SQL查询组合，无重复

我需要一个可以在（或作为）函数中使用的查询，并检索n值的所有组合。我需要长度为k的所有组合，其中k = 1..n。

扩展的样本输入和结果，因此输入具有3个值而不是2个-但是，输入值的数量可能从1到n不等。

示例：输入：具有多行一列中的值的表

Value  (nvarchar(500))
------
Ann
John
Mark

输出＃1：表，其值串联在一栏中

    Ann
    John
    Mark
    Ann,John
    John,Mark
    Ann,Mark
    Ann,John,Mark

对于相对较小的值n（在此示例中为20），可以使用利用自然整数是位的组合这一事实的方法。

T-SQL解决方案

样本数据：

DECLARE @Sample AS TABLE 
(
    item_id     tinyint IDENTITY(1,1) PRIMARY KEY NONCLUSTERED,
    item        nvarchar(500) NOT NULL,
    bit_value   AS 
                CONVERT
                (
                    integer, 
                    POWER(2, item_id - 1)
                )
                PERSISTED UNIQUE CLUSTERED
);    

INSERT @Sample
    (item)
VALUES
    (N'Ann'),
    (N'Bob'),
    (N'Charles'),
    (N'Darren'),
    (N'Eric'),
    (N'Fred'),
    (N'George'),
    (N'Harry'),
    (N'Ian'),
    (N'John'),
    (N'Keith'),
    (N'Larry'),
    (N'Mark'),
    (N'Nathan'),
    (N'Owen'),
    (N'Paul'),
    (N'Quentin'),
    (N'Ryan'),
    (N'Steve'),
    (N'Terry');

解：

-- Maximum integer we need
-- for all combinations of 'n' bits
DECLARE 
    @max integer = 
    POWER(2,
        (
            SELECT COUNT(*) 
            FROM @Sample AS s
        )
    ) - 1;

SELECT
    combination =
        STUFF
        (
            (
                -- Choose items where the bit is set
                -- and concatenate all matches
                SELECT ',' + s.item 
                FROM @Sample AS s
                WHERE
                    n.n & s.bit_value = s.bit_value
                ORDER BY
                    s.bit_value
                FOR XML 
                    PATH (''),
                    TYPE                    
            ).value('(./text())[1]', 'varchar(8000)'), 1, 1, ''
        )
-- A standard numbers table
-- (single column, integers from 1 to 1048576, indexed)
FROM dbo.Numbers AS N
WHERE
    N.n BETWEEN 1 AND @max;

输出样本：

╔════════════════════════╗
║      combination       ║
╠════════════════════════╣
║ Ann                    ║
║ Bob                    ║
║ Ann,Bob                ║
║ Charles                ║
║ Ann,Charles            ║
║ Bob,Charles            ║
║ Ann,Bob,Charles        ║
║ Darren                 ║
║ Ann,Darren             ║
║ Bob,Darren             ║
║ Ann,Bob,Darren         ║
║ Charles,Darren         ║
║ Ann,Charles,Darren     ║
║ Bob,Charles,Darren     ║
║ Ann,Bob,Charles,Darren ║
║ ...                    ║
╚════════════════════════╝

执行计划：

将1,048,576个组合写入我的笔记本电脑上的变量需要41秒。通过强制并行，我能够将执行时间减少到13秒。DOP 8

如果您需要数字表，这是产生一个数字表的快速方法：

SELECT TOP (1048576)
    n = ISNULL(CONVERT(integer, ROW_NUMBER() OVER (ORDER BY (SELECT NULL))), 0)
INTO dbo.Numbers
FROM sys.columns AS c
CROSS JOIN sys.columns AS c2
CROSS JOIN sys.columns AS c3;

CREATE UNIQUE CLUSTERED INDEX cuq
ON dbo.Numbers (n)
WITH (MAXDOP = 1, SORT_IN_TEMPDB = ON);

SQLCLR解决方案

在SQLCLR（SQL Server 2005及更高版本）中可以实现更有效的实现：

CREATE ASSEMBLY [Demo]
    AUTHORIZATION [dbo]
    FROM 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GO
CREATE FUNCTION dbo.Combinations
(
    @ElementsCSV nvarchar (4000)
)
RETURNS TABLE
(
    Combination nvarchar (4000) NULL
)
AS EXTERNAL NAME Demo.UserDefinedFunctions.Permute;

用法示例：

SELECT 
    f.Combination
FROM dbo.Combinations('A,B,C,D') AS f;

输出：

╔═════════════╗
║ Combination ║
╠═════════════╣
║ A           ║
║ B           ║
║ A,B         ║
║ C           ║
║ A,C         ║
║ B,C         ║
║ A,B,C       ║
║ D           ║
║ A,D         ║
║ B,D         ║
║ A,B,D       ║
║ C,D         ║
║ A,C,D       ║
║ B,C,D       ║
║ A,B,C,D     ║
╚═════════════╝

使用之前的20个元素集：

DECLARE @Sample AS TABLE 
(
    item_id     tinyint IDENTITY(1,1) PRIMARY KEY CLUSTERED,
    item        nvarchar(50) NOT NULL
);

INSERT @Sample
    (item)
VALUES
    (N'Ann'),
    (N'Bob'),
    (N'Charles'),
    (N'Darren'),
    (N'Eric'),
    (N'Fred'),
    (N'George'),
    (N'Harry'),
    (N'Ian'),
    (N'John'),
    (N'Keith'),
    (N'Larry'),
    (N'Mark'),
    (N'Nathan'),
    (N'Owen'),
    (N'Paul'),
    (N'Quentin'),
    (N'Ryan'),
    (N'Steve'),
    (N'Terry');

SQLCLR解决方案：

-- Create CSV input
DECLARE 
    @Elements nvarchar(4000) =
        STUFF
        (
            (
                SELECT ',' + s.item 
                FROM @Sample AS s
                ORDER BY
                    s.item_id
                FOR XML 
                    PATH (''),
                    TYPE                    
            ).value('(./text())[1]', 'varchar(8000)'), 1, 1, ''
        );

DECLARE
    @bitbucket nvarchar(4000);

SELECT
    @bitbucket = combination
FROM dbo.Combinations(@Elements);

我的笔记本电脑上的1,048,576组合的执行时间为2.5秒。DOP 1

创建CSV输入：

查找组合：

C＃源代码：

using System;
using System.Collections;
using Microsoft.SqlServer.Server;

public partial class UserDefinedFunctions
{
    [SqlFunction
        (
        DataAccess=DataAccessKind.None, 
        SystemDataAccess=SystemDataAccessKind.None,
        FillRowMethodName="FillRow",
        TableDefinition="Permutation nvarchar(4000)"
        )
    ]
    public static IEnumerable Permute(string ElementsCSV)
    {
        // Split CSV
        string[] elements = ElementsCSV.Split(new char[] { ',' });

        // Highest integer needed
        int count = (int)Math.Pow(2, elements.Length) - 1;

        // Pre-computed array of 2^n values
        int[] powers = new int[elements.Length];

        for (int i = 0; i < powers.Length; i++)
        {
            powers[i] = (int)Math.Pow(2, i);
        }

        // Test integers
        for (int i = 1; i <= count; i++)
        {
            // Reset output
            string s = string.Empty;

            // Test each bit that could be set
            for (int bit = 0; bit < powers.Length && i >= powers[bit]; bit++)
            {
                if ((i & powers[bit]) == powers[bit])
                {
                    // Add the element corresponding to the set bit
                    s += elements[bit] + ',';
                }
            }

            // Return a row via enumeration
            yield return s.Substring(0, s.Length - 1);
        }
    }

    // Called by SQL Server to fetch a row
    public static void FillRow(object o, out string Permutation)
    {
        Permutation = (string)o;
    }
}