Question

我有一个复杂的oracle视图，它返回在返回的行中具有逻辑副本的数据。我的目标是在基于两列（文本和日期时间）找到重复项时仅检索一行，但是要确定要返回哪一个重复项将基于第三列（日期时间）。

我已将下面的结果集模拟到带有存根数据的表中（在SQLFiddle上找到here）：

CREATE TABLE TimeTable (
  ID number NOT NULL,
  NAME VARCHAR2(20) NOT NULL,      -- Grouped by this first
  TARGETVALUE INT,                 -- ultimate target value to be returned (no precedence from this value)
  NOTE VARCHAR2(20) NULL,          -- Just a note for the developer on StackOverflow
  BEGIN_DATE TIMESTAMP NOT NULL,   -- Grouped by this 2nd (down to the minute, not seconds) 
  APPROVAL_DATE TIMESTAMP NOT NULL -- Decides the ties for duplicates

 );

 insert into TimeTable (ID, NAME, TARGETVALUE, NOTE, BEGIN_DATE, APPROVAL_DATE) values 
(1, 'Alpha', 5,  'Duplicate First', '08-MAR-14 09.43.00.000000000', 
                                    '09-MAR-14 09.43.00.000000000');

 insert into TimeTable (ID, NAME, TARGETVALUE, NOTE, BEGIN_DATE, APPROVAL_DATE) values 
(2, 'Alpha', 2,  'Duplicate Middle', '08-MAR-14 09.43.00.000000000', 
                                     '09-MAR-14 09.43.00.000000000');


 insert into TimeTable (ID, NAME, TARGETVALUE, NOTE, BEGIN_DATE, APPROVAL_DATE) values 
(3, 'Alpha', 3, 'Final Target', '08-MAR-14 09.43.00.000000000', 
                                '09-MAR-14 10.00.00.000000000');

-- Same time as alpha, but not related.
 insert into TimeTable (ID, NAME, TARGETVALUE, NOTE, BEGIN_DATE, APPROVAL_DATE) values 
(4, 'Beta', 4, 'Only Target', '08-MAR-14 09.43.30.000000000', 
                              '09-MAR-14 11.00.30.000000000');

需要的结果集是2行

3, 'Alpha', 3, '08-MAR-14 09.43.00.000000000', '09-MAR-14 10.00.00.000000000'
4, 'Beta', 4, '08-MAR-14 09.43.30.000000000' '09-MAR-14 11.00.30.000000000'

如果我在数据库中有这个值，请注意澄清

5, 'Alpha', 8, '09-MAR-14 09.43.00.000000000', '12-MAR-14 10.00.00.000000000'

然后该Alpha集将是唯一的并且也返回，因为由于不同的BEGIN_DATE（即3月9日而不是8日），它不被视为重复。

以下是遵循的规则

NAME与数据相关。
BEGIN_DATE是第二种关系，其中直到分钟的确切时间将有重复，需要将其淘汰以基于＃3。
如果每个＃1和＃2都有重复项，则会根据最新APPROVAL_DATE确定删除它们，这些<{1}}将在之前的日期赢取。

Answer 1

根据所提到的规则聚合数据应该是ANALYTICS的简单实现。

您希望每组MAX中的APPROVAL DATE NAME, BEGIN_DATE。所以，你需要做的就是：

MAX(APPROVAL_DATE) OVER(PARTITION BY NAME, BEGIN_DATE ORDER BY APPROVAL_DATE DESC) max_appr_dt

并且，在您的外部查询中，只需使用DUPLICATES中的WHERE APPROVAL_DATE = max_aapr_dt过滤掉PREDICATE。

注意从PERFORMANCE的角度来看，此方法仅执行一次TABLE SCAN 。因此，比加入表格和进行多表扫描

的其他方法要好得多
更新按照评论中的要求添加完整的测试用例

使用分析有两种方法：

<强> 1.MAX

SQL> SELECT * 2 FROM 3 (SELECT A.*, 4 MAX(APPROVAL_DATE) OVER(PARTITION BY NAME, BEGIN_DATE ORDER BY APPROVAL_DATE DESC) max_appr_dt 5 FROM TIMETABLE A 6 ) 7 WHERE approval_date = max_appr_dt 8 / ID NAME TARGETVALUE NOTE BEGIN_DATE APPROVAL_DATE MAX_APPR_DT ---------- -------------------- ----------- -------------------- ------------------------------ ------------------------------ ------------------------------ 3 Alpha 3 Final Target 08-MAR-14 09.43.00.000000 AM 09-MAR-14 10.00.00.000000 AM 09-MAR-14 10.00.00.000000 AM 4 Beta 4 Only Target 08-MAR-14 09.43.30.000000 AM 09-MAR-14 11.00.30.000000 AM 09-MAR-14 11.00.30.000000 AM

<强> 2.ROW_NUMBER（）

SQL> SELECT * 2 FROM 3 (SELECT a.*, 4 row_number() OVER(PARTITION BY NAME, BEGIN_DATE ORDER BY APPROVAL_DATE DESC) AS "RNK" 5 FROM TIMETABLE A 6 ) 7 WHERE rnk =1 8 / ID NAME TARGETVALUE NOTE BEGIN_DATE APPROVAL_DATE RNK ---------- -------------------- ----------- -------------------- ------------------------------ ------------------------------ ---------- 3 Alpha 3 Final Target 08-MAR-14 09.43.00.000000 AM 09-MAR-14 10.00.00.000000 AM 1 4 Beta 4 Only Target 08-MAR-14 09.43.30.000000 AM 09-MAR-14 11.00.30.000000 AM 1

两个查询的执行计划：

SQL> EXPLAIN PLAN FOR 2 SELECT * 3 FROM 4 (SELECT A.*, 5 MAX(APPROVAL_DATE) OVER(PARTITION BY NAME, BEGIN_DATE ORDER BY APPROVAL_DATE DESC) max_appr_dt 6 FROM TIMETABLE A 7 ) 8 WHERE approval_date = max_appr_dt 9 / Explained. SQL> SQL> select * from table(dbms_xplan.display) 2 / PLAN_TABLE_OUTPUT ---------------------------------------------------------------------------------------------------- Plan hash value: 2691156688 --------------------------------------------------------------------------------- | Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time | --------------------------------------------------------------------------------- | 0 | SELECT STATEMENT | | 4 | 356 | 3 (0)| 00:00:01 | |* 1 | VIEW | | 4 | 356 | 3 (0)| 00:00:01 | | 2 | WINDOW SORT | | 4 | 304 | 3 (0)| 00:00:01 | | 3 | TABLE ACCESS FULL| TIMETABLE | 4 | 304 | 3 (0)| 00:00:01 | --------------------------------------------------------------------------------- PLAN_TABLE_OUTPUT ---------------------------------------------------------------------------------------------------- Predicate Information (identified by operation id): --------------------------------------------------- 1 - filter("APPROVAL_DATE"="MAX_APPR_DT") Note ----- - dynamic statistics used: dynamic sampling (level=2) 19 rows selected. SQL> SQL> EXPLAIN PLAN FOR 2 SELECT * 3 FROM 4 (SELECT a.*, 5 row_number() OVER(PARTITION BY NAME, BEGIN_DATE ORDER BY APPROVAL_DATE DESC) AS "RNK" 6 FROM TIMETABLE A 7 ) 8 WHERE rnk =1 9 / Explained. SQL> SQL> select * from table(dbms_xplan.display) 2 / PLAN_TABLE_OUTPUT ---------------------------------------------------------------------------------------------------- Plan hash value: 3768566268 -------------------------------------------------------------------------------------- | Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time | -------------------------------------------------------------------------------------- | 0 | SELECT STATEMENT | | 4 | 356 | 3 (0)| 00:00:01 | |* 1 | VIEW | | 4 | 356 | 3 (0)| 00:00:01 | |* 2 | WINDOW SORT PUSHED RANK| | 4 | 304 | 3 (0)| 00:00:01 | | 3 | TABLE ACCESS FULL | TIMETABLE | 4 | 304 | 3 (0)| 00:00:01 | -------------------------------------------------------------------------------------- PLAN_TABLE_OUTPUT ---------------------------------------------------------------------------------------------------- Predicate Information (identified by operation id): --------------------------------------------------- 1 - filter("RNK"=1) 2 - filter(ROW_NUMBER() OVER ( PARTITION BY "NAME","BEGIN_DATE" ORDER BY INTERNAL_FUNCTION("APPROVAL_DATE") DESC )<=1) Note ----- - dynamic statistics used: dynamic sampling (level=2) 21 rows selected.

Answer 2

我知道您使用的是Oracle DB。但是，我使用SQL服务器测试了这个。 SQL应该适用于所有DB。尝试我的查询。我不确定这是否是最有效的方法。如果这有帮助，请告诉我。

select t.ID, t.name, t.targetvalue, t.begin_date, t.approval_date
from
(
select name, begin_date, max(approval_date) as approval_date
from timetable
group by name, begin_date
) as mx
inner join timetable as t
on mx.name = t.name and
mx.begin_date = t.begin_date and
mx.approval_date = t.approval_date

额外查询 - 如果要在SQL Server中的问题中创建表 -

CREATE TABLE TimeTable (
  ID int NOT NULL,
  NAME VARCHAR(20) NOT NULL,      
  TARGETVALUE INT,                
  NOTE VARCHAR(20) NULL,          
  BEGIN_DATE datetime NOT NULL,  
  APPROVAL_DATE datetime NOT NULL 

 );

 insert into TimeTable (ID, NAME, TARGETVALUE, NOTE, BEGIN_DATE, APPROVAL_DATE) values 
(1, 'Alpha', 5,  'Duplicate First', '08-03-14 09:43:00', 
                                    '09-03-14 09:43:00');

 insert into TimeTable (ID, NAME, TARGETVALUE, NOTE, BEGIN_DATE, APPROVAL_DATE) values 
(2, 'Alpha', 2,  'Duplicate Middle', '08-03-14 09:43:00', 
                                     '09-03-14 09:43:00');


 insert into TimeTable (ID, NAME, TARGETVALUE, NOTE, BEGIN_DATE, APPROVAL_DATE) values 
(3, 'Alpha', 3, 'Final Target', '08-03-14 09:43:00', 
                                '09-03-14 10:00:00');

-- Same time as alpha, but not related:
 insert into TimeTable (ID, NAME, TARGETVALUE, NOTE, BEGIN_DATE, APPROVAL_DATE) values 
(4, 'Beta', 4, 'Only Target', '08-03-14 09:43:30', 
                              '09-03-14 11:00:30');

基于三列删除结果集重复项

2 个答案: