如何从连接表中选择数据作为表类型?

时间:2016-11-15 10:51:15

标签: sql database oracle oracle11g

我想构建处理表类型中参考数据的非规范化视图。

create table reftab (id number, name varchar2(40), details varchar2(1000));
/
create table basictab (id number, name varchar2(300), contract varchar2 (20));
/
create or replace type reftype is object (id number, name varchar2(40), details varchar2(1000));
/
create or replace type reftypetab as table of reftype;
/
insert into basictab values (1, 'aaa', 'c1');
insert into basictab values (2, 'aab', 'c1');
insert into basictab values (3, 'aaa', 'c2');
insert into basictab values (4, 'aaa', 'c3');

insert into reftab values (1, 'asd', 'aaa');
insert into reftab values (1, 'asg', 'ass');
insert into reftab values (1, 'ash', 'add');
insert into reftab values (1, 'asf', 'agg');
insert into reftab values (3, 'asd', 'aaa');
insert into reftab values (3, 'ad', 'aa');
insert into reftab values (4, 'asd', 'aaa');
insert into reftab values (4, 'as', 'a');
insert into  values (4, 'ad', 'aa');
/

对于此类数据,我希望包含4行basictab的视图,其中包含reftypetab的其他列,并包含id上加入的所有参考数据。

我知道我可以通过以下方式获得它:

CREATE OR REPLACE FUNCTION pipef (p_id IN NUMBER) RETURN reftypetab PIPELINED AS
BEGIN
  FOR x IN (select * from reftab where id = p_id) LOOP
    PIPE ROW(reftype(x.id, x.name, x.details));   
  END LOOP;

  RETURN;
END;
/

SELECT id, pipef(id)
FROM  reftab
group BY id;
/

但是有没有更好的方法来获得结果?

2 个答案:

答案 0 :(得分:2)

您当前的设置得到:

SELECT id, pipef(id) as result
FROM  reftab
group BY id;

        ID RESULT(ID, NAME, DETAILS)                                                                                               
---------- ------------------------------------------------------------------------------------------------------------------------
         1 REFTYPETAB(REFTYPE(1, 'asd', 'aaa'), REFTYPE(1, 'asg', 'ass'), REFTYPE(1, 'ash', 'add'), REFTYPE(1, 'asf', 'agg'))      
         4 REFTYPETAB(REFTYPE(4, 'asd', 'aaa'), REFTYPE(4, 'as', 'a'), REFTYPE(4, 'ad', 'aa'))                                     
         3 REFTYPETAB(REFTYPE(3, 'asd', 'aaa'), REFTYPE(3, 'ad', 'aa'))                                                            

您可以使用the collect() function来简化:

select id, cast(collect(reftype(id, name, details)) as reftypetab) as result
from reftab
group by id;

        ID RESULT(ID, NAME, DETAILS)                                                                                               
---------- ------------------------------------------------------------------------------------------------------------------------
         1 REFTYPETAB(REFTYPE(1, 'asd', 'aaa'), REFTYPE(1, 'asf', 'agg'), REFTYPE(1, 'ash', 'add'), REFTYPE(1, 'asg', 'ass'))      
         3 REFTYPETAB(REFTYPE(3, 'asd', 'aaa'), REFTYPE(3, 'ad', 'aa'))                                                            
         4 REFTYPETAB(REFTYPE(4, 'asd', 'aaa'), REFTYPE(4, 'ad', 'aa'), REFTYPE(4, 'as', 'a'))                                     

如果您还需要来自basictab的信息,可以使用a multiset operator

select bt.id, bt.name,
  cast(multiset(select reftype(rt.id, rt.name, rt.details)
    from reftab rt where rt.id = bt.id) as reftypetab) as result
from basictab bt;

        ID NAME       RESULT(ID, NAME, DETAILS)                                                                                               
---------- ---------- ------------------------------------------------------------------------------------------------------------------------
         1 aaa        REFTYPETAB(REFTYPE(1, 'asd', 'aaa'), REFTYPE(1, 'asg', 'ass'), REFTYPE(1, 'ash', 'add'), REFTYPE(1, 'asf', 'agg'))      
         2 aab        REFTYPETAB()                                                                                                            
         3 aaa        REFTYPETAB(REFTYPE(3, 'asd', 'aaa'), REFTYPE(3, 'ad', 'aa'))                                                            
         4 aaa        REFTYPETAB(REFTYPE(4, 'asd', 'aaa'), REFTYPE(4, 'as', 'a'), REFTYPE(4, 'ad', 'aa'))                                     

答案 1 :(得分:0)

进一步的研究给了我另一种方法:

SELECT id, CAST(COLLECT(reftype(r.id, r.name, r.details)) AS reftypetab) AS customer_ids
    FROM reftab r group by id;

这看起来好多了,但我仍然会问是否有任何其他方法可以做到这一点。

修改

也许这不是我所询问的,但是光标表达可以给出类似的结果。

 SELECT id, cursor(select reftype(r.id, r.name, r.details) from reftab r where r.id = b.id)  AS customer_ids
    FROM basictab b;