listunagg功能？(listunagg function?)

有没有像listunagg功能甲骨文这样的事情？ 举例来说，如果我有一个数据，如：

------------------------------------------------------------
| user_id | degree_fi    | degree_en       | degree_sv       |
--------------------------------------------------------------
| 3601464 | 3700         |  1600           |  2200           |
|  1020   | 100          |  0              |   0             |
| 3600520 | 100,3200,400 | 1300, 800, 3000 | 1400, 600, 1500 |
| 3600882 |  0           |   100           |  200            |
--------------------------------------------------------------

我想显示这样的数据：

-----------------------------------------------
| user_id | degree_fi | degree_en | degree_sv |
-----------------------------------------------
| 3601464 | 3700      |  1600     |  2200     |
|  1020   | 100       |  0        |   0       |
| 3600520 |  100      | 1300      |  1400     |
| 3600882 |  0        |   100     |  200      |
| 3600520 |  3200     |   800     |  600      |
| 3600520 |  400      | 3000      |  1500     |
-----------------------------------------------

我试图找到像LISTAGG相反一些功能，但找不到任何。 提前致谢 ：-）

Answer 1:

作为@be现在这里的评论已经指出的Oracle不提供这样的功能。 因此，作为一个快速的解决方法，你可以写类似的查询：

with t1(user_id, degree_fi, degree_en, degree_sv) as
(
  select 3601464, '3700', '1600', '2200' from dual union all
  select 1020   , '100' , '0'   , '0'    from dual union all
  select 3600520, '100,3200,400', '1300, 800, 3000', '1400, 600, 1500'  from dual union all
  select 3600882, '0',    '100',  '200'  from dual
),
Occurence(ocr) as(
  select Level as ocr
    from (select max(greatest(regexp_count(degree_fi, '[^,]+')
                             , regexp_count(degree_en, '[^,]+')
                             , regexp_count(degree_sv, '[^,]+')
                             )
                    ) mx
            from t1    
          ) 
    connect by level <= mx
)
select *
  from (
select User_id
     , regexp_substr(degree_fi, '[^,]+', 1, o.ocr) as degree_fi
     , regexp_substr(degree_en, '[^,]+', 1, o.ocr) as degree_en
     , regexp_substr(degree_sv, '[^,]+', 1, o.ocr) as degree_sv
   from t1 t
  cross join Occurence o
)
where degree_fi is not null
  or degree_en is not null 
  or degree_sv is not null

结果：

User_Id   Degree_Fi  Degree_En  Degree_Sv
------------------------------------------------------------ 
3601464   3700       1600       2200 
1020      100        0          0 
3600520   100        1300       1400 
3600882   0          100        200 
3600520   3200       800        600 
3600520   400        3000       1500 

Answer 2:

有listunagg功能提供OraOpenSource utils的包集合。 这也是行之有效的。

Answer 3:

要unagg列表考虑一下汤姆Oracle的说“问汤姆”看到http://www.oracle.com/technetwork/issue-archive/2007/07-mar/o27asktom-084983.html代码3或4。

汤姆不讨论我的优先选择有利于短字符串（<34个字符）。 我使用Oracle DBMS_UTILITY.comma_to_table功能。 例：

SET SERVEROUTPUT ON
DECLARE
/** test data **/   
  L_LIST1   VARCHAR2(500) := '"A","B","C","Pierre - Andre","D","E","OFVampFVapos;CBryan","F","G","H","I","J"';
  l_list2   VARCHAR2(500);
  l_tablen  BINARY_INTEGER;
  l_tab     DBMS_UTILITY.uncl_array;
BEGIN
  DBMS_OUTPUT.put_line('l_list1 : ' || l_list1);

  DBMS_UTILITY.comma_to_table (
     list   => l_list1,
     tablen => l_tablen,
     tab    => l_tab);

  FOR i IN 1 .. l_tablen LOOP
    DBMS_OUTPUT.put_line(i || ' : ' || l_tab(i));
  END LOOP;

  DBMS_UTILITY.table_to_comma (
     tab    => l_tab,
     tablen => l_tablen,
     list   => l_list2);

  DBMS_OUTPUT.put_line('l_list2 : ' || l_list2);
end;

文章来源: listunagg function?