使用SQL执行计划基线可以保证SQL的性能不下降，但实际生产中默认没有开启，这里是姚远老师在给OCM的学员授课中关于SQL执行计划基线的一个案例，大家可以借鉴一下。
01
—
修改配置，采集SQL执行计划基线
Oracle 19c与SQL执行计划基线相关的默认参数值如下：
代码语言：javascript复制SQL> show parameter baseline
NAME TYPE VALUE
------------------------------------ ----------- ------------------------------
optimizer_capture_sql_plan_baselines boolean FALSE
optimizer_use_sql_plan_baselines boolean TRUE
使用存储过程DBMS_SPM.CONFIGURE修改配置，自动捕捉TPCC用户执行的SQL，并创建基线：
代码语言：javascript复制EXEC DBMS_SPM.CONFIGURE('AUTO_CAPTURE_PARSING_SCHEMA_NAME','TPCC',true);
alter system set optimizer_capture_sql_plan_baselines=true;
修改后的参数存放在视图DBA_SQL_MANAGEMENT_CONFIG中，检查一下：
代码语言：javascript复制COL PARAMETER_NAME FORMAT a32
COL PARAMETER_VALUE FORMAT a32
SELECT PARAMETER_NAME, PARAMETER_VALUE FROM DBA_SQL_MANAGEMENT_CONFIG ;
PARAMETER_NAME PARAMETER_VALUE
-------------------------------- --------------------------------
AUTO_CAPTURE_ACTION
AUTO_CAPTURE_MODULE
AUTO_CAPTURE_PARSING_SCHEMA_NAME parsing_schema IN (TPCC)
AUTO_CAPTURE_SQL_TEXT
AUTO_SPM_EVOLVE_TASK OFF
AUTO_SPM_EVOLVE_TASK_INTERVAL 3600 --
AUTO_SPM_EVOLVE_TASK_MAX_RUNTIME 1800
PLAN_RETENTION_WEEKS 53 -- 53不用的计划会被清理
SPACE_BUDGET_PERCENT 10 -- 占用SYSAUX的空间不超过10%，超过在alert中报警
9 rows selected.
02
—
查看SQL执行计划基线的应用
先将一个索引改成不可见：
代码语言：javascript复制SQL> alter index tpcc.CUSTOMER_I1 invisible;
Index altered.应用运行一段时间后，检查已经创建的SQL基线：
代码语言：javascript复制SQL> select count(distinct sql_handle),count(distinct plan_name),count(distinct SIGNATURE) from DBA_SQL_PLAN_BASELINES;
COUNT(DISTINCTSQL_HANDLE) COUNT(DISTINCTPLAN_NAME) COUNT(DISTINCTSIGNATURE)
------------------------- ------------------------ ------------------------
30 30 30
可以看到为30个SQL建立了基线，都是ACCEPTED，因为每个SQL只有一个执行计划。
检查与这个索引相关的SQL的执行情况：
代码语言：javascript复制SQL> select SQL_ID,EXECUTIONS,OPTIMIZER_COST,SQL_PLAN_BASELINE from v$sql where sql_id='arykx3hpq9xsa';
SQL_ID EXECUTIONS OPTIMIZER_COST SQL_PLAN_BASELINE
------------- ---------- -------------- ------------------------------
arykx3hpq9xsa 5102 1945 SQL_PLAN_2v1cfx8jds3vt3a6ea7ea
可以看到它用到了一个SQL基线的执行计划，成本是1945，查看这个SQL基线的执行计划：
代码语言：javascript复制SELECT * FROM TABLE(DBMS_XPLAN.DISPLAY_SQL_PLAN_BASELINE('SQL_2d858eea22dc0f79','SQL_PLAN_2v1cfx8jds3vt3a6ea7ea','basic') );
PLAN_TABLE_OUTPUT
--------------------------------------------------------------------------------
SQL handle: SQL_2d858eea22dc0f79
SQL text: UPDATE CUSTOMER SET C_BALANCE = C_BALANCE :B1 WHERE C_W_ID = :B4
AND C_D_ID = :B3 AND C_ID = :B2
--------------------------------------------------------------------------------
--------------------------------------------------------------------------------
Plan name: SQL_PLAN_2v1cfx8jds3vt3a6ea7ea Plan id: 980330474
Enabled: YES Fixed: NO Accepted: YES Origin: AUTO-CAPTURE
Plan rows: From dictionary
--------------------------------------------------------------------------------
Plan hash value: 3529770744
----------------------------------------
| Id | Operation | Name |
----------------------------------------
| 0 | UPDATE STATEMENT | |
| 1 | UPDATE | CUSTOMER |
| 2 | INDEX SKIP SCAN| CUSTOMER_I2 |
----------------------------------------
22 rows selected.
可以看到这个SQL执行中使用了CUSTOMER_I2 索引，没有使用CUSTOMER_I1索引，因为CUSTOMER_I1这个索引被修改成了不可见。这种检查SQL执行计划的方法和在游标中查询SQL执行计划的方法得到同样的结果：
代码语言：javascript复制set pagesize 200
SELECT PLAN_TABLE_OUTPUT FROM TABLE(DBMS_XPLAN.DISPLAY_CURSOR('arykx3hpq9xsa'));将这个索引改成可见：
代码语言：javascript复制alter index tpcc.CUSTOMER_I1 visible;第二次执行应用程序，然后再检查这个SQL的执行情况：
代码语言：javascript复制SQL> select SQL_ID,EXECUTIONS,OPTIMIZER_COST,SQL_PLAN_BASELINE from v$sql where sql_id='arykx3hpq9xsa';
SQL_ID EXECUTIONS OPTIMIZER_COST SQL_PLAN_BASELINE
------------- ---------- -------------- ------------------------------
arykx3hpq9xsa 2376 1945 SQL_PLAN_2v1cfx8jds3vt3a6ea7ea
发现这个SQL的执行成本和使用执行计划基线仍然没有发生变化，检查这个SQL对应的执行计划基线：
代码语言：javascript复制col plan_name form a30
col signature forma 99999999999999999999999
select sql_handle,plan_name,signature,accepted,optimizer_cost from DBA_SQL_PLAN_BASELINES
where SIGNATURE=(select EXACT_MATCHING_SIGNATURE from v$sql where sql_id='arykx3hpq9xsa');
SQL_HANDLE PLAN_NAME SIGNATURE ACC OPTIMIZER_COST
------------------------------ ------------------------------ ------------------------ --- --------------
SQL_2d858eea22dc0f79 SQL_PLAN_2v1cfx8jds3vt341d91fc 3280185039867613049 NO 3
SQL_2d858eea22dc0f79 SQL_PLAN_2v1cfx8jds3vt3a6ea7ea 3280185039867613049 YES 1945
发现这个SQL对应了两个基线，期中成本小到3的基线居然是没有被接受的！
03
—
手工进化基线
因为SQL基线的进化任务要到晚上维护窗口时才会执行，新的基线没有进化成可接受的，所以SQL执行时不会选择这个基线，我们可以手工对这个SQL基线进行进化：
代码语言：javascript复制VARIABLE cnt NUMBER
VARIABLE tk_name VARCHAR2(50)
VARIABLE exe_name VARCHAR2(50)
VARIABLE evol_out CLOB
begin
:tk_name := DBMS_SPM.CREATE_EVOLVE_TASK(
sql_handle => 'SQL_2d858eea22dc0f79',
plan_name => 'SQL_PLAN_2v1cfx8jds3vt341d91fc');
end;
/
SELECT :tk_name FROM DUAL;
SQL> SELECT :tk_name FROM DUAL;
:TK_NAME
-----------------------
TASK_1551
EXECUTE :exe_name :=DBMS_SPM.EXECUTE_EVOLVE_TASK(task_name=>:tk_name);
SELECT :exe_name FROM DUAL;
:EXE_NAME
--------------------------------
EXEC_3452
检查这个进化计划的执行报告：
代码语言：javascript复制EXECUTE :evol_out := DBMS_SPM.REPORT_EVOLVE_TASK( task_name=>:tk_name, execution_name=>:exe_name );
SELECT :evol_out FROM DUAL;
SQL> SELECT :evol_out FROM DUAL;
:EVOL_OUT
--------------------------------------------------------------------------------
GENERAL INFORMATION SECTION
----------------------------------------------------
Task Information:
---------------------------------------------
Task Name : TASK_1551
Task Owner : SYS
Execution Name : EXEC_3452
Execution Type : SPM EVOLVE
Scope : COMPREHENSIVE
Status : COMPLETED
Started : 09/08/2023 15:21:50
Finished : 09/08/2023 15:21:50
Last Updated : 09/08/2023 15:21:50
Global Time Limit : 2147483646
Per-Plan Time Limit : UNUSED
Number of Errors : 0
-----------------------------------------
SUMMARY SECTION
--------------------------------------------------
7 Number of plans processed : 1
Number of findings : 1
Number of recommendations : 1
Number of errors : 0
--------------------------------------------------------
-------------------------------------
DETAILS SECTION
---------------------------------------------
----------------------------------------
--------
Object ID : 2
Test Plan Name : SQL_PLAN_2v1cfx8jds3vt341d91fc
Base Plan Name : SQL_PLAN_2v1cfx8jds3vt3a6ea7ea
SQL Handle : SQL_2d858eea22dc0f79
Parsing Schema : TPCC
Test Plan Creator : SYS
SQL Text : UPDATE CUSTOMER SET C_BALANCE = C_BALANCE :B1 WHERE
C_W_ID = :B4 AND C_D_ID= :B3 AND C_ID = :B2
Bind Variables:
-----------------------------
2 - (NUMBER): 3
3 - (NUMBER): 1
4 - (NUMBER): 2285
Execution Statistics:
-----------------------------
Base Plan
Test Plan
--------------------
-------- ----------------------------
Elapsed Time (s): .00064 .000002
CPU Time (s): .000638 .000002
Buffer Gets: 117 0
Optimizer Cost: 1945 3
Disk Reads: 0 0
Direct Writes: 0 0
Rows Processed: 0 0
Executions: 10 10
FINDINGS SECTION
-----------------------------------------
Findings (1):
-----------------------------
1. The plan was verified in 0.12000 seconds. It passed the benefit criterion
because its verified performance was 390.90001 times better than that of
the baseline plan.
Recommendation:
-----------------------------
Consider accepting the plan. Execute
dbms_spm.accept_sql_plan_baseline(task_name => 'TASK_1551', object_id => 2,
task_owner => 'SYS');
EXPLAIN PLANS SECTION
---------------------------------------------
Baseline Plan
-----------------------------
Plan Id : 9217
Plan Hash Value : 980330474
-------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost | Time |
---------------------------------------------------
------------------------
| 0 | UPDATE STATEMENT | | 1 | 15 | 1945 | 00:00:01 |
| 1 | UPDATE | CUSTOMER | | | | |
| * 2 | INDEX SKIP SCAN | CUSTOMER_I2 | 1 | 15 | 1944 | 00:00:01 |
-----------------------------------------------
Predicate Information (identified by operation id)
:
------------------------------------------
* 2 - access("C_W_ID"=:B4 AND "C_D_ID"=:B3 AND "C_ID"=:B2)
* 2 - filter("C_ID"=:B2 AND "C_W_ID"=:B4 AND "C_D_ID"=:B3)
Test Plan
-----------------------------
Plan Id : 9218
Plan Hash Value : 874353148
----------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost | Time |
--------------------------------------------------------
| 0 | UPDATE STATEMENT | | 1 | 15| 3 | 00:00:01 |
| 1 | UPDATE | CUSTOMER | | | | |
| * 2 | INDEX UNIQUE SCAN | CUSTOMER_I1 | 1 | 15 | 2 | 00:00:01 |
--------------------------------------------
Predicate Information (identified by operation id):
------------------------------------------
* 2 - access("C_W_ID"=:B4 AND "C_D_ID"=:B3 AND "C_ID"=:B2)
--------------------------------------------------
根据报告中的建议，执行下面的SQL接受第二个基线：
代码语言：javascript复制SQL> exec dbms_spm.accept_sql_plan_baseline(task_name => 'TASK_1551', object_id => 2, task_owner => 'SYS');
PL/SQL procedure successfully completed.
再次检查这个SQL对应的执行计划基线：
代码语言：javascript复制col plan_name form a30
col signature forma 99999999999999999999999
select sql_handle,plan_name,signature,accepted,optimizer_cost from DBA_SQL_PLAN_BASELINES
2 where SIGNATURE=(select EXACT_MATCHING_SIGNATURE from v$sql where sql_id='arykx3hpq9xsa');
SQL_HANDLE PLAN_NAME SIGNATURE ACC OPTIMIZER_COST
------------------------------ ------------------------------ ------------------------ --- --------------
SQL_2d858eea22dc0f79 SQL_PLAN_2v1cfx8jds3vt341d91fc 3280185039867613049 YES 3
SQL_2d858eea22dc0f79 SQL_PLAN_2v1cfx8jds3vt3a6ea7ea 3280185039867613049 YES 1945
发现两个基线都是接受的。
第三次执行应用后，然后检查SQL的执行情况：
代码语言：javascript复制SQL> select SQL_ID,EXECUTIONS,OPTIMIZER_COST,SQL_PLAN_BASELINE from v$sql where sql_id='arykx3hpq9xsa';
SQL_ID EXECUTIONS OPTIMIZER_COST SQL_PLAN_BASELINE
------------- ---------- -------------- ------------------------------
arykx3hpq9xsa 6849 3 SQL_PLAN_2v1cfx8jds3vt341d91fc
发现应用的SQL采用了第二个基线，执行成本从1945降低到3。
04
—
删除SQL基线
使用dbms_spm.drop_sql_plan_baseline可以删除SQL基线，但不能一次删除所有的SQL基线，如果要删除所有的SQL基线，需要用游标进行循环删除 ，相关程序如下（参见Doc ID 790039.1）
代码语言：javascript复制declare
pgn number;
sqlhdl varchar2(30);
cursor hdl_cur is
select distinct sql_handle from dba_sql_plan_baselines;
begin
open hdl_cur;
loop
fetch hdl_cur into sqlhdl;
exit when hdl_cur%NOTFOUND;
pgn := dbms_spm.drop_sql_plan_baseline(sql_handle=>sqlhdl);
end loop;
close hdl_cur;
commit;
end;
/
PL/SQL procedure successfully completed.
SQL> select count(*) from dba_sql_plan_baselines;
COUNT(*)
----------
0
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