SQL查询以将单个值中的多个ID与逗号分隔符组合起来

Question

我有 2 个表 Employee and Skills 包含以下字段：

员工：Id,Name,Location

技能：Id,Skills

员工数据：

Id    Name   Location
7     Abc     London
8     Xyz     London
9     Xyz     USA
10    Abc     USA

技能：

Id   skills     EmployeeId
12    AAA Xyz     7
13    BBB         7
14    CCC         7
15    AAA         8
16    BBB         8
17    CCC         8
18    AAA         9
19    BBB         9
20    CCC         9
21    AAA         10
22    BBB         10
20    CCC         10

现在我想获得那些Name is Abc,Location is London and Skills is both AAA and BBB only但我想要如下单一输出的员工的技能：

Skills
10,11

这是我的查询：

select S.Id from Employee E inner join
Skills S on E.Id=s.EmployeeId
where (E.Name='Abc' and E.Location='London'
and s.skills='AAA') and (E.Name='Abc' and E.Location='London'
and s.skills='BBB')

但是我得到了错误的输出。

Answer 1

您可以为此使用STUFF：

Select  Stuff
(
    (
        Select  ',' + Convert(Varchar, S.Id)
        From    Employee    E
        Join    Skills      S   On  S.EmployeeId = E.Id
        Where   E.Name = 'Abc'
        And     E.Location = 'London'
        And     S.Skills In ('AAA', 'BBB')
        For Xml Path ('')
    ), 1, 1, ''
) As Skills

编辑：SQL Fiddle 目前似乎不想与我合作，但使用以下数据（来自 OP），它确实得到了正确的结果：

Create Table Skills
(
  Id int,
  skills varchar (3),
  employeeid int
);

Create Table Employee
(
  Id int,
  Name varchar (10),
  location varchar (10)
  );

insert skills values 
(10    ,'AAA',      1),
(11    ,'BBB',      1),
(13    ,'CCC',      1),
(14    ,'DDD',      1);

insert employee values
(    1     ,'Abc' ,    'London'),
(    2     ,'Xyz',     'Uk');


Select  Stuff
(
    (
        Select  ',' + Convert(Varchar, S.Id)
        From    Employee    E
        Join    Skills      S   On  S.EmployeeId = E.Id
        Where   E.Name = 'Abc'
        And     E.Location = 'London'
        And     S.Skills In ('AAA', 'BBB')
        For Xml Path ('')
    ), 1, 1, ''
) As Skills

编辑 #2 新数据；相同的查询：

Create Table Skills
(
  Id int,
  skills varchar (3),
  employeeid int
);

Create Table Employee
(
  Id int,
  Name varchar (10),
  location varchar (10)
  );

insert skills values
(12    ,'AAA',      7   ),
(13    ,'BBB',      7   ),
(14    ,'CCC',      7   ),
(15    ,'AAA',      8   ),
(16    ,'BBB',      8   ),
(17    ,'CCC',      8   ),
(18    ,'AAA',      9   ),
(19    ,'BBB',      9   ),
(20    ,'CCC',      9   ),
(21    ,'AAA',      10  ),
(22    ,'BBB',      10  ),
(20    ,'CCC',      10  );

insert employee values
(7     ,'Abc',     'London'  ),
(8     ,'Xyz',     'London'  ),
(9     ,'Xyz',     'USA'       ),
(10    ,'Abc',     'USA'       );


Select  Stuff
(
    (
        Select  ',' + Convert(Varchar, S.Id)
        From    Employee    E
        Join    Skills      S   On  S.EmployeeId = E.Id
        Where   E.Name = 'Abc'
        And     E.Location = 'London'
        And     S.Skills In ('AAA', 'BBB')
        For Xml Path ('')
    ), 1, 1, ''
) As Skills

这次的结果是12,13——根据你的情况是正确的：

员工7 是唯一拥有London 中的Abc 的员工，而12 和13 是AAA 和BBB 技能。