Question

我有心理障碍，希望有人在这里伸出援助之手。表的DDL如下。

text_locate(text, "replacement")
#>   text           before             instance              after            
#> 1 1                     be advisd  replacment   coming tomorow             
#> 2 2    …u get the email aobut the  repalcement  tomorro                    
#> 3 3    …been avdised of a posible  replacement                             
#> 4 4                    there is a  replacement  coming tomorrow            
#> 5 5     what time tommorow is the  replacment   coming

表格中的数据如下。

CREATE TABLE TEST_QUERY
   (   APPROVER VARCHAR2(20 BYTE), 
   STEP VARCHAR2(20 BYTE), 
   APPROVED VARCHAR2(20 BYTE), 
   DOCUMENT VARCHAR2(20 BYTE)
   )

对于此表中的每个文档，这些是按顺序排列的步骤列。它从1 ...无穷大。这里的关键是BK1和BK2条目将始终是具有最大步骤的表格中的条目。我正在尝试编写的查询是一个查询，用于标识将所有较小步骤（按文档）设置为“Y”但Bk1和BK2仍为空的文档。

因此使用样本数据的此查询的预期输出将是。

Insert into TEST_QUERY (APPROVER,STEP,APPROVED,DOCUMENT) values ('JHON','1','Y','A1');
Insert into TEST_QUERY (APPROVER,STEP,APPROVED,DOCUMENT) values ('PAT','2','Y','A1');
Insert into TEST_QUERY (APPROVER,STEP,APPROVED,DOCUMENT) values ('FRANK','3','Y','A1');
Insert into TEST_QUERY (APPROVER,STEP,APPROVED,DOCUMENT) values ('BK1','4',null,'A1');
Insert into TEST_QUERY (APPROVER,STEP,APPROVED,DOCUMENT) values ('BK2','5',null,'A1');
Insert into TEST_QUERY (APPROVER,STEP,APPROVED,DOCUMENT) values ('JHON','1',null,'A2');
Insert into TEST_QUERY (APPROVER,STEP,APPROVED,DOCUMENT) values ('PAT','2',null,'A2');
Insert into TEST_QUERY (APPROVER,STEP,APPROVED,DOCUMENT) values ('BK1','3',null,'A2');

A2文档不会显示在此查询中，因为仍然存在小于设置为null的最大步骤的步骤。关于如何完成这个的任何想法？

这是我到目前为止所做的事情。（这显然是关闭的）

Approver     Step     Document
BK1          4        A1
BK2          5        A1

感谢。

Answer 1

您可以使用group by：

select document
from test_query
group by document
having sum(case when approver in ('BK1', 'BK2') and approved is not null then 1 else 0 end) = 0 and
       sum(case when approver not in ('BK1', 'BK2') and approved is not null then 1 else 0 end) = count(*) - 2;

如果您需要详细数据，则应将其与原始行组合使用：

select tq.*
from test_query tq
where approved is null and
      document in (select document
                   from test_query
                   group by document
                   having sum(case when approver in ('BK1', 'BK2') and approved is not null then 1 else 0 end) = 0 and
                          sum(case when approver not in ('BK1', 'BK2') and approved is not null then 1 else 0 end) = count(*) - 2
                  );

仅在以前的记录已设置为Y时选择最大记录

1 个答案: