Question

我有两个表，在第一个表中存储了课程ID，在第二个表中，课程ID和不同的主题区域描述存储如下所示。

Table PA_CPNT
CPNT_ID( Course ID)    Course Title 
06201826               AAAA
06201827               BBBB
06201828               CCCC

Table PA_CPNT_SUBJ
CPNT_ID     SUBJ_ID 
06201826    PLNT_DEV
06201826    WRKS_COUN 
06201827    WRKS_COUN1
06201827    WRKS_COUN2
06201827    WRKS_COUN3
06201828    WRKS_COUN 

My requirement is to have an output in the below format


CPNT_ID   COUrse Title   SUBJ_ID1    SUBJ_ID2     SUBJ_ID3
06201826  AAAA           PLNT_DEV    WRKS_COUN 
06201827  BBBB           WRKS_COUN1  WRKS_COUN2   WRKS_COUN3 
06201828  CCCC           WRKS_COUN

我编写了以下代码，如何修改此代码以达到上述要求。

select distinct CPNT_ID,
cpnt_desc,
SUBJ_ID1,
SUBJ_ID2,
SUBJ_ID3
from
(
 select a.cpnt_id, 
 a.cpnt_desc,
 b.subj_id as subj_id1, 
 c.subj_id as subj_id2,
 d.subj_id as subj_id3
 from PA_CPNT a 
 inner join PA_CPNT_SUBJ b
  on a.cpnt_id=b.cpnt_id
  inner join PA_CPNT_SUBJ c
 on a.cpnt_id=c.cpnt_id
 inner join PA_CPNT_SUBJ d
 on a.cpnt_id=d.cpnt_id
   ) X
where subj_id1 ! = subj_id2
and subj_id2 ! = subj_id3
and subj_id3 ! = subj_id1

请帮忙

Answer 1

select DISTINCT
        a.cpnt_id, 
        a.cpnt_desc,
        b.subj_id as subj_id1, 
        c.subj_id as subj_id2,
        d.subj_id as subj_id3
from PA_CPNT a 
left join PA_CPNT_SUBJ b
    on a.cpnt_id=b.cpnt_id
left join PA_CPNT_SUBJ c
    on a.cpnt_id=c.cpnt_id and b.subj_id < c.subj_id
left join PA_CPNT_SUBJ d
    on a.cpnt_id=d.cpnt_id and c.subj_id < d.subj_id

使用<而不是!=可以防止它生成包含主题所有不同排列的重复项。

Answer 2

您可以使用此查询...

 select a.cpnt_id, 
 a.cpnt_desc,
 b.subj_id as subj_id1, 
 c.subj_id as subj_id2,
 d.subj_id as subj_id3
 from PA_CPNT a 
 left join PA_CPNT_SUBJ b on a.cpnt_id=b.cpnt_id and b.subj_id='PLNT_DEV'
 left join PA_CPNT_SUBJ c on a.cpnt_id=c.cpnt_id and c.subj_id='WRKS_COUN'
 left join PA_CPNT_SUBJ d on a.cpnt_id=d.cpnt_id and d.subj_id='WRKS_COUN1'

Answer 3

不需要多次连接表pa_cpnt_subj。

您可以按cpnt_id分组。第一个值当然是最小的subj_id，第二个（如果有一个以上的主题）可以计算（假设从不超过三个值，我们可以用having子句保证），最后一个（以防万一）有超过2个科目）是最大的subj_id。

select 
  pa_cpnt.cpnt_id,
  pa_cpnt.cpnt_desc,
  min(pa_cpnt_subj.subj_id) as subj_id1,
  case when count(*) > 1 then
    avg(pa_cpnt_subj.subj_id) * count(*) 
     - min(pa_cpnt_subj.subj_id) 
     - max(pa_cpnt_subj.subj_id)
  end as subj_id2,
  case when count(*) > 2 then
    max(pa_cpnt_subj.subj_id)
  end as subj_id3
from pa_cpnt 
left outer join pa_cpnt_subj on pa_cpnt_subj.cpnt_id = pa_cpnt.cpnt_id
group by pa_cpnt.cpnt_id, pa_cpnt.cpnt_desc
having count(*) <= 3;

Answer 4

您可以使用row_number为课程中的每个科目提供一个数字，然后显示科目＃1，＃2和＃3。

select
  pa_cpnt.cpnt_id,
  pa_cpnt.cpnt_desc,
  min(case when subj.rn = 1 then subj.subj_id end) as subj_id1,
  min(case when subj.rn = 2 then subj.subj_id end) as subj_id2,
  min(case when subj.rn = 3 then subj.subj_id end) as subj_id3
from pa_cpnt
left outer join 
(
  select 
    cpnt_id, 
    subj_id, 
    row_number() over (partition by cpnt_id order by subj_id) as rn
  from pa_cpnt_subj 
) subj on subj.cpnt_id = pa_cpnt.cpnt_id
group by pa_cpnt.cpnt_id, pa_cpnt.cpnt_desc;

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4 个答案: