Question

我有以下选择查询，它返回以下数据：

查询：

SELECT DISTINCT ALLVW.GP_PAYGROUP,
                ALLVW.PRD_END_DT,
                TMP.PRD_END_DT
FROM PS_AZ_DFN_ALPRD_VW ALLVW,
     PS_AZ_DFMPRP_TMP4 TMP
WHERE ALLVW.GP_PAYGROUP = TMP.GP_PAYGROUP
  AND ALLVW.PRD_END_DT < TMP.PRD_END_DT

结果：

PAYGROUP    PRD_END_DT      PRD_END_DT_1

AMZ_PG_T1   31-JAN-12   31-DEC-12
AMZ_PG_T1   29-FEB-12   31-DEC-12
AMZ_PG_T1   31-MAR-12   31-DEC-12
AMZ_PG_T1   30-NOV-12   31-DEC-12
AMZ_PG_T2   31-JAN-12   31-OCT-12
AMZ_PG_T2   30-SEP-12   31-OCT-12
AMZ_PG_T2   31-MAR-12   31-OCT-12
AMZ_PG_T2   30-APR-12   31-OCT-12

现在，我希望每个PRD_END_DT的{{1}}是PAYGROUP之前的'x'个句点，例如x = 1，我的查询应该返回：< / p>

PRD_END_DT_1

我厌倦了以下但没有全部帮助：

 PAYGROUP    PRD_END_DT      PRD_END_DT_1
 AMZ_PG_T1   30-NOV-12       31-DEC-12
 AMZ_PG_T2   30-SEP-12       31-OCT-12

请你建议可行的方法。感谢您的意见。

Answer 1

SELECT * FROM (
  your original SELECT DISTINCT... query
) t where datediff(month, PRD_END_DT, PRD_END_DT_1) = x

Answer 2

SELECT DISTINCT ALLVW.GP_PAYGROUP, ALLVW.PRD_END_DT, TMP.PRD_END_DT
FROM PS_AZ_DFN_ALPRD_VW ALLVW
JOIN PS_AZ_DFMPRP_TMP4 TMP ON ALLVW.GP_PAYGROUP = TMP.GP_PAYGROUP
WHERE ALLVW.PRD_END_DT < TMP.PRD_END_DT
  AND add_months(trunc(ALLVW.PRD_END_DT, 'month'), 1) >= trunc(TMP.PRD_END_DT, 'month')

Answer 3

尝试并使用dense_rank功能：

E.g。

Select a, b, c
From ( select a, b, c, dense_rank() over (partition by a order by b desc) r
       from table)
Where r < :x

使用SELECT，COUNT和子查询的SQL查询连接不返回行

3 个答案: