string - listunagg函数？

Question

甲骨文中有像listunagg函数这样的东西吗？例如，如果我有类似的数据:

------------------------------------------------------------
| 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            |
--------------------------------------------------------------

我想显示如下数据:

-----------------------------------------------
| 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     |
-----------------------------------------------

我试图找到一些与listagg相反的功能，但找不到任何功能。
提前致谢 :-)

Answer 1

正如@be现在已经在注释中指出的那样，Oracle不提供此类功能。因此，作为一种快速的解决方法，您可以编写类似的查询:

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