Как использовать подзапросы в Oracle?

Question

У меня есть следующая таблица:

table1 
------------------------- 
date  | ID | name 
------------------------- 
13-jul-15 | 1 | abc 
13-jul-15 | 2 | abc 
14-jul-15 | 1 | def 
13-jul-15 | 3 | abc 
15-jul-15 | 3 | def 
... 

То, что я хочу сделать, это соответствовать идентификатору и представлять, как показано ниже:

date1  | name | date2  | name | ID 
------------------------------------------ 
13-jul-15 | abc | 14-jul-15 | def | 1 
13-jul-15 | abc |   |  | 2 
13-jul-15 | abc | 15-jul-15 | def | 3 
... 

Я использовал следующий код, но не получение результата.

CREATE PROCEDURE get_details (oresults1 OUT SYS_REFCURSOR) AS 
BEGIN 
    SELECT * 
    FROM ((SELECT date, ID FROM table1 WHERE name= "abc") T1 
      UNION ALL 
      (SELECT date, ID FROM table1 WHERE name= "def") T2 
     ) 
    WHERE T1.ID= T2.ID 
    ORDER BY ID; 
END; 

Что я сделал не так?

Answer 1

Вот несколько альтернатив:

with table1 as (select to_date('13/07/2015', 'dd/mm/yyyy') dt, 1 id, 'abc' name from dual union all 
       select to_date('13/07/2015', 'dd/mm/yyyy') dt, 2 id, 'abc' name from dual union all 
       select to_date('14/07/2015', 'dd/mm/yyyy') dt, 1 id, 'def' name from dual union all 
       select to_date('13/07/2015', 'dd/mm/yyyy') dt, 3 id, 'abc' name from dual union all 
       select to_date('15/07/2015', 'dd/mm/yyyy') dt, 3 id, 'def' name from dual) 
-- end of mimicking your table1. See below for the query 
select t1.dt date1, 
     t1.name name1, 
     t2.dt date2, 
     t2.name name2, 
     t1.id 
from table1 t1 
     left outer join table1 t2 on (t1.id = t2.id and t1.name = 'abc' and t2.name = 'def') 
where t1.name = 'abc' 
order by t1.id; 

DATE1  NAME1 DATE2  NAME2   ID 
---------- ----- ---------- ----- ---------- 
13/07/2015 abc 14/07/2015 def   1 
13/07/2015 abc        2 
13/07/2015 abc 15/07/2015 def   3 

with table1 as (select to_date('13/07/2015', 'dd/mm/yyyy') dt, 1 id, 'abc' name from dual union all 
       select to_date('13/07/2015', 'dd/mm/yyyy') dt, 2 id, 'abc' name from dual union all 
       select to_date('14/07/2015', 'dd/mm/yyyy') dt, 1 id, 'def' name from dual union all 
       select to_date('13/07/2015', 'dd/mm/yyyy') dt, 3 id, 'abc' name from dual union all 
       select to_date('15/07/2015', 'dd/mm/yyyy') dt, 3 id, 'def' name from dual) 
-- end of mimicking your table1. See below for the query 
select t1.dt date1, 
     t1.name name1, 
     t2.dt date2, 
     t2.name name2, 
     t1.id 
from (select id, dt, name from table1 where name = 'abc') t1 
     left outer join (select id, dt, name from table1 where name = 'def') t2 on (t1.id = t2.id) 
where t1.name = 'abc' 
order by t1.id; 

DATE1  NAME1 DATE2  NAME2   ID 
---------- ----- ---------- ----- ---------- 
13/07/2015 abc 14/07/2015 def   1 
13/07/2015 abc        2 
13/07/2015 abc 15/07/2015 def   3 

Answer 2

Вы также можете использовать функцию поворота, доступные в Oracle

WITH table_(date#, id#, name#) AS 
(SELECT to_date('13-jul-15', 'dd-mon-yy'), 1, 'abc' FROM dual UNION all 
SELECT to_date('13-jul-15', 'dd-mon-yy'), 2, 'abc' FROM dual UNION all 
SELECT to_date('14-jul-15', 'dd-mon-yy'), 1, 'def' FROM dual UNION all 
SELECT to_date('13-jul-15', 'dd-mon-yy'), 3, 'abc' FROM dual UNION all 
SELECT to_date('15-jul-15', 'dd-mon-yy'), 3, 'def' FROM dual) 
-------- 
-- End of data preparation 
-------- 
SELECT * 
    FROM table_ 
PIVOT (MIN(date#) AS date#, MIN(name#) AS NAME# FOR name# IN ('abc' AS ABC, 'def' AS DEF)); 

Выход

| ID# |    ABC_DATE# | ABC_NAME# |    DEF_DATE# | DEF_NAME# | 
|-----|------------------------|-----------|------------------------|-----------| 
| 1 | July, 13 2015 00:00:00 |  abc | July, 14 2015 00:00:00 |  def | 
| 2 | July, 13 2015 00:00:00 |  abc |      |   | 
| 3 | July, 13 2015 00:00:00 |  abc | July, 15 2015 00:00:00 |  def |