SQL查询连接不返回行

Question

我有一个返回以下数据下面的选择查询：

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 

所以现在，我希望PRD_END_DT对于每个PAYGROUP，例如x = 1的PRD_END_DT_1之前的'x'个周期数，我的查询应返回：

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 

我厌倦了下面的，但没有帮助所有：

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 
    AND 1 = 
    (SELECT COUNT(XVW.PRD_END_DT) 
    FROM PS_AZ_DFN_ALPRD_VW XVW 
    WHERE XVW.GP_PAYGROUP = TMP.GP_PAYGROUP 
     AND XVW.PRD_END_DT < TMP.PRD_END_DT) 

你可以建议可能的方法吧。欣赏你的投入。

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

尝试使用dense_rank功能：

例如

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