Как правильно использовать listagg с кейсами

Question

Я пытаюсь использовать listagg, но получаю неправильный вывод. Обычно я бы использовал каждое выражение case отдельно, но тогда как бы я использовал listagg?

Таблица A:

 JAN     FEB    MAR    APR    Tag
 C                            102
          D            T      100
          D                   101
 C        D      B     T      103                  

Таблица B:

 Name       Tag
 Ally       100
 Ben        101
 Missy      102
 Noah       103

Желаемый результат:

 Ally  Dog,Turtle
 Ben   Dog
 Missy Chicken
 Noah  Chicken,Dog,Bird,Turtle

Мой попытанный (неправильный) код:

 select listagg(
   nvl(
     case when a.jan = 'C' then 'Chicken'
          when a.feb = 'D' then 'Dog'
          when a.mar = 'B' then 'Bird'
          when a.apr = 'T' then 'Turtle'
      end,','),'none')
  within group (order by a.tag)
      from a where a.tag = b.tag 

Answer 1

Вы можете построить запрос путем отмены запроса в подзапросе:

select b.Name, a.*
  from b
  join
  (
  select tag, listagg( case
                       when val = 'C' then
                        'Chicken'
                       when val = 'D' then
                        'Dog'
                       when val = 'B' then
                        'Bird'
                       when val = 'T' then
                        'Turtle'
                       end,',') within group (order by 1) as animals
    from a
   unpivot ( val for name in (jan as 'JAN', feb as 'FEB', mar as 'MAR', apr as 'APR') )
   group by tag ) a
     on b.tag = a.tag   
  order by a.tag  

Демо

Answer 2

with
-- sample data
taba (jan, feb, mar, apr, tag) as
  (select 'c' , null, null, null, 102 from dual union all
   select null, 'd' , null, 't' , 100 from dual union all
   select null, 'd' , null, null, 101 from dual union all
   select 'c' , 'd' , 'b' , 't' , 103 from dual
  ),
tabb (name, tag) as
  (select 'ally' , 100 from dual union all
   select 'ben'  , 101 from dual union all
   select 'missy', 102 from dual union all
   select 'noah' , 103 from dual
  ),
  tabab as (
   --select a.tag, b.name, nvl(a.jan,'na') as jan, nvl(a.feb,'na') as feb, nvl(a.mar,'na') as mar, nvl(a.apr,'na') as apr
   select 
      a.tag, b.name, 
      decode(a.jan,'b','bird','c','chicken','d','dog','t','turtle',null) as jan,
      decode(a.feb,'b','bird','c','chicken','d','dog','t','turtle',null) as feb,
      decode(a.mar,'b','bird','c','chicken','d','dog','t','turtle',null) as mar,
      decode(a.apr,'b','bird','c','chicken','d','dog','t','turtle',null) as apr
   from taba a   
   join tabb b
     on a.tag = b.tag
  )
  select 
      name,
      listagg(val,',') WITHIN GROUP (ORDER by decode(mon, 'JAN',1,'FEB',2,'MAR',3,'APR',4,null)) as listval
   from tabab
   unpivot
   (
     val
       for mon in (jan,feb,mar,apr)
   )
   group by name
   order by 1
;

Answer 3

Вот один вариант, который использует несколько самостоятельных объединений.

• строки # 1 - 14 представляют ваши примеры данных
• anima CTE здесь для упрощения кода; лучше иметь его в таблице (даже если это CTE), чем использовать CASE
• конечный результат, строки # 24 - 34, объединять имена животных
  • trim + regexp_replace используется для удаления лишних запятых

Здесь вы go:

SQL> with
  2  -- sample data
  3  taba (jan, feb, mar, apr, tag) as
  4    (select 'c' , null, null, null, 102 from dual union all
  5     select null, 'd' , null, 't' , 100 from dual union all
  6     select null, 'd' , null, null, 101 from dual union all
  7     select 'c' , 'd' , 'b' , 't' , 103 from dual
  8    ),
  9  tabb (name, tag) as
 10    (select 'ally' , 100 from dual union all
 11     select 'ben'  , 101 from dual union all
 12     select 'missy', 102 from dual union all
 13     select 'noah' , 103 from dual
 14    ),
 15  --
 16  -- replace animal codes with their names
 17  anima (code, name) as
 18    (select 'c', 'chicken' from dual union all
 19     select 'd', 'dog'     from dual union all
 20     select 'b', 'bird'    from dual union all
 21     select 't', 'turtle'  from dual
 22    )
 23  -- the final result
 24  select b.name,
 25    trim(',' from regexp_replace(n1.name ||','|| n2.name ||','|| n3.name ||','|| n4.name,
 26                                 '( *, *){2,}', ','
 27                                )
 28        ) result
 29  from tabb b join taba a on a.tag = b.tag
 30  left join anima n1 on n1.code = a.jan
 31  left join anima n2 on n2.code = a.feb
 32  left join anima n3 on n3.code = a.mar
 33  left join anima n4 on n4.code = a.apr
 34  order by b.name;

NAME  RESULT
----- ------------------------------
ally  dog,turtle
ben   dog
missy chicken
noah  chicken,dog,bird,turtle

SQL>