从连接字符串中删除重复值

时间：2014-07-03 12:30:23

标签： sql sql-server sql-server-2008

我有下表：

Object  Field  Values
------------------------------------
1       1      A;A;A;B;A;A
2       1      A;B;C;C
2       2      X
3       1      X;Y;Z
3       2      V;V;V;V;V;V;V;V;V;V;V

如何从此表中仅选择连接values中的唯一值？所以：

Object  Field  Values
---------------------
1       1      A;B
2       1      A;B;C
2       2      X
3       1      X;Y;Z
3       2      V

在任何脚本语言中，我都会遍历来自Values的值，在;上爆炸并循环遍历该数组，并使用一些逻辑过滤掉重复项。但是，我只需要使用SQL（Server 2008）来执行此操作。

有人可以告诉我是否以及如何做到这一点？

非常感谢任何帮助： - ）

3 个答案:

答案 0 :(得分：2)

要执行此操作，请首先创建拆分功能。这是我使用的那个，但是如果你在网上搜索（或者甚至是SO）＆＃34; SQL Server分割功能＆＃34;如果你不喜欢这个，你会发现很多选择：

ALTER FUNCTION [dbo].[Split](@StringToSplit NVARCHAR(MAX), @Delimiter NCHAR(1))
RETURNS TABLE
AS
RETURN
(   
    SELECT  ID = ROW_NUMBER() OVER(ORDER BY n.Number),
            Position = Number,
            Value = SUBSTRING(@StringToSplit, Number, CHARINDEX(@Delimiter, @StringToSplit + @Delimiter, Number) - Number)
    FROM    (   SELECT  TOP (LEN(@StringToSplit) + 1) Number = ROW_NUMBER() OVER(ORDER BY a.object_id)
                FROM    sys.all_objects a
                        CROSS JOIN sys.all_objects b
            ) n
    WHERE   SUBSTRING(@Delimiter + @StringToSplit + @Delimiter, n.Number, 1) = @Delimiter
);

然后你可以分割你的领域，所以运行：

SELECT  t.Object, t.Field, s.Value
FROM    T
        CROSS APPLY dbo.Split(t.[Values], ';') AS s

转过来：

Object  Field  Values
------------------------------------
1       1      A;A;A;B;A;A

成：

Object  Field  Values
------------------------------------
1       1      A
1       1      A
1       1      A
1       1      B
1       1      A
1       1      A

然后您可以应用DISTINCT运营商：

SELECT  DISTINCT t.Object, t.Field, s.Value
FROM    T
        CROSS APPLY dbo.Split(t.[Values], ';') AS s;

给予：

Object  Field  Values
------------------------------------
1       1      A
1       1      B

然后，您可以将行连接回一个列，提供最终查询：

SELECT  t.Object, t.Field, [Values] = STUFF(x.value('.', 'NVARCHAR(MAX)'), 1, 1, '')
FROM    T
        CROSS APPLY 
        (   SELECT  DISTINCT ';' + s.Value
            FROM    dbo.Split(t.[Values], ';') AS s
            FOR XML PATH(''), TYPE
        ) AS s (x)

SQL Fiddle似乎已关闭，但是一旦你创建了Split功能，下面就是一个完整的例子：

CREATE TABLE #T (Object INT, Field INT, [Values] VARCHAR(MAX));
INSERT #T
VALUES
    (1, 1, 'A;A;A;B;A;A'),
    (2, 1, 'A;B;C;C'),
    (2, 2, 'X'),
    (3, 1, 'X;Y;Z'),
    (3, 2, 'V;V;V;V;V;V;V;V;V;V;V');

SELECT  t.Object, t.Field, [Values] = STUFF(x.value('.', 'NVARCHAR(MAX)'), 1, 1, '')
FROM    #T AS T
        CROSS APPLY 
        (   SELECT  DISTINCT ';' + s.Value
            FROM    dbo.Split(t.[Values], ';') AS s
            FOR XML PATH(''), TYPE
        ) AS s (x);

修改

根据您的评论，您无法创建表格或修改DDL，我想我会考虑您无法创建功能的情况。您可以将上面的拆分功能扩展到您的查询中，因此您实际上并不需要创建一个功能：

CREATE TABLE #T (Object INT, Field INT, [Values] VARCHAR(MAX));
INSERT #T
VALUES
    (1, 1, 'A;A;A;B;A;A'),
    (2, 1, 'A;B;C;C'),
    (2, 2, 'X'),
    (3, 1, 'X;Y;Z'),
    (3, 2, 'V;V;V;V;V;V;V;V;V;V;V');

SELECT  t.Object,
        t.Field,
        [Values] = STUFF(x.value('.', 'NVARCHAR(MAX)'), 1, 1, '')
FROM    #T AS T
        CROSS APPLY 
        (   SELECT  DISTINCT ';' + SUBSTRING(t.[Values], Number, CHARINDEX(';', t.[Values] + ';', Number) - Number)
            FROM    (   SELECT  TOP (LEN(t.[Values]) + 1) Number = ROW_NUMBER() OVER(ORDER BY a.object_id)
                        FROM    sys.all_objects a
                                CROSS JOIN sys.all_objects b
                    ) n
            WHERE   SUBSTRING(';' + t.[Values] + ';', n.Number, 1) = ';'
            FOR XML PATH(''), TYPE
        ) AS s (x);

答案 1 :(得分：2)

这是一个独立的解决方案：

DECLARE @t table(Object int, Field int, [Values] varchar(max))
INSERT @t values
(1, 1, 'A;A;A;B;A;A'),
(2, 1, 'A;B;C;C'),
(3, 1, 'X'),
(4, 1, 'X;Y;Z'),
(5, 1, 'V;V;V;V;V;V;V;V;V;V;V')

SELECT t.Object, t.Field, x.[NewValues]
FROM @t t
CROSS APPLY
(
  SELECT STUFF((
     SELECT distinct ';'+t.c.value('.', 'VARCHAR(2000)') value
     FROM (
         SELECT x = CAST('<t>' + 
               REPLACE([Values], ';', '</t><t>') + '</t>' AS XML)
     ) a
     CROSS APPLY x.nodes('/t') t(c)
        for xml path(''), type 
    ).value('.', 'varchar(max)'), 1, 1, '') [NewValues] 
) x

结果：

Object Field  NewValues
1      1      A;B
2      1      A;B;C
3      1      X
4      1      X;Y;Z
5      1      V

根据@ GarethD的评论，这可能会表现得很慢。

测试数据：

create table #t(Object int identity(1,1), Field int, [Values] varchar(max))
INSERT #t values
(1, 'A;A;A;B;A;A'),(1, 'A;B;C;C'), (1, 'X'), (1, 'X;Y;Z'),(1, 'V;V;V;V;V;V;V;V;V;V;V')

insert #t select field, [values] from #t union all select field, [values] from #t union all select field, [values] from #t
insert #t select field, [values] from #t union all select field, [values] from #t union all select field, [values] from #t
insert #t select field, [values] from #t union all select field, [values] from #t union all select field, [values] from #t
insert #t select field, [values] from #t union all select field, [values] from #t union all select field, [values] from #t
insert #t select field, [values] from #t union all select field, [values] from #t union all select field, [values] from #t
insert #t select field, [values] from #t union all select field, [values] from #t union all select field, [values] from #t

性能测试我的脚本：

SELECT t.Object, t.Field, x.[NewValues]
FROM #t t
CROSS APPLY
(
  SELECT STUFF((
     SELECT distinct ';'+t.c.value('.', 'VARCHAR(2000)') value
     FROM (
         SELECT x = CAST('<t>' + 
               REPLACE([Values], ';', '</t><t>') + '</t>' AS XML)
     ) a
     CROSS APPLY x.nodes('/t') t(c)
        for xml path(''), type 
    ).value('.', 'varchar(max)'), 1, 1, '') [NewValues] 
) x

结果不到1秒

性能测试Garath脚本

（必须编辑testdata以获取所有行。相同的行被视为1行）：

WITH CTE AS
(   SELECT  DISTINCT t.Object, t.Field, s.Value
    FROM    #T AS T
            CROSS APPLY 
            (   SELECT  ID = ROW_NUMBER() OVER(ORDER BY n.Number),
                        Position = Number,
                        Value = SUBSTRING(t.[Values], Number, CHARINDEX(';', t.[Values] + ';', Number) - Number)
                FROM    (   SELECT  TOP (LEN(t.[Values]) + 1) Number = ROW_NUMBER() OVER(ORDER BY a.object_id)
                            FROM    sys.all_objects a
                                    CROSS JOIN sys.all_objects b
                        ) n
                WHERE   SUBSTRING(';' + t.[Values] + ';', n.Number, 1) = ';'
            ) AS s
)
SELECT  Object,
        Field,
        [Values] = STUFF((SELECT ';' + Value
                        FROM CTE AS T2
                        WHERE T2.Object = T.Object
                        AND T2.Field = T.Field
                        FOR XML PATH(''), TYPE
                        ).value('.', 'VARCHAR(MAX)'), 1, 1, '')

FROM    CTE AS T
GROUP BY Object, Field;

结果6秒

如果任何行的值为null，则此脚本也会崩溃。

答案 2 :(得分：0)

就像没有CTE的标量值函数替代...... 

ALTER FUNCTION [SplitRemoveDupes] (
@String VARCHAR(MAX)
,@Delimiter VARCHAR(5)
)
RETURNS VARCHAR(MAX)
AS
BEGIN

DECLARE @SplitLength INT 
DECLARE @DedupedValues VARCHAR(MAX) 

DECLARE @SplittedValues TABLE 
( 
  OccurenceId SMALLINT IDENTITY(1,1), 
  SplitValue VARCHAR(200) 
)  

    WHILE LEN(@String) > 0
    BEGIN
        SELECT @SplitLength = (
                CASE CHARINDEX(@Delimiter, @String)
                    WHEN 0
                        THEN LEN(@String)
                    ELSE CHARINDEX(@Delimiter, @String) - 1
                    END
                )

    INSERT INTO @SplittedValues
    SELECT SUBSTRING(@String, 1, @SplitLength)

    SELECT @String = (
            CASE (LEN(@String) - @SplitLength)
                WHEN 0
                    THEN ''
                ELSE RIGHT(@String, LEN(@String) - @SplitLength - 1) END)

END

SET @DedupedValues=(SELECT DISTINCT STUFF((
            SELECT DISTINCT (@Delimiter + SplitValue)
            FROM @SplittedValues s
            ORDER BY (@Delimiter +  SplitValue)
                        FOR XML PATH('')
            ), 1, 1, '') AS a
FROM @SplittedValues ss)

RETURN @DedupedValues 
END

将其称为内联... 

SELECT Object, Field, [dbo].[SplitRemoveDupes](Values,';') From Table
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