最好的方法来粘合＆＃34;列一起

时间：2014-12-16 23:04:35

标签： sql-server

我需要在一个大表中组合大约15个表的列。以下内容有效。但是，当CPU使用率达到100％时，运行需要很长时间，这会引起担忧。任何建议都将受到高度赞赏。

declare @t1 table (empid int)
declare @t2 table (empid int, phone varchar(50))
declare @t3 table (empid int, license varchar(50))
declare @t4 table (empid int, email varchar(100))

insert into @t1 values (1)
insert into @t1 values (2)
insert into @t1 values (3)
insert into @t2 values (1, '5551234')
insert into @t2 values (2, '5553333')
insert into @t2 values (2, 'ttt2222')
insert into @t3 values (2, 'L4455')
insert into @t3 values (3, 'L7890')
insert into @t4 values (2, 'xxx@abc')

SELECT t1.empid, t2.phone, t3.license, t4.email
FROM
    @t1 t1
    LEFT OUTER JOIN
    (SELECT empid, phone, row_number() over (partition by empid order by phone) as rn 
    FROM @t2) t2 
    ON t2.empid = t1.empid
    FULL OUTER JOIN
    (SELECT empid, license, row_number() over (partition by empid order by license) as rn 
    FROM @t3) t3 
    ON t3.empid=t1.empid and (t2.rn is null or t3.rn = t2.rn)
    FULL OUTER JOIN
    (SELECT empid, email, row_number() over (partition by empid order by email) as rn 
    FROM @t4) t4
    ON t4.empid=t1.empid and t4.rn=coalesce(t2.rn, t3.rn) --image how long this coalesce clause is going to be for the 15th table?
order by t1.empid, t2.rn

2 个答案:

答案 0 :(得分：2)

您的问题不是很清楚，如果您包含预期结果会有所帮助。让我猜猜你想要什么...

我将为您的示例中的表提供更有意义的名称，并添加更多行以突出显示该问题。在现实生活中，这些表将是真正的表，当然，不是变量，但我会坚持使用变量来使这个示例脚本易于运行和尝试。我在这个例子中使用SQL Server 2008。

declare @TMain table (empid int);
declare @TPhones table (empid int, phone varchar(50));
declare @TLicenses table (empid int, license varchar(50));
declare @TEmails table (empid int, email varchar(100));

insert into @TMain values (1);
insert into @TMain values (2);
insert into @TMain values (3);
insert into @TMain values (4);

insert into @TPhones values (1, '5551234');
insert into @TPhones values (2, '5551111');
insert into @TPhones values (2, '5552222');
insert into @TPhones values (2, '5553333');
insert into @TPhones values (2, '5554444');

insert into @TLicenses values (2, 'L4455');
insert into @TLicenses values (3, 'L7890');

insert into @TEmails values (2, 'xxx@abc');
insert into @TEmails values (2, 'yyy@abc');
insert into @TEmails values (2, 'zzz@abc');

简单变体

有一种快速，有效和错误的天真方法：

SELECT
    Main.empid
    ,Phones.phone
    ,Licenses.license
    ,Emails.email
FROM
    @TMain AS Main
    LEFT JOIN @TPhones AS Phones ON Phones.empid = Main.empid
    LEFT JOIN @TLicenses AS Licenses ON Licenses.empid = Main.empid
    LEFT JOIN @TEmails AS Emails ON Emails.empid = Main.empid
ORDER BY Main.empid, phone, license, email;

它生成所有行的笛卡尔积并复制行。这是上面查询的结果集。您可以看到empid = 2返回了12行，即4个电话乘以3封电子邮件和1个许可证。我的猜测是你只想看到empid = 2的4行。换句话说，对于每个empid，结果应该具有最小可能的行数（我将在最后显示正确的结果集）。

empid   phone   license email
1   5551234 NULL    NULL
2   5551111 L4455   xxx@abc
2   5551111 L4455   yyy@abc
2   5551111 L4455   zzz@abc
2   5552222 L4455   xxx@abc
2   5552222 L4455   yyy@abc
2   5552222 L4455   zzz@abc
2   5553333 L4455   xxx@abc
2   5553333 L4455   yyy@abc
2   5553333 L4455   zzz@abc
2   5554444 L4455   xxx@abc
2   5554444 L4455   yyy@abc
2   5554444 L4455   zzz@abc
3   NULL    L7890   NULL
4   NULL    NULL    NULL

长变种

我不确定下面提出的方法是否比你的更有效。您必须同时尝试并比较数据的性能。

我们需要一张数字表。 SQL, Auxiliary table of numbers http://web.archive.org/web/20150411042510/http://sqlserver2000.databases.aspfaq.com/why-should-i-consider-using-an-auxiliary-numbers-table.html http://dataeducation.com/you-require-a-numbers-table/

同样，在现实生活中，你会有一个合适的数字表，但是对于这个例子，我将使用以下内容：

declare @TNumbers table (Number int);
insert into @TNumbers values (1);
insert into @TNumbers values (2);
insert into @TNumbers values (3);
insert into @TNumbers values (4);
insert into @TNumbers values (5);

我的方法背后的主要思想是首先制作一个包含每个empid的正确行数的辅助表，然后使用该表有效地获得结果。

我们将从计算每个empid的电话号码，许可证和电子邮件开始：

WITH
CTE_Rows
AS
(
    SELECT Phones.empid, COUNT(*) AS EmpRows
    FROM @TPhones AS Phones
    GROUP BY Phones.empid

    UNION ALL

    SELECT Licenses.empid, COUNT(*) AS EmpRows
    FROM @TLicenses AS Licenses
    GROUP BY Licenses.empid

    UNION ALL

    SELECT Emails.empid, COUNT(*) AS EmpRows
    FROM @TEmails AS Emails
    GROUP BY Emails.empid
)

然后我们计算每个empid的最大行数：

,CTE_MaxRows
AS
(
    SELECT
        CTE_Rows.empid
        ,MAX(CTE_Rows.EmpRows) AS MaxEmpRows
    FROM CTE_Rows
    GROUP BY CTE_Rows.empid
)

上面的CTE对每个empid都有一行：empid本身以及这个empid的最大数量的电话，许可证和电子邮件。现在我们需要扩展此表并为每个empid生成给定的行数。我在这里使用Numbers表：

,CTE_RowNumbers
AS
(
SELECT
    CTE_MaxRows.empid
    ,Numbers.Number AS rn
FROM
    CTE_MaxRows
    CROSS JOIN @TNumbers AS Numbers
WHERE
    Numbers.Number <= CTE_MaxRows.MaxEmpRows
)

然后我们需要向所有带有数据的表中添加行号，我们稍后会用它们加入：

,CTE_Phones
AS
(
    SELECT
        Phones.empid
        ,ROW_NUMBER() OVER (PARTITION BY Phones.empid ORDER BY phone) AS rn
        ,Phones.phone
    FROM @TPhones AS Phones
)
,CTE_Licenses
AS
(
    SELECT
        Licenses.empid
        ,ROW_NUMBER() OVER (PARTITION BY Licenses.empid ORDER BY license) AS rn
        ,Licenses.license
    FROM @TLicenses AS Licenses
)
,CTE_Emails
AS
(
    SELECT
        Emails.empid
        ,ROW_NUMBER() OVER (PARTITION BY Emails.empid ORDER BY email) AS rn
        ,Emails.email
    FROM @TEmails AS Emails
)

现在我们已准备好加入所有这些。 CTE_RowNumbers具有我们需要的确切行数，因此此处不需要复杂的FULL JOINs，简单LEFT JOIN就足够了：

,CTE_Data
AS
(
    SELECT
        CTE_RowNumbers.empid
        ,CTE_Phones.phone
        ,CTE_Licenses.license
        ,CTE_Emails.email
    FROM
        CTE_RowNumbers
        LEFT JOIN CTE_Phones ON CTE_Phones.empid = CTE_RowNumbers.empid AND CTE_Phones.rn = CTE_RowNumbers.rn
        LEFT JOIN CTE_Licenses ON CTE_Licenses.empid = CTE_RowNumbers.empid AND CTE_Licenses.rn = CTE_RowNumbers.rn
        LEFT JOIN CTE_Emails ON CTE_Emails.empid = CTE_RowNumbers.empid AND CTE_Emails.rn = CTE_RowNumbers.rn
)

我们差不多完成了。我猜，主表有可能没有任何相关数据（没有电话，没有虱子，没有电子邮件），比如样本数据中的empid = 4。为了在结果集中获得这些empid，我将把CTE_Data连接到主表：

SELECT
    Main.empid
    ,CTE_Data.phone
    ,CTE_Data.license
    ,CTE_Data.email
FROM
    @TMain AS Main
    LEFT JOIN CTE_Data ON CTE_Data.empid = Main.empid
ORDER BY Main.empid, phone, license, email;

要获取完整脚本，只需将此帖子中的所有代码块按照此处显示的顺序放在一起。

这是结果集：

empid   phone   license email
1   5551234 NULL    NULL
2   5551111 L4455   xxx@abc
2   5552222 NULL    yyy@abc
2   5553333 NULL    zzz@abc
2   5554444 NULL    NULL
3   NULL    L7890   NULL
4   NULL    NULL    NULL

答案 1 :(得分：-1)

您是否有理由在

上使用row_number（）

看起来你想要这样的东西...... 

SELECT t1.empid, t2.phone, t3.license, t4.email
FROM @t1 t1
LEFT JOIN @t2 t2 ON t1.EmpId = t2.EmpId
LEFT JOIN @t3 t3 ON t1.EmpId = t3.EmpId
LEFT JOIN @t4 t4 ON t1.EmpId = t4.EmpId
相关问题
最新问题