基于三列删除结果集重复

Question

我有一个复杂的oracle视图，它返回在返回的行内具有逻辑重复的数据。我的目标是当基于两列（文本和日期时间）找到重复项时只检索一行，但要根据第三列（日期时间）决定要返回哪一个重复项。

我有模拟的结果低于设定到表中与存根数据（如发现here上SQLFiddle）：

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'); 

其中需要将是2行的

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' 

结果集澄清的注意事项，如果我在数据库中有这个值

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

然后t帽子阿尔法设置将是唯一的，并且也返回，因为它不被认为是重复的，因为不同的BEGIN_DATE（这是3月9日而不是8日）。

---

这里有规律可循

1. 的NAME涉及的数据。
2. BEGIN_DATE是第二个关系，其中准确时间到分钟将有重复，需要根据＃3删除重复。
3. 如果每个＃1和＃2有重复，那么它们将被删除，由最近的APPROVAL_DATE确定，这将在早期日期赢得。

Answer 1

这应该是一个简单的实现的ANALYTICS聚集基于提到规则的数据。

您需要NAME, BEGIN_DATE的每个组中的MAXAPPROVAL DATE。所以，你需要做的是：

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

而且，在你的外部查询，只是滤除DUPLICATES在PREDICATE使用WHERE APPROVAL_DATE = max_aapr_dt。

注意 从PERFORMANCE的角度来看，这种做法会做一个TABLE SCAN只有一次。因此，除了加入表，并具有如意见要求多次表扫描

更新添加完整的测试用例的另一种方法更好

有使用分析方法有两种：

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 

2.ROW_NUMBE R（）

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 

执行计划两种查询：

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 Server测试了这一点。 SQL应该适用于所有数据库。尝试我的查询，但。我不确定这是否是最有效的方法。让我知道这是否有帮助。

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 

额外的查询 - 如果你想在SQL服务器内的问题，以创建表 -

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');