我最近在oracle数据库中将一段代码移到了生产环境中,其中一位经验丰富的开发人员提到了我提到的exists
和not exists
语句过多,而且应该有一个删除它们的方法,但它已经太久了,因为他不得不使用它并且不记得它是如何工作的。目前,我正在回过头来使代码片段更易于维护,因为随着业务逻辑/需求的变化,它可能会在未来几年内多次更改,我希望继续优化它,同时使其更易于维护
我已经尝试查找了,但我能找到的建议是将not in
替换为not exists
并且不返回实际结果。
因此,我想知道可以采取哪些措施来优化exists
/ not exists
,或者是否有办法编写exists
/ not exists
以便oracle将在内部对其进行优化(可能比我更好)。
例如,如何优化以下内容?
UPDATE
SCOTT.TABLE_N N
SET
N.VALUE_1 = 'Data!'
WHERE
N.VALUE_2 = 'Y'
AND
EXISTS
(
SELECT
1
FROM
SCOTT.TABLE_Q Q
WHERE
N.ID = Q.N_ID
)
AND
NOT EXISTS
(
SELECT
1
FROM
SCOTT.TABLE_W W
WHERE
N.ID = W.N_ID
)
答案 0 :(得分:8)
你的陈述对我来说似乎很好。
在任何优化任务中,不要考虑模式。不要认为,“(not) exists
是坏的,慢的,(not) in
非常酷而且快”。
想想,数据库在每个步骤上做了多少工作以及如何衡量它?
一个简单的例子:
- 不在:
23:59:41 HR@sandbox> alter system flush buffer_cache;
System altered.
Elapsed: 00:00:00.03
23:59:43 HR@sandbox> set autotrace traceonly explain statistics
23:59:49 HR@sandbox> select country_id from countries where country_id not in (select country_id from locations);
11 rows selected.
Elapsed: 00:00:00.02
Execution Plan
----------------------------------------------------------
Plan hash value: 1748518851
------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 6 | 4 (0)| 00:00:01 |
|* 1 | FILTER | | | | | |
| 2 | NESTED LOOPS ANTI SNA| | 11 | 66 | 4 (75)| 00:00:01 |
| 3 | INDEX FULL SCAN | COUNTRY_C_ID_PK | 25 | 75 | 1 (0)| 00:00:01 |
|* 4 | INDEX RANGE SCAN | LOC_COUNTRY_IX | 13 | 39 | 0 (0)| 00:00:01 |
|* 5 | TABLE ACCESS FULL | LOCATIONS | 1 | 3 | 3 (0)| 00:00:01 |
------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter( NOT EXISTS (SELECT 0 FROM "LOCATIONS" "LOCATIONS" WHERE
"COUNTRY_ID" IS NULL))
4 - access("COUNTRY_ID"="COUNTRY_ID")
5 - filter("COUNTRY_ID" IS NULL)
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
11 consistent gets
8 physical reads
0 redo size
446 bytes sent via SQL*Net to client
363 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
11 rows processed
- NOT EXISTS
23:59:57 HR@sandbox> alter system flush buffer_cache;
System altered.
Elapsed: 00:00:00.17
00:00:02 HR@sandbox> select country_id from countries c where not exists (select 1 from locations l where l.country_id = c.country_id );
11 rows selected.
Elapsed: 00:00:00.30
Execution Plan
----------------------------------------------------------
Plan hash value: 840074837
-------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
-------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 11 | 66 | 1 (0)| 00:00:01 |
| 1 | NESTED LOOPS ANTI| | 11 | 66 | 1 (0)| 00:00:01 |
| 2 | INDEX FULL SCAN | COUNTRY_C_ID_PK | 25 | 75 | 1 (0)| 00:00:01 |
|* 3 | INDEX RANGE SCAN| LOC_COUNTRY_IX | 13 | 39 | 0 (0)| 00:00:01 |
-------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - access("L"."COUNTRY_ID"="C"."COUNTRY_ID")
Statistics
----------------------------------------------------------
0 recursive calls
0 db block gets
5 consistent gets
2 physical reads
0 redo size
446 bytes sent via SQL*Net to client
363 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
11 rows processed
此示例中的NOT IN读取两倍的数据库块并执行更复杂的过滤 - 问问自己,为什么要在NOT EXISTS上选择它?
答案 1 :(得分:2)
当您需要时,没有理由避免使用EXISTS或NOT EXISTS。在您给出的示例中,这可能正是您要使用的内容。
典型的困境是使用IN / NOT IN还是EXISTS / NOT EXISTS。它们的评估方式完全不同,根据您的具体情况,可能会更快或更慢。
有关详细信息,请参阅here。
答案 2 :(得分:1)
我不知道它是否更快,但这是一种不用EXISTS
/ NOT EXISTS
来编写它的方法:
MERGE INTO TABLE_N T
USING (
SELECT N.ID, 'Data!' AS NEW_VALUE_1
FROM SCOTT.TABLE_N N
INNER JOIN SCOTT.TABLE_Q Q
ON Q.N_ID = N.ID
LEFT JOIN SCOTT.TABLE_W W
ON W.N_ID = N.ID
WHERE N.VALUE_2 = 'Y'
AND W.ID IS NULL
) X
ON ( T.ID = X.ID )
WHEN MATCHED THEN UPDATE
SET T.VALUE_1 = X.NEW_VALUE_1;