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 ：-）
最佳答案
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