列表功能?
oracle 中是否有类似 listunagg 函数的东西?例如,如果我有这样的数据:
is there such thing in oracle like listunagg function? For example, if I have a data like:
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 |
我试图找到一些与 listagg
相反的函数,但找不到.
I tried to find some function like opposite of listagg
but couldn't find any.
推荐答案
正如@be here 现在已经在评论中指出的,Oracle 不提供这样的功能.因此,作为一种快速解决方法,您可以编写类似的查询:
As @be here now has already noted in the comment Oracle doesn't provide such a function. So as a quick workaround you could write similar query:
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
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