以下是架构
+---------+---------+
| 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 |
+---------+--------+
我尝试了连接，但它提供了多余的行。
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 |
| | | | |
+---------+---------+--------------+---------+
答案 0 :(得分：3)
看起来你正试图分配贷款＆＃34;顺序＆＃34;到族表中的行。解决这个问题的方法是首先获得正确的行，然后将贷款分配给行。
右行（和前三列）是：
select f.EmpId, e.Name, f.Relationship
from family f join
Employee e
on f.empid = e.empid;
请注意，这不会在重复值的列中添加连字符，而是放入实际值。虽然您可以在SQL中安排连字符，但这是一个坏主意。 SQL结果采用表格的形式，表格是无序集合，每个列和每行都有值。当你开始使用连字符时，你取决于顺序。
现在问题是加入贷款。实际上，使用row_number()添加join密钥非常简单：
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，这并不保证给定员工的贷款分配顺序。您的数据似乎没有足够的信息来处理更一致的分配。
答案 1 :(得分：2)
下面概述的方法可以轻松地连接＆＃34;更多表到结果集。它不仅限于两个表格。
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。
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');
我们需要一张数字表。
SQL, Auxiliary table of numbers
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正确行数的辅助表，然后使用该表有效地获得结果。
我们首先计算每个EmpId的关系和贷款数量：
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
)
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：EmpId本身都有一行，并为此EmpId提供最大数量的关系或贷款。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表格：
,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添加行号，我们将在以后加入这些表。您可以使用表格中的其他列来订购行号。对于这个例子，没有太多选择。
,CTE_Relationships
AS
(
SELECT
Relationships.EmpId
,ROW_NUMBER() OVER (PARTITION BY Relationships.EmpId ORDER BY Relationships.Relationship) AS rn
,Relationships.Relationship
FROM @TFamily AS Relationships
)
,CTE_Loans
AS
(
SELECT
Loans.EmpId
,ROW_NUMBER() OVER (PARTITION BY Loans.EmpId ORDER BY Loans.LoanId) AS rn
,Loans.LoanId
FROM @TLoan AS Loans
)
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，因此简单LEFT JOIN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表可能有一些EmpIds没有任何相关数据，例如样本数据中的EmpId = 3。要在结果集中获取这些EmpIds，我会将CTE_Data加入主表，并用短划线替换NULLs：
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 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
答案 2 :(得分：0)
弗拉基米尔·巴拉诺夫已经写了一个很好的解决方案，但它相当长（并且有一个小问题：你想要Husband-L3和Child-L4，但这个解决方案会返回Child-L3和Husband-L4）。
Gordon Linoff写了一个较短的解决方案，但它无法正常工作。
我可以修复戈登的解决方案，如下：
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)
但是，我宁愿说问题不正确，因为它要求我们将家庭成员任意地与特定贷款相匹配（当没有真正的关系时）。
我宁愿说正确的问题是具有以下答案的问题：
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