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CBO与RBO下的IN/EXISTS
发表于：2007-07-02来源：作者：点击数：
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晚上抽空看了看ask tom的RSS，发现两篇应该说很入门的关于IN/EXISTS的文章：http://asktom.oracle.com/pls/ask/f?p=4950:8:::::F4950_P8_DISPLAYID:953229842074，http://asktom.oracle.com/pls/ask/f?p=4950:8:::::F4950_P8_DISPLAYID:442029737684 文章很
晚上抽空看了看ask tom的RSS，发现两篇应该说很入门的关于IN/EXISTS的文章：http://asktom.oracle.com/pls/ask/f?p=4950:8:::::F4950_P8_DISPLAYID:953229842074，http://asktom.oracle.com/pls/ask/f?p=4950:8:::::F4950_P8_DISPLAYID:442029737684 文章很长，也没有仔细看完，不过有些东西还是很有意思的，动手实验了一下。大家随便看看咯，Tom当初回答问题的环境我已经没办法测试了，在10.1.0.4下做了些测试，CBO和RBO（使用hints）下还是很大区别的。SQL> select * from scott.emp; EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- 7369 SMITH CLERK 7902 17-12月-80 800 20 7499 ALLEN SALESMAN 7698 20-2月 -81 1600 300 30 7521 WARD SALESMAN 7698 22-2月 -81 1250 500 30 7566 JONES MANAGER 7839 02-4月 -81 2975 20 7654 MARTIN SALESMAN 7698 28-9月 -81 1250 1400 30 7698 BLAKE MANAGER 7839 01-5月 -81 2850 30 7782 CLARK MANAGER 7839 09-6月 -81 2450 10 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 7839 KING PRESIDENT 17-11月-81 5000 10 7844 TURNER SALESMAN 7698 08-9月 -81 1500 0 30 7876 ADAMS CLERK 7788 23-5月 -87 1100 20 7900 JAMES CLERK 7698 03-12月-81 950 30 7902 FORD ANALYST 7566 03-12月-81 3000 20 7934 MILLER CLERK 7782 23-1月 -82 1300 10已选择14行。执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=3 Card=14 Bytes=1218) 1 0 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) (Cost=3 Card=14 Bytes=1218)SQL> create table tmp_emp as select * from scott.emp where ename like @#S%@#;表已创建。SQL> select * from tmp_emp; EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- 7369 SMITH CLERK 7902 17-12月-80 800 20 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=3 Card=2 Bytes=174) 1 0 TABLE ACCESS (FULL) OF @#TMP_EMP@# (TABLE) (Cost=3 Card=2 Bytes=174)先测试一下IN：SQL> select * from tmp_emp where ename in (select ename from scott.emp); EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- 7369 SMITH CLERK 7902 17-12月-80 800 20 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=7 Card=2 Bytes=188) 1 0 HASH JOIN (SEMI) (Cost=7 Card=2 Bytes=188) 2 1 TABLE ACCESS (FULL) OF @#TMP_EMP@# (TABLE) (Cost=3 Card=2 Bytes=174) 3 1 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) (Cost=3 Card=14 Bytes=98)表没有做过分析，使用了HASH JOIN (SEMI)。Tom在回答问题的时候提到：Select * from T1 where x in ( select y from T2 )is typically processed as:select * from t1, ( select distinct y from t2 ) t2 where t1.x = t2.y;The subquery is evaluated, distincted, indexed (or hashed or sorted) and then joined to the original table -- typically.显然第二个查询是很奇怪的。也正是这个促使我打开了数据库测试，难道一个IN还需要先DISTINCT一下？没见过IN产生排序操作啊。SQL> ed已写入 file afiedt.buf 1 select * from tmp_emp,(select distinct ename from scott.emp) t 2* where tmp_emp.ename=t.enameSQL> / EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO ENAME---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- ---------- 7369 SMITH CLERK 7902 17-12月-80 800 20 SMITH 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 SCOTT执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=8 Card=2 Bytes=188) 1 0 VIEW (Cost=8 Card=2 Bytes=188) 2 1 SORT (UNIQUE) (Cost=8 Card=2 Bytes=202) 3 2 HASH JOIN (Cost=7 Card=2 Bytes=202) 4 3 TABLE ACCESS (FULL) OF @#TMP_EMP@# (TABLE) (Cost=3 Card=2 Bytes=188) 5 3 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) (Cost=3 Card=14 Bytes=98)显然不一样了。那么改用RULE模式：SQL> ed已写入 file afiedt.buf 1 select /*+ rule */ * from tmp_emp,(select distinct ename from scott.emp) t 2* where tmp_emp.ename=t.enameSQL> / EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO ENAME---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- ---------- 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 SCOTT 7369 SMITH CLERK 7902 17-12月-80 800 20 SMITH执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=HINT: RULE 1 0 MERGE JOIN 2 1 VIEW 3 2 SORT (UNIQUE) 4 3 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) 5 1 SORT (JOIN) 6 5 TABLE ACCESS (FULL) OF @#TMP_EMP@# (TABLE)再试试加上索引：SQL> alter table tmp_emp add constraints pk_tmpemp primary key(ename);表已更改。分析一下：SQL> analyze table tmp_emp compute statistics for table for all indexed columns;表已分析。SQL> analyze table scott.emp compute statistics for table for all columns;表已分析。SQL> select * 2 from tmp_emp e,(select distinct ename from scott.emp) t 3 where e.ename=t.ename; EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO ENAME---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- ---------- 7369 SMITH CLERK 7902 17-12月-80 800 20 SMITH 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 SCOTT执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=6 Card=2 Bytes=188) 1 0 VIEW (Cost=6 Card=2 Bytes=188) 2 1 SORT (UNIQUE) (Cost=6 Card=2 Bytes=88) 3 2 NESTED LOOPS (Cost=5 Card=2 Bytes=88) 4 3 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) (Cost=3 Card=14 Bytes=70) 5 3 TABLE ACCESS (BY INDEX ROWID) OF @#TMP_EMP@# (TABLE) (Cost=1 Card=1 Bytes=39) 6 5 INDEX (UNIQUE SCAN) OF @#PK_TMPEMP@# (INDEX (UNIQUE)) (Cost=0 Card=1)SQL> ed已写入 file afiedt.buf 1 select /*+rule*/ * 2 from tmp_emp e,(select distinct ename from scott.emp) t 3* where e.ename=t.enameSQL> / EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO ENAME---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- ---------- 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 SCOTT 7369 SMITH CLERK 7902 17-12月-80 800 20 SMITH执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=HINT: RULE 1 0 NESTED LOOPS 2 1 VIEW 3 2 SORT (UNIQUE) 4 3 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) 5 1 TABLE ACCESS (BY INDEX ROWID) OF @#TMP_EMP@# (TABLE) 6 5 INDEX (UNIQUE SCAN) OF @#PK_TMPEMP@# (INDEX (UNIQUE))SQL> select * from tmp_emp where ename in (select ename from scott.emp); EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 7369 SMITH CLERK 7902 17-12月-80 800 20执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=5 Card=1 Bytes=44) 1 0 NESTED LOOPS (Cost=5 Card=1 Bytes=44) 2 1 SORT (UNIQUE) (Cost=3 Card=14 Bytes=70) 3 2 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) (Cost=3 Card=14 Bytes=70) 4 1 TABLE ACCESS (BY INDEX ROWID) OF @#TMP_EMP@# (TABLE) (Cost=1 Card=1 Bytes=39) 5 4 INDEX (UNIQUE SCAN) OF @#PK_TMPEMP@# (INDEX (UNIQUE)) (Cost=0 Card=1)SQL> select /*+rule*/ * from tmp_emp where ename in (select ename from scott.emp); EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 7369 SMITH CLERK 7902 17-12月-80 800 20执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=HINT: RULE 1 0 NESTED LOOPS 2 1 VIEW OF @#VW_NSO_1@# (VIEW) 3 2 SORT (UNIQUE) 4 3 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) 5 1 TABLE ACCESS (BY INDEX ROWID) OF @#TMP_EMP@# (TABLE) 6 5 INDEX (UNIQUE SCAN) OF @#PK_TMPEMP@# (INDEX (UNIQUE))这次确实接近了很多，总体上看和Tom说的情况差不多。不过RBO这种模式对后边的表对应列选择性低时应该很好，而其他情况恐怕不见得是优化的。下面看看EXISTS：SQL> select * from tmp_emp t where exists(select null from scott.emp e where t.ename=e.ename); EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20 7369 SMITH CLERK 7902 17-12月-80 800 20执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=5 Card=1 Bytes=44) 1 0 NESTED LOOPS (Cost=5 Card=1 Bytes=44) 2 1 SORT (UNIQUE) (Cost=3 Card=14 Bytes=70) 3 2 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) (Cost=3 Card=14 Bytes=70) 4 1 TABLE ACCESS (BY INDEX ROWID) OF @#TMP_EMP@# (TABLE) (Cost=1 Card=1 Bytes=39) 5 4 INDEX (UNIQUE SCAN) OF @#PK_TMPEMP@# (INDEX (UNIQUE)) (Cost=0 Card=1)看着和用IN一样哦。SQL> select /*+rule*/ * from tmp_emp t where exists(select null from scott.emp e where t.ename=e.ename); EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- 7369 SMITH CLERK 7902 17-12月-80 800 20 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=HINT: RULE 1 0 FILTER 2 1 TABLE ACCESS (FULL) OF @#TMP_EMP@# (TABLE) 3 1 TABLE ACCESS (FULL) OF @#EMP@# (TABLE)用RBO就不同咯！SQL> select t.* from tmp_emp t,scott.emp e where t.ename=e.ename; EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO---------- ---------- --------- ---------- -------------- ---------- ---------- ---------- 7369 SMITH CLERK 7902 17-12月-80 800 20 7788 SCOTT ANALYST 7566 19-4月 -87 3000 20执行计划---------------------------------------------------------- 0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=5 Card=2 Bytes=88) 1 0 NESTED LOOPS (Cost=5 Card=2 Bytes=88) 2 1 TABLE ACCESS (FULL) OF @#EMP@# (TABLE) (Cost=3 Card=14 Bytes=70) 3 1 TABLE ACCESS (BY INDEX ROWID) OF @#TMP_EMP@# (TABLE) (Cost=1 Card=1 Bytes=39) 4 3 INDEX (UNIQUE SCAN) OF @#PK_TMPEMP@# (INDEX (UNIQUE)) (Cost=0 Card=1)这样看来在CBO下，使用内关联、IN、EXISTS很可能得到同一个执行计划（更多的情况就不测试了），优化器会发现三者的语义是相同的；而在较早的RBO下EXISTS采用FILTER而IN相当于对子查询先DISTINCT后关联，内关联则是直接关联就行了。Tom在他最新的回复中这样说：Use the RBO and see what you see. way back when I wrote this, that was the "more popular" of the two perhapstoday in 2005, what I said years ago using the RBO does not apply to the CBO. the cbo is smart enough to recognize these two things are effectively the same.IN相当于对子查询先DISTINCT后关联这一条真的没有想通，RBO为什么做这样的事呢？实在是没有普适性，我个人的理解就是设计的时候认为IN后面是跟一个值列表的情况居多，当然先把值算出来，然后NL就可以了，如果后面是一个表里的值那也就一样处理；而且使用IN的时候也许大多是子查询对外层查询的筛选性高，即外层的表较大，而子查询的返回值较少。看看不同情况的不同执行计划，Oracle在CBO上确实还是花了点心思的，赫赫。结论：在RBO下，使用IN还是EXISTS需要视情况而定，只要记住使用IN存在排序和DISTINCT这一步骤应该就不难判断；CBO下优化器会为你选择，怎么写就只是习惯问题了。
