Oracle 使用connect by PREVIOR连接字段

具有下表“我的表”：

M01 |   1
M01 |   2
M02 |   1

我想查询一下，以便获得：

M01 |   1,2
M02 |   1

我通过以下查询成功接近：

with my_tabe as
(
    select 'M01' as scycle, '1' as sdate from dual union
    select 'M01' as scycle, '2' as sdate from dual union
    select 'M02' as scycle, '1' as sdate from dual
)
SELECT scycle, ltrim(sys_connect_by_path(sdate, ','), ',')
FROM
(
    select scycle, sdate, rownum rn
    from my_tabe
    order by 1 desc
)
START WITH rn = 1
CONNECT BY PRIOR rn = rn - 1

屈服：

SCYCLE      |   RES
M02         |   1,2,1
M01         |   1,2

这是错误的。看起来我很接近，但我不知道下一步该怎么做

---

有什么提示吗？

您需要将您的

connect by

限制为相同的

scycle

值，还需要计算匹配数并对其进行筛选，以避免看到中间结果

with my_tabe as
(
    select 'M01' as scycle, '1' as sdate from dual union
    select 'M01' as scycle, '2' as sdate from dual union
    select 'M02' as scycle, '1' as sdate from dual
)
select scycle, ltrim(sys_connect_by_path(sdate, ','), ',')
from
(
    select distinct sdate,
        scycle,
        count(1) over (partition by scycle) as cnt,
        row_number() over (partition by scycle order by sdate) as rn
    from my_tabe
)
where rn = cnt
start with rn = 1
connect by prior rn + 1 = rn
and prior scycle = scycle
/

SCYCLE LTRIM(SYS_CONNECT_BY_PATH(SDATE,','),',')
------ -----------------------------------------
M01    1,2
M02    1

如果您使用的是11g，则可以使用内置功能：

with my_tabe as
(
    select 'M01' as scycle, '1' as sdate from dual union
    select 'M01' as scycle, '2' as sdate from dual union
    select 'M02' as scycle, '1' as sdate from dual
)
select scycle, listagg (sdate, ',') 
within group (order by sdate) res
from my_tabe
group by scycle
/ 

这两种方法（和其他方法）都显示出来