Question

我的表格有4列，分别由两组(ID_PROCESSO, ID_INSTANCIA_BPM)和(ID_INSTANCIA_BPM_PAI, ID_PROCESSO_PAI)隔开。

如果第二组(ID_INSTANCIA_BPM_PAI AND ID_PROCESSO_PAI)为null，并且该记录的第一组不作为记录的第二组，其中ID_PROCESSO == ID_PROCESSO_PAI，则必须将其返回。

如果第二组(ID_INSTANCIA_BPM_PAI AND ID_PROCESSO_PAI)是not null和ID_PROCESSO <> ID_PROCESSO_PAI，则必须将其返回。

对于sqlfiddle示例，返回值应遵循类似的规则

#|ID_PROCESSO|ID_INSTANCIA_BPM|ID_INSTANCIA_BPM_PAI|ID_PROCESSO_PAI
1   3   2957    (null)  (null)  --Returned
2   5   2958    2957    3   --Returned
3   4   2960    (null)  (null)  --Returned
4   4   2975    (null)  (null)  --Returned
5   3   2976    (null)  (null)  --Removed by #10
6   3   2977    (null)  (null)  --Removed by #9
7   5   2978    2977    3   --Removed by #8
8   5   2979    2978    5   --Returned
9   3   2980    2977    3   --Returned
10  3   2981    2976    3   --Returned
11  4   2984    (null)  (null)  --Returned

所以结果必须是：

#|ID_PROCESSO|ID_INSTANCIA_BPM|ID_INSTANCIA_BPM_PAI|ID_PROCESSO_PAI
1   3   2957    (null)  (null)  --Returned
2   5   2958    2957    3   --Returned
3   4   2960    (null)  (null)  --Returned
4   4   2975    (null)  (null)  --Returned
8   5   2979    2978    5   --Returned
9   3   2980    2977    3   --Returned
10  3   2981    2976    3   --Returned
11  4   2984    (null)  (null)  --Returned

http://sqlfiddle.com/#!18/32c86/1

有人知道只有选择才能做到这一点，还是我需要一个游标才能实现此结果？

Answer 1

如果我理解您说的没错，

select *
from processes p1
where (p1.ID_INSTANCIA_BPM_PAI is null AND p1.ID_PROCESSO_PAI is null
      and
      not exists (select *
                  from processes p2
                  where p2.ID_INSTANCIA_BPM_PAI = p1.ID_INSTANCIA_BPM
                    and p2.ID_PROCESSO_PAI = p1.ID_PROCESSO 
                    and p2.ID_PROCESSO = p1.ID_PROCESSO)
      )
      or
      (p1.ID_INSTANCIA_BPM_PAI is not null AND ID_PROCESSO_PAI is not null 
      and
      not exists (select *
                  from processes p2
                  where p2.ID_INSTANCIA_BPM_PAI = p1.ID_INSTANCIA_BPM
                    and p2.ID_PROCESSO_PAI = p1.ID_PROCESSO 
                    )
      )

在sqlfiddle的输入行中，您错过了1行：

insert into processes values (5,    '2979', '2978', 5);

根据同一张表的两列进行过滤

1 个答案: