恒等查询条件改变执行计划——SQL优化之Everything is Possible

有的时候开发人员为了方便会在WHERE语句后面添加一个1=1，这样在处理页面传入的条件是就可用不用判断直接添加AND 条件。
一直认为添加一个恒等条件，不会对Oracle的查询造成什么影响，最多不过影响一下Oracle的性能，但是今天突然发现，这个恒等的查询条件居然可以影响Oracle的执行计划。

首先看一个简单的例子：
SQL> CREATE TABLE T1 AS SELECT * FROM DBA_OBJECTS
2 WHERE OWNER = 'SYS'
3 AND OBJECT_TYPE NOT LIKE '%BODY'
4 AND OBJECT_TYPE NOT LIKE 'JAVA%';
Table created.
SQL> CREATE TABLE T2 AS SELECT * FROM DBA_SEGMENTS WHERE OWNER = 'SYS';
Table created.
SQL> CREATE TABLE T3 AS SELECT * FROM DBA_INDEXES WHERE OWNER = 'SYS';
Table created.
SQL> ALTER TABLE T1 ADD CONSTRAINT PK_T1 PRIMARY KEY (OBJECT_NAME);
Table altered.
SQL> CREATE INDEX IND_T2_SEGNAME ON T2(SEGMENT_NAME);
Index created.
SQL> CREATE INDEX IND_T3_TABNAME ON T3(TABLE_NAME);
Index created.
SQL> EXEC DBMS_STATS.GATHER_TABLE_STATS(USER, 'T1', METHOD_OPT => 'FOR ALL INDEXED COLUMNS SIZE 100', CASCADE => TRUE)
PL/SQL procedure successfully completed.
SQL> EXEC DBMS_STATS.GATHER_TABLE_STATS(USER, 'T2', METHOD_OPT => 'FOR ALL INDEXED COLUMNS SIZE 100', CASCADE => TRUE)
PL/SQL procedure successfully completed.
SQL> EXEC DBMS_STATS.GATHER_TABLE_STATS(USER, 'T3', METHOD_OPT => 'FOR ALL INDEXED COLUMNS SIZE 100', CASCADE => TRUE)
PL/SQL procedure successfully completed.
SQL> SET AUTOT TRACE EXP
SQL> SELECT T1.OBJECT_NAME, T1.OBJECT_TYPE, T2.TABLESPACE_NAME
2 FROM T1, T2
3 WHERE T1.OBJECT_NAME = T2.SEGMENT_NAME
4 AND NOT EXISTS
5 (
6 SELECT 1 FROM T3
7 WHERE T3.INDEX_NAME = T1.OBJECT_NAME
8 AND T3.TABLE_NAME = 'OBJ$'
9 )
10 ;
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=12 Card=668 Bytes=58784)
1 0 HASH JOIN (ANTI) (Cost=12 Card=668 Bytes=58784)
2 1 HASH JOIN (Cost=9 Card=668 Bytes=39412)
3 2 TABLE ACCESS (FULL) OF 'T2' (Cost=2 Card=668 Bytes=21376)
4 2 TABLE ACCESS (FULL) OF 'T1' (Cost=6 Card=3806 Bytes=102762)
5 1 TABLE ACCESS (BY INDEX ROWID) OF 'T3' (Cost=2 Card=2 Bytes=58)
6 5 INDEX (RANGE SCAN) OF 'IND_T3_TABNAME' (NON-UNIQUE) (Cost=1 Card=2)

SQL> SELECT T1.OBJECT_NAME, T1.OBJECT_TYPE, T2.TABLESPACE_NAME
2 FROM T1, T2
3 WHERE T1.OBJECT_NAME = T2.SEGMENT_NAME
4 AND NOT EXISTS
5 (
6 SELECT 1 FROM T3
7 WHERE 1 = 1
8 AND T3.INDEX_NAME = T1.OBJECT_NAME
9 AND T3.TABLE_NAME = 'OBJ$'
10 )
11 ;
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=12 Card=668 Bytes=50768)
1 0 HASH JOIN (ANTI) (Cost=12 Card=668 Bytes=50768)
2 1 HASH JOIN (Cost=9 Card=668 Bytes=39412)
3 2 TABLE ACCESS (FULL) OF 'T2' (Cost=2 Card=668 Bytes=21376)
4 2 TABLE ACCESS (FULL) OF 'T1' (Cost=6 Card=3806 Bytes=102762)
5 1 VIEW OF 'VW_SQ_1' (Cost=2 Card=2 Bytes=34)
6 5 TABLE ACCESS (BY INDEX ROWID) OF 'T3' (Cost=2 Card=2 Bytes=58)
7 6 INDEX (RANGE SCAN) OF 'IND_T3_TABNAME' (NON-UNIQUE) (Cost=1 Card=2)

观察两个查询语句会发现，两个查询语句唯一的区别就是第二个查询语句中的NOT EXISTS子查询中包含了一个恒等查询条件1 = 1。
仅仅是这一点的区别，却造成了两个SQL语句的执行计划出现了差异。对于第二个查询Oracle居然多生成了一个临时的VIEW。也许有人认为这个执行计划没有太大的区别，基本上可以认为等价。但是由于多生成了一个VIEW的步骤，必然会造成性能的下降，更为关键的是：这说明Oracle认为两个SQL语句是不同的，而且处理方式也是不同的。
如果一个简单的VIEW步骤还不能说明什么问题的话，那么看看下面这个在实际运行中碰到的问题：
SQL> SET AUTOT TRACE EXP
SQL> SELECT A.ID,
2 B.NAME_CHN,
3 B.SPEC,
4 A.MANUFACTURE_ID
5 FROM CAT_PRODUCT A,
6 CAT_DRUG B
7 WHERE A.MEDICAL_ID = B.ID
8 AND A.CHECK_FLAG = 1
9 AND NOT EXISTS
10 (
11 SELECT 1
12 FROM PROJECT_SUPPLY_PRODUCT C
13 WHERE C.DATA_PRODUCT = A.ID
14 AND C.ENABLE_FLAG = '1'
15 AND C.PROJECT_ID = 'MRBR00000000000000709824'
16 )
17 ;
Elapsed: 00:00:00.01
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=483 Card=1 Bytes=213)
1 0 NESTED LOOPS (ANTI) (Cost=483 Card=1 Bytes=213)
2 1 NESTED LOOPS (Cost=481 Card=1 Bytes=160)
3 2 VIEW OF 'index$_join$_001' (Cost=480 Card=1 Bytes=77)
4 3 HASH JOIN
5 4 HASH JOIN
6 5 INDEX (FAST FULL SCAN) OF 'PK_CAT_PRODUCT' (UNIQUE) (Cost=33 Card=1 Bytes=77)
7 5 INDEX (FAST FULL SCAN) OF 'TU_CAT_PRODUCT_MED_CHECK' (NON-UNIQUE) (Cost=33 Card=1 Bytes=77)
8 4 INDEX (FAST FULL SCAN) OF 'TU_CAT_PRODUCT_MANUFACT_ID' (NON-UNIQUE) (Cost=33 Card=1 Bytes=77)
9 2 TABLE ACCESS (BY INDEX ROWID) OF 'CAT_DRUG' (Cost=1 Card=1 Bytes=83)
10 9 INDEX (UNIQUE SCAN) OF 'PK_CAT_DRUG' (UNIQUE)
11 1 TABLE ACCESS (BY INDEX ROWID) OF 'PROJECT_SUPPLY_PRODUCT' (Cost=2 Card=1 Bytes=53)
12 11 INDEX (RANGE SCAN) OF 'IDX_PROJECT_SUPPLY_PRODUCT_PJ' (NON-UNIQUE) (Cost=1 Card=1)

SQL> SELECT A.ID,
2 B.NAME_CHN,
3 B.SPEC,
4 A.MANUFACTURE_ID
5 FROM CAT_PRODUCT A,
6 CAT_DRUG B
7 WHERE A.MEDICAL_ID = B.ID
8 AND A.CHECK_FLAG = 1
9 AND NOT EXISTS
10 (
11 SELECT 1
12 FROM PROJECT_SUPPLY_PRODUCT C
13 WHERE 1 = 1
14 AND C.DATA_PRODUCT = A.ID
15 AND C.ENABLE_FLAG = '1'
16 AND C.PROJECT_ID = 'MRBR00000000000000709824'
17 )
18 ;
Elapsed: 00:00:00.00
Execution Plan
----------------------------------------------------------
0 SELECT STATEMENT Optimizer=CHOOSE (Cost=484 Card=1 Bytes=186)
1 0 HASH JOIN (ANTI) (Cost=484 Card=1 Bytes=186)
2 1 NESTED LOOPS (Cost=481 Card=1 Bytes=160)
3 2 VIEW OF 'index$_join$_001' (Cost=480 Card=1 Bytes=77)
4 3 HASH JOIN
5 4 HASH JOIN
6 5 INDEX (FAST FULL SCAN) OF 'PK_CAT_PRODUCT' (UNIQUE) (Cost=33 Card=1 Bytes=77)
7 5 INDEX (FAST FULL SCAN) OF 'TU_CAT_PRODUCT_MED_CHECK' (NON-UNIQUE) (Cost=33 Card=1 Bytes=77)
8 4 INDEX (FAST FULL SCAN) OF 'TU_CAT_PRODUCT_MANUFACT_ID' (NON-UNIQUE) (Cost=33 Card=1 Bytes=77)
9 2 TABLE ACCESS (BY INDEX ROWID) OF 'CAT_DRUG' (Cost=1 Card=1 Bytes=83)
10 9 INDEX (UNIQUE SCAN) OF 'PK_CAT_DRUG' (UNIQUE)
11 1 VIEW OF 'VW_SQ_1' (Cost=2 Card=1 Bytes=26)
12 11 TABLE ACCESS (BY INDEX ROWID) OF 'PROJECT_SUPPLY_PRODUCT' (Cost=2 Card=1 Bytes=53)
13 12 INDEX (RANGE SCAN) OF 'IDX_PROJECT_SUPPLY_PRODUCT_PJ' (NON-UNIQUE) (Cost=1 Card=1)
SQL语句仍然是只相差了NOT EXISTS语句中的一个1 = 1的恒等条件，但是如果不加这个恒等条件，Oracle用的是NESTED LOOP ANTI，而加上了这个恒等条件，则Oracle选择了HASH JOIN ANTI。
这次执行计划的改变足以使SQL的执行时间发生成百上千倍的变化。
一个小小的恒等查询条件居然可以使查询计划发生变化，这是我从来没有想到的。
通过这个问题，我总结了如下三点：
要敢于怀疑任何事情，没有什么是不可能的，一切以实践的结果为准；
不要在SQL上添加一些没有必要的小零碎，一些看似无害的东西在一些情况下可能会引发严重的问题。
建立良好的编码风格，以及严格SQL语句的编写制度的必要性。