一次針對Not in的優化--如何讓其選擇hash join anti

首先來看兩個sql，返回結果相同，但是耗時差別很大

SQL> select count(*)
  2             fromjustin_good r
  3            where notexists
  4                 (select 'x' from justin_count pc where pc.id =r.justin_good_id)
  5  ;

  COUNT(*)
----------
      7229

Executed in 3.437 seconds

SQL> select count(*)
  2             fromjustin_good r
  3            wherer.justin_good_id not in
  4                 (select pc.id from justin_count pc)
  5  ;

  COUNT(*)
----------
      7229

Executed in 128.203 seconds

再來看一下它們的執行計劃
使用not exist的語句cost爲3452，而not in 的卻達到14216
SQL> explain plan for select count(*)
  2             fromjustin_good r
  3            where notexists
  4                 (select 'x' from justin_count pc where pc.id =r.justin_good_id);

Explained.

SQL> select * from table(dbms_xplan.display);

PLAN_TABLE_OUTPUT
-------------------------------------------------------------------------------------------------------
Plan hash value: 1087925722

Id	Operation	Name	Rows	Bytes	Cost(%CPU)	Time
0	SELECT STATEMENT		1	9	3452(2)	00:00:42
1	SORT AGGREGATE		1	9
2	HASH JOIN RIGHT ANTI		59	531	3452(2)	00:00:42
3	INDEX FAST FULL SCAN	PK11_1	4562	22810	4(0)	00:00:01
4	TABLE ACCESS FULL	justin_good	602K	2355K	3440(2)	00:00:42

PLAN_TABLE_OUTPUT
-------------------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
2 -access("PC"."ID"="R"."justin_good_id")
16 rows selected.

SQL> explain plan for select count(*)
  2             fromjustin_good r
  3            wherer.justin_good_id not in
  4                 (select pc.id from justin_count pc);

Explained.

SQL> select * from table(dbms_xplan.display);

PLAN_TABLE_OUTPUT
-------------------------------------------------------------------------------------------------------Planhash value: 4119029611

Id	Operation	Name	Rows	Bytes	Cost(%CPU)	Time
0	SELECT STATEMENT		1	4	14216   (2)	00:00:42
1	SORT AGGREGATE		1	4
2	FILTER					00:00:42
3	TABLE ACCESS FULL	PK11_1	602K	2355K	3442   (2)	00:00:01
4	INDEX FULL SCAN	justin_good	1	5	11   (0)	00:00:42

PLAN_TABLE_OUTPUT
-------------------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
 2 - filter( NOT EXISTS (SELECT 0 FROM "justin"."justin_count""PC" WHERE
             LNNVL("PC"."ID"<>:B1)))
 4 - filter(LNNVL("PC"."ID"<>:B1))
18 rows selected.

 

當使用 where a not in (select …)的時候，必須確保該列a的返回值沒有null,否則無法進行hash join anti.
解決辦法：
1、  爲該列加not null約束
2、  改寫sql,如not exists

 

當oracle解析where a not in (select )之類的sql時，會查看列a是否存在null，只有當確保不存在的時候，纔會考慮將Not in轉換爲anti-join；然後進行一系列諸如nest loop join,merge join的cost對比，最終敲定執行計劃