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在SQL Server中水平连接两个以上的表

sql sql-server-2008 tsql

在SQL Server中水平连接两个以上的表,sql,sql-server-2008,tsql,outer-join,Sql,Sql Server 2008,Tsql,Outer Join,下面是模式 +---------+---------+ | Employee Table | +---------+---------+ | EmpId | Name | +---------+---------+ | 1 | John | | 2 | Lisa | | 3 | Mike | | | | +---------+---------+ +---------+-----------

下面是模式

+---------+---------+
| Employee Table    |
+---------+---------+
| EmpId   | Name    | 
+---------+---------+
| 1       | John    |
| 2       | Lisa    |
| 3       | Mike    |
|         |         |
+---------+---------+

+---------+-----------------+
| Family   Table            |
+---------+-----------------+
| EmpId   | Relationship    | 
+---------+-----------------+
| 1       | Father          |
| 1       | Mother          |
| 1       | Wife            |
| 2       | Husband         |
| 2       | Child           |
+---------+-----------------+

+---------+---------+
| Loan  Table       |
+---------+--------+
| LoanId  | EmpId  | 
+---------+--------+
| L1      | 1      |
| L2      | 1      |
| L3      | 2      |
| L4      | 2      |
| L5      | 3      |
+---------+--------+
  • Employee表和Family表具有一对多关系
  • 员工表和贷款表有一对多的关系
我尝试过联接，但它会产生多余的行

现在，所需的输出将是

+---------+---------+--------------+---------+
| EmpId   | Name    | RelationShip | Loan    | 
+---------+---------+--------------+---------+
| 1       | John    | Father       | L1      |
| -       | -       | Mother       | L2      |
| -       | -       | Wife         | -       |
| 2       | Lisa    | Husband      | L3      |
| -       | -       | Child        | L4      |
| 3       | Mike    | -            | L5      |
|         |         |              |         |
+---------+---------+--------------+---------+
下面概述的方法允许轻松地将更多表“连接”到结果集。它不限于两张桌子

我将使用表变量来说明解决方案。在现实生活中，这些表将是真实的表，当然不是变量，但我将坚持使用变量，以使此示例脚本易于运行和尝试

declare @TEmployee table (EmpId int, Name varchar(50));
declare @TFamily table (EmpId int, Relationship varchar(50));
declare @TLoan table (EmpId int, LoanId varchar(50));

insert into @TEmployee values (1, 'John');
insert into @TEmployee values (2, 'Lisa');
insert into @TEmployee values (3, 'Mike');

insert into @TFamily values (1, 'Father');
insert into @TFamily values (1, 'Mother');
insert into @TFamily values (1, 'Wife');
insert into @TFamily values (2, 'Husband');
insert into @TFamily values (2, 'Child');

insert into @TLoan values (1, 'L1');
insert into @TLoan values (1, 'L2');
insert into @TLoan values (2, 'L3');
insert into @TLoan values (2, 'L4');
insert into @TLoan values (3, 'L5');
我们需要一个数字表

同样，在现实生活中，你会有一个合适的数字表，但对于这个例子，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创建一个helper表，该表首先包含每个
EmpId
的正确行数，然后使用该表高效地获得结果

我们将从计算每个
EmpId
的关系和贷款数量开始：

WITH
CTE_Rows
AS
(
    SELECT Relationships.EmpId, COUNT(*) AS EmpRows
    FROM @TFamily AS Relationships
    GROUP BY Relationships.EmpId

    UNION ALL

    SELECT Loans.EmpId, COUNT(*) AS EmpRows
    FROM @TLoan AS Loans
    GROUP BY Loans.EmpId
)

,CTE_MaxRows
AS
(
    SELECT
        CTE_Rows.empid
        ,MAX(CTE_Rows.EmpRows) AS MaxEmpRows
    FROM CTE_Rows
    GROUP BY CTE_Rows.empid
)
然后计算每个
EmpId
的最大行数：

WITH
CTE_Rows
AS
(
    SELECT Relationships.EmpId, COUNT(*) AS EmpRows
    FROM @TFamily AS Relationships
    GROUP BY Relationships.EmpId

    UNION ALL

    SELECT Loans.EmpId, COUNT(*) AS EmpRows
    FROM @TLoan AS Loans
    GROUP BY Loans.EmpId
)

,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为每个
EmpId
：
EmpId
本身有一行，以及此
EmpId
的最大关系数或贷款数。现在我们需要展开这个表，为每个
EmpId
生成给定数量的行。这里我使用的是
数字表：


,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
)


现在，我们已经准备好加入这一切
CTE_RowNumbers
具有我们所需的精确行数，因此简单的
左连接就足够了：


,CTE_Data
AS
(
    SELECT
        CTE_RowNumbers.empid
        ,CTE_Relationships.Relationship
        ,CTE_Loans.LoanId
    FROM
        CTE_RowNumbers
        LEFT JOIN CTE_Relationships ON CTE_Relationships.EmpId = CTE_RowNumbers.EmpId AND CTE_Relationships.rn = CTE_RowNumbers.rn
        LEFT JOIN CTE_Loans ON CTE_Loans.EmpId = CTE_RowNumbers.EmpId AND CTE_Loans.rn = CTE_RowNumbers.rn
)


我们差不多完成了。主
Employee
表中可能有一些
EmpId
没有任何相关数据，例如示例数据中的
EmpId=3
。要在结果集中获得这些
empid
，我将把
CTE_数据
加入主表，并用破折号替换
null
：


SELECT
    Employees.EmpId
    ,Employees.Name
    ,ISNULL(CTE_Data.Relationship, '-') AS Relationship
    ,ISNULL(CTE_Data.LoanId, '-') AS LoanId
FROM
    @TEmployee AS Employees
    LEFT JOIN CTE_Data ON CTE_Data.EmpId = Employees.EmpId
ORDER BY Employees.EmpId, Relationship, LoanId;


要获得完整的脚本，只需将本文中的所有代码块按此处显示的顺序放在一起

这是结果集：


EmpId   Name   Relationship   LoanId
1       John   Father         L1
1       John   Mother         L2
1       John   Wife           -
2       Lisa   Child          L3
2       Lisa   Husband        L4
3       Mike   -              L5



看起来您正试图“按顺序”将贷款分配给族表中的行。解决这个问题的方法是首先获得正确的行，然后获得分配给行的贷款

右行（和前三列）为：

请注意，这不会在重复值的列中放置连字符，而是放置实际值。虽然可以在SQL中安排连字符，但这不是一个好主意。SQL结果以表的形式出现，表是无序的集合，每列和每行都有值。在开始放置连字符时，取决于顺序

现在的问题是加入贷款。这实际上非常简单，使用
行编号（）
添加
连接
键：


select f.EmpId, e.Name, f.Relationship, l.LoanId
from Employee e left join
     (select f.*, row_number() over (partition by f.EmpId order by (select NULL)) as seqnum
      from family f
     ) f 
     on f.empid = e.empid left join
     (select l.*, row_number() over (partition by l.EmpId order by (select NULL)) as seqnum
      from Loan l
     ) l
     on f.EmpId = l.EmpId and f.seqnum = l.seqnum;



请注意，这并不保证为给定员工分配贷款的顺序。您的数据似乎没有足够的信息来处理更一致的分配。
弗拉基米尔·巴拉诺夫已经编写了一个很好的解决方案，但它相当长（并且有一个小问题：您想要丈夫-L3和孩子-L4，但此解决方案返回孩子-L3和丈夫-L4）

Gordon Linoff编写了一个较短的解决方案，但它无法正常工作

我可以修复Gordon的解决方案，如下所示：


SELECT e.EmpId, e.Name, f.Relationship, l.LoanId
FROM @TEmployee e
LEFT JOIN (
    SELECT f.*, ROW_NUMBER() OVER (PARTITION BY f.EmpId ORDER BY (SELECT NULL)) AS seqnum
    FROM @TFamily f
) f ON f.empid = e.empid 
LEFT JOIN (
    SELECT l.*, ROW_NUMBER() OVER (PARTITION BY l.EmpId ORDER BY (SELECT NULL)) AS seqnum
    FROM @TLoan l
) l ON l.EmpId = e.EmpId AND (f.seqnum = l.seqnum OR f.seqnum IS NULL)


然而，我宁愿说这个问题是不正确的，因为它要求我们任意地将家庭成员与特定的贷款相匹配（当不存在真正的关系时）

我宁愿说正确的问题是有以下答案的问题：


SELECT e.EmpId, e.Name,
    SUBSTRING((
        SELECT ', '+f.Relationship AS '*'
        FROM @TFamily f
        WHERE f.EmpId=e.EmpId
        FOR XML PATH(''), TYPE
    ).value('.','nvarchar(4000)'),3,4000) AS FamilyMembers,
    SUBSTRING((
        SELECT ', '+l.LoanId AS '*'
        FROM @TLoan l
        WHERE l.EmpId=e.EmpId
        FOR XML PATH(''), TYPE
    ).value('.','nvarchar(4000)'),3,4000) AS Loans
FROM @TEmployee e



贷款属于员工（1:N），家庭关系也属于员工（1:N）。如何将贷款L2和L4归于非员工？此信息不包含在数据中？select没有设置的顺序，也没有生成此信息的排序。对现有列没有办法做到这一点。@VladimirBaranov。现在应该是了。我用错误的表启动了
左join
s链。