 26;
    }

    return number;
}

- Noldorin

---

请注意，只有在所给出的“列名”（name）是大写字母时才能按预期工作。（请将name[i]替换为name.ToUpper()[i]以修复此问题。） - Marcus Mangelsdorf

你可以通过使用霍纳方法来加速运行并消除pow变量的需求。这意味着你可以使用正常的i = 0；i < L；i ++循环，这很好。 - Ethan Maness

Answer 2

我几个月前就遇到了这个问题。反向操作-从列索引到列名称-也很有趣，如果你尝试使用零基索引解决它而没有意识到这会使事情变得复杂，那么结果将变得非常混乱。如果它是一个正常的多项数字系统，那么它可能会变得非常简单...

以下是我解决该问题的简化版本，作为扩展方法，没有错误处理和所有那些东西。

public static Int32 ToOneBasedIndex(this String name)
{
    return name.ToUpper().
       Aggregate(0, (column, letter) => 26 * column + letter - 'A' + 1);
}

- Daniel Brückner

Answer 3

我已经使用这个技术一段时间了，发现它在处理超出A-Z甚至AA-ZZ的列时非常有效。它通过将字符串中的每个字符分解，并递归调用自身来推导ASCII字符的DEC值（减去64），然后乘以26^n来实现。当n>4时，使用long类型的返回值可以克服潜在的限制。

    public long columnNumber(String columnName)
    {
        char[] chars = columnName.ToUpper().ToCharArray();

        return (long)(Math.Pow(26, chars.Count() - 1)) * 
            (System.Convert.ToInt32(chars[0]) - 64) + 
            ((chars.Count() > 2) ? columnNumber(columnName.Substring(1, columnName.Length - 1)) : 
            ((chars.Count() == 2) ? (System.Convert.ToInt32(chars[chars.Count() - 1]) - 64) : 0));
    }

此外，如果您想获得反向操作（即传入列号并获取列名），以下是一些可实现此操作的代码。

    public String columnName(long columnNumber)
    {
        StringBuilder retVal = new StringBuilder();
        int x = 0;

        for (int n = (int)(Math.Log(25*(columnNumber + 1))/Math.Log(26)) - 1; n >= 0; n--)
        {
            x = (int)((Math.Pow(26,(n + 1)) - 1) / 25 - 1);
            if (columnNumber > x)
                retVal.Append(System.Convert.ToChar((int)(((columnNumber - x - 1) / Math.Pow(26, n)) % 26 + 65)));
        }

        return retVal.ToString();
    }

- Sam Martinez

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伟大的算法。完美运行！ - Gašper Sladič

Answer 4

源代码：

namespace XLS
{
/// <summary>
/// Represents a single cell in a excell sheet
/// </summary>
public struct Cell
{
    private long row;
    private long column;
    private string columnAddress;
    private string address;
    private bool dataChange;

    /// <summary>
    /// Initializes a new instance of the XLS.Cell 
    /// class with the specified row and column of excel worksheet
    /// </summary>
    /// <param name="row">The row index of a cell</param>
    /// <param name="column">The column index of a cell</param>
    public Cell(long row, long column)
    {
        this.row = row;
        this.column = column;
        dataChange = true;
        address = string.Empty;
        columnAddress = string.Empty;
    }

    /// <summary>
    /// Initializes a new instance of the XLS.Cell
    /// class with the specified address of excel worksheet
    /// </summary>
    /// <param name="address">The adress of a cell</param>
    public Cell(string address)
    {
        this.address = address;
        dataChange = false;
        row = GetRow(address);
        columnAddress = GetColumnAddress(address);
        column = GetColumn(columnAddress);
    }

    /// <summary>
    /// Gets or sets the row of this XLS.Cell
    /// </summary>
    public long Row
    {
        get { return row <= 0 ? 1 : row; }
        set { row = value; dataChange = true; }
    }

    /// <summary>
    /// Gets or sets the column of this XLS.Cell
    /// </summary>
    public long Column
    {
        get { return column <= 0 ? 1 : column; }
        set { column = value; dataChange = true; }
    }

    /// <summary>
    /// Gets or sets the address of this XLS.Cell
    /// </summary>
    public string Address
    {
        get { return dataChange ? ToAddress() : address; }
        set
        {
            address = value;
            row = GetRow(address);
            column = GetColumn(address);
        }
    }

    /// <summary>
    /// Gets the column address of this XLS.Cell
    /// </summary>
    public string ColumnAddress
    {
        get { return GetColumnAddress(Address); }
        private set { columnAddress = value; }
    }

    #region Private Methods

    private static long GetRow(string address)
    {
        return long.Parse(address.Substring(GetStartIndex(address)));
    }

    private static string GetColumnAddress(string address)
    {
        return address.Substring(0, GetStartIndex(address)).ToUpperInvariant();
    }

    private static long GetColumn(string columnAddress)
    {
        char[] characters = columnAddress.ToCharArray();
        int sum = 0;
        for (int i = 0; i < characters.Length; i++)
        {
            sum *= 26;
            sum += (characters[i] - 'A' + 1);
        }

        return (long)sum;
    }

    private static int GetStartIndex(string address)
    {
        return address.IndexOfAny("123456789".ToCharArray());
    }

    private string ToAddress()
    {
        string indexToString = string.Empty;

        if (Column > 26)
        {
            indexToString = ((char)(65 + (int)((Column - 1) / 26) - 1)).ToString();
        }

        indexToString += (char)(65 + ((Column - 1) % 26));

        dataChange = false;
        return indexToString + Row;
    }

    #endregion
}

}

- Arijit Pal Chaudhury

Answer 5

public static string GetColumnName(int index)
{
    const string letters = "ZABCDEFGHIJKLMNOPQRSTUVWXY";

    int NextPos = (index / 26);
    int LastPos = (index % 26);
    if (LastPos == 0) NextPos--;

    if (index > 26)
        return GetColumnName(NextPos) + letters[LastPos];
    else
        return letters[LastPos] + "";
}

- singapore saravanan

Answer 6

O24有一个列号，你想要一个名称：

=LEFT(RIGHT(ADDRESS(1,O24),LEN(ADDRESS(1,O24))-1),FIND("$",RIGHT((ADDRESS(1,O24)),LEN(ADDRESS(1,O24))-1))-1)

O37有一个列名，你想要一个数字：

=COLUMN(INDIRECT(O37&1))

- George K.