Question

假设我有以下内容：

if object_id('tempdb..#Dependents') is not null
    drop table #Dependents

create table #Dependents
(EmpNo varchar(12), DepID varchar(12), Relation varchar(12));
go

if object_id('tempdb..#Dep_Benefits') is not null
    drop table #Dep_Benefits

create table #Dep_Benefits
(DepID varchar(12), PlanCode varchar(12), Coverage varchar(25));
go


INSERT INTO #Dependents VALUES('001', '111', 'SON');
INSERT INTO #Dependents VALUES('001', '222', 'SON');
INSERT INTO #Dependents VALUES('001', '333', 'DAUGHTER');
INSERT INTO #Dependents VALUES('002', '666', 'SON');
INSERT INTO #Dependents VALUES('002', '777', 'DAUGHTER');

INSERT INTO #Dep_Benefits VALUES('111', 'AAAA', 'MEDICAL');
INSERT INTO #Dep_Benefits VALUES('111', 'BBBB', 'DENTAL');
INSERT INTO #Dep_Benefits VALUES('111', 'CCCC', 'VISION');
INSERT INTO #Dep_Benefits VALUES('111', 'DDDD', 'DISABL');
INSERT INTO #Dep_Benefits VALUES('222', 'AAAA', 'MEDICAL');
INSERT INTO #Dep_Benefits VALUES('222', 'BBBB', 'DENTAL');
INSERT INTO #Dep_Benefits VALUES('222', 'CCCC', 'VISION');
INSERT INTO #Dep_Benefits VALUES('333', 'AAAA', 'MEDICAL');
INSERT INTO #Dep_Benefits VALUES('333', 'BBBB', 'DENTAL');
INSERT INTO #Dep_Benefits VALUES('666', 'AAAA', 'MEDICAL');
INSERT INTO #Dep_Benefits VALUES('666', 'BBBB', 'DENTAL');
INSERT INTO #Dep_Benefits VALUES('666', 'CCCC', 'VISION');


SELECT * FROM #Dependents;
SELECT * FROM #Dep_Benefits;

我怎样才能得到这个的输出：

Employee   | Dependent  | Plan 1   | Plan 2   | Plan 3   
001          111          AAAA       BBBB       DDDD
001          222          AAAA       BBBB       CCCC
001          333          AAAA       BBBB       
002          666          AAAA       BBBB       CCCC

我只关心3个计划，即使有更多分配给依赖者。这是来自客户的固定请求（只有3个计划）所以请不要评论为什么这可能是一个坏主意:)我已经告诉他们，如果一个受抚养者超过3，我们无法保证可以返回3个计划

我查看了一个数据透视表，但似乎无法使其正常工作。

Answer 1

使用row_number()对计划进行编号，并使用条件聚合：

select 
    d.empno
  , d.depid
  , Plan_1 = max(case when rn = 1 then PlanCode end)
  , Plan_2 = max(case when rn = 2 then PlanCode end)
  , Plan_3 = max(case when rn = 3 then PlanCode end)
  , Plan_4 = max(case when rn = 4 then PlanCode end)
from (
  select d.empNo, db.*
    , rn = row_number() over(partition by db.depid order by plancode)
  from #Dependents d
  inner join #Dep_Benefits db
    on d.depid = db.depid
  ) d
group by d.empno, d.depid

rextester演示：http://rextester.com/FNLH5237

返回：

+-------+-------+--------+--------+--------+--------+
| empno | depid | Plan_1 | Plan_2 | Plan_3 | Plan_4 |
+-------+-------+--------+--------+--------+--------+
|   001 |   111 | AAAA   | BBBB   | CCCC   | DDDD   |
|   001 |   222 | AAAA   | BBBB   | CCCC   | NULL   |
|   001 |   333 | AAAA   | BBBB   | NULL   | NULL   |
|   002 |   666 | AAAA   | BBBB   | CCCC   | NULL   |
+-------+-------+--------+--------+--------+--------+

使用条件聚合按类型转动计划：

select 
    d.empno
  , d.depid
  , Medical    = max(case when db.Coverage = 'Medical' then PlanCode end)
  , Dental     = max(case when db.Coverage = 'Dental' then PlanCode end)
  , Vision     = max(case when db.Coverage = 'Vision' then PlanCode end)
  , Disability = max(case when db.Coverage = 'disabl' then PlanCode end)
from #Dependents d
  inner join #Dep_Benefits db
    on d.depid = db.depid
group by d.empno, d.depid

返回：

+-------+-------+---------+--------+--------+------------+
| empno | depid | Medical | Dental | Vision | Disability |
+-------+-------+---------+--------+--------+------------+
|   001 |   111 | AAAA    | BBBB   | CCCC   | DDDD       |
|   001 |   222 | AAAA    | BBBB   | CCCC   | NULL       |
|   001 |   333 | AAAA    | BBBB   | NULL   | NULL       |
|   002 |   666 | AAAA    | BBBB   | CCCC   | NULL       |
+-------+-------+---------+--------+--------+------------+

SQL：行到列数据

1 个答案: