Oracle JSON_ARRAYAGG 不支持 DISTINCT 关键字

Question

SELECT key1, key2, JSON_ARRAYAGG(foo) foo, JSON_ARRAYAGG(bar) bar FROM (
select 1 as key1, 2 as key2, '1.0' as foo, 'A' as bar from dual
UNION 
select 1, 2, '2.0' , 'A' as bar from dual
UNION 
select 3, 4, '2.0' , 'A' as bar from dual
UNION 
select 3, 4, '2.0' , 'B' as bar from dual
UNION 
select 3, 4, '2.0' , 'B' as bar from dual) z
GROUP BY key1, key2

查询返回如下结果：

1   2   ["1.0","2.0"]   ["A","A"]
3   4   ["2.0","2.0"]   ["A","B"]

我在期待

1   2   ["1.0","2.0"]   ["A"]
3   4   ["2.0"]         ["A","B"]

JSON_ARRAYAGG 似乎不支持 DISTINCT，有什么建议吗？

Answer 1

是的，那还没有实现。解决方法是提前去重数据，eg

SQL> with x as (
  2      select 1 as i, 'A' as j from dual union all select 1, 'A' from dual union all select 1, 'A' from dual
  3      union all
  4      select 2, 'D' from dual union all select 2, 'D' from dual union all select 2, 'E' from dual
  5      union all
  6      select 3, 'G' from dual union all select 3, 'H' from dual union all select 3, 'H' from dual
  7      )
  8  select x.i,
  9      listagg(distinct x.j, ', ') within group (order by x.j) as X_LIST,
 10      json_arrayagg( x.j order by x.j) as X_JSON
 11  from  x
 12  group by x.i;

         I X_LIST                         X_JSON
---------- ------------------------------ ------------------------------
         1 A                              ["A","A","A"]
         2 D, E                           ["D","D","E"]
         3 G, H                           ["G","H","H"]

SQL>
SQL> with x as (
  2      select 1 as i, 'A' as j from dual union all select 1, 'A' from dual union all select 1, 'A' from dual
  3      union all
  4      select 2, 'D' from dual union all select 2, 'D' from dual union all select 2, 'E' from dual
  5      union all
  6      select 3, 'G' from dual union all select 3, 'H' from dual union all select 3, 'H' from dual
  7      )
  8  select x.i,
  9      listagg(distinct x.j, ', ') within group (order by x.j) as X_LIST,
 10      json_arrayagg( x.j order by x.j) as X_JSON
 11  from ( select distinct i,j from x order by i,j ) x
 12  group by x.i;

         I X_LIST                         X_JSON
---------- ------------------------------ ------------------------------
         1 A                              ["A"]
         2 D, E                           ["D","E"]
         3 G, H                           ["G","H"]

Answer 2

除了 Connor 写的替代解决方法之外

aggregate function + row_number for de-duplication (

rn

可以移动到

where clause

而不是

decode

视情况而定，但在这种情况下

distinct

更可取)

select 
    x.i,
    listagg(decode(rn, 1, x.j), ', ') within group (order by x.j) as X_LIST,
    json_arrayagg(decode(rn, 1, x.j) order by x.j) as X_JSON
from (select x.*, row_number() over (partition by i, j order by '') rn from x) x
group by x.i;

相关标量中的嵌套聚合（请记住，每个相关标量实际上是一个隐式

join

）

select 
    xx.*,
    (select listagg(min(x.j), ', ') within group (order by x.j) from x where x.i = xx.i group by x.j) as X_LIST, 
    (select json_arrayagg(min(x.j) order by x.j) from x where x.i = xx.i group by x.j) as X_LIST
from (select distinct i from x) xx;