Bog*_*mac 6 oracle performance
考虑将更改应用于聚合表的问题.必须更新存在的行,同时必须插入新行.我的方法如下:
SQL(带有SORT MERGE提示)如下所示(没什么特别的):
merge /*+ USE_MERGE(t s) */
into F_SCREEN_INSTANCE t
using F_SCREEN_INSTANCE_BUF s
on (s.DAY_ID = t.DAY_ID and s.PARTIAL_ID = t.PARTIAL_ID)
when matched then update set
t.ACTIVE_TIME_SUM = t.ACTIVE_TIME_SUM + s.ACTIVE_TIME_SUM,
t.IDLE_TIME_SUM = t.IDLE_TIME_SUM + s.IDLE_TIME_SUM
when not matched then insert values (
s.DAY_ID, s.PARTIAL_ID, s.ID, s.AGENT_USER_ID, s.COMPUTER_ID, s.RAW_APPLICATION_ID, s.APP_USER_ID, s.APPLICATION_ID, s.USER_ID, s.RAW_MODULE_ID, s.MODULE_ID, s.START_TIME, s.RAW_SCREEN_NAME, s.SCREEN_ID, s.SCREEN_TYPE, s.ACTIVE_TIME_SUM, s.IDLE_TIME_SUM)
Run Code Online (Sandbox Code Playgroud)
该F_SCREEN_INSTANCE表具有(DAY_ID, PARTIAL_ID)主键,也是IOT(索引组织表).这使其成为合并连接的理想候选者:行按查找键进行物理排序.
到现在为止还挺好.我已经开始了一个基准测试,初始时间看起来不错,合并时间为10秒.但大约一个小时后,合并大约需要4分钟,因为tempdb使用率很高(每个合并4GB).下面的查询计划显示F_SCREEN_INSTANCE在合并之前重新排序,即使表已经理想地排序.当然,随着表的增长,将需要更多的tempdb,整个方法就会崩溃.

好的,为什么要重新排序表呢?它变成了合并连接实现的限制:第二个表总是排序的.
如果存在索引,则数据库可以避免对第一个数据集进行排序.但是,无论索引如何,数据库始终会对第二个数据集进行排序.
O ... K,那么我可以将主表放在第一位,将缓冲放在第二位吗?不,那也不可能.无论我如何列出USE_MERGE提示中的表,源表始终是第一个.
最后,这是我的问题:我错过了什么吗?是否有可能使这种SORT MERGE方法有效?
以下是一些解决您可能会问的问题的详细信息:
无论提示如何,排序合并外连接总是将外连接表放在第二位.添加额外的内连接允许控制连接顺序,然后可以使用ROWID再次连接到大表.希望两个好的连接比一个坏的连接更好.
假设
此答案假定排序合并连接是最快的连接,并且手册是正确的,第二个数据集始终排序.如果没有关于数据的更多信息,就很难测试这些假设.
示例模式
下面是一些类似的表,使用虚假统计信息使优化器认为它们有500M行和100K行.
create table F_SCREEN_INSTANCE(DAY_ID number, PARTIAL_ID number, ID number, AGENT_USER_ID number,COMPUTER_ID number, RAW_APPLICATION_ID number, APP_USER_ID number, APPLICATION_ID number, USER_ID number, RAW_MODULE_ID number,MODULE_ID number, START_TIME date, RAW_SCREEN_NAME varchar2(100), SCREEN_ID number, SCREEN_TYPE number, ACTIVE_TIME_SUM number, IDLE_TIME_SUM number,
constraint f_screen_instance_pk primary key (day_id, partial_id)
) organization index;
create table F_SCREEN_INSTANCE_BUF(DAY_ID number, PARTIAL_ID number, ID number, AGENT_USER_ID number,COMPUTER_ID number, RAW_APPLICATION_ID number, APP_USER_ID number,APPLICATION_ID number, USER_ID number, RAW_MODULE_ID number, MODULE_ID number, START_TIME date, RAW_SCREEN_NAME varchar2(100), SCREEN_ID number, SCREEN_TYPE number, ACTIVE_TIME_SUM number, IDLE_TIME_SUM number,
constraint f_screen_instance_buf_pk primary key (day_id, partial_id)
);
begin
dbms_stats.set_table_stats(user, 'F_SCREEN_INSTANCE', numrows => 500000000);
dbms_stats.set_table_stats(user, 'F_SCREEN_INSTANCE_BUF', numrows => 100000);
end;
/
Run Code Online (Sandbox Code Playgroud)
问题
使用内部联接时,可以使用LEADING提示实现所需的连接和连接顺序.较小的表F_SCREEN_INSTANCE_BUF是第二个表.
explain plan for
select /*+ use_merge(t s) leading(t s) */ *
from f_screen_instance_buf s
join f_screen_instance t
on (s.DAY_ID = t.DAY_ID and s.PARTIAL_ID = t.PARTIAL_ID);
select * from table(dbms_xplan.display(format => '-predicate'));
Plan hash value: 563239985
-----------------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes |TempSpc| Cost (%CPU)| Time |
-----------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 100K| 19M| | 6898 (66)| 00:00:01 |
| 1 | MERGE JOIN | | 100K| 19M| | 6898 (66)| 00:00:01 |
| 2 | INDEX FULL SCAN | F_SCREEN_INSTANCE_PK | 500M| 46G| | 4504 (100)| 00:00:01 |
| 3 | SORT JOIN | | 100K| 9765K| 26M| 2393 (1)| 00:00:01 |
| 4 | TABLE ACCESS FULL| F_SCREEN_INSTANCE_BUF | 100K| 9765K| | 34 (6)| 00:00:01 |
-----------------------------------------------------------------------------------------------------
Run Code Online (Sandbox Code Playgroud)
更改为左连接时,LEADING提示不起作用.
explain plan for
select /*+ use_merge(t s) leading(t s) */ *
from f_screen_instance_buf s
left join f_screen_instance t
on (s.DAY_ID = t.DAY_ID and s.PARTIAL_ID = t.PARTIAL_ID);
select * from table(dbms_xplan.display(format => '-predicate'));
Plan hash value: 1472690071
-----------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes |TempSpc| Cost (%CPU)| Time |
-----------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 100K| 19M| | 16M (1)| 00:10:34 |
| 1 | MERGE JOIN OUTER | | 100K| 19M| | 16M (1)| 00:10:34 |
| 2 | TABLE ACCESS BY INDEX ROWID| F_SCREEN_INSTANCE_BUF | 100K| 9765K| | 826 (0)| 00:00:01 |
| 3 | INDEX FULL SCAN | F_SCREEN_INSTANCE_BUF_PK | 100K| | | 26 (0)| 00:00:01 |
| 4 | SORT JOIN | | 500M| 46G| 131G| 16M (1)| 00:10:34 |
| 5 | INDEX FAST FULL SCAN | F_SCREEN_INSTANCE_PK | 500M| 46G| | 2703 (100)| 00:00:01 |
-----------------------------------------------------------------------------------------------------------------
Run Code Online (Sandbox Code Playgroud)
据我所知,这个限制没有记录.我尝试使用+outline设置DBMS_XPLAN查看完整的提示,然后更改它们.但我没有做任何事情可以使LEFT JOIN版本的连接顺序更改.也许其他人可以让这个工作.
select * from table(dbms_xplan.display(format => '-predicate +outline'));
...
Outline Data
-------------
/*+
BEGIN_OUTLINE_DATA
USE_MERGE(@"SEL$0E991E55" "T"@"SEL$1")
LEADING(@"SEL$0E991E55" "S"@"SEL$1" "T"@"SEL$1")
INDEX_FFS(@"SEL$0E991E55" "T"@"SEL$1" ("F_SCREEN_INSTANCE"."DAY_ID" "F_SCREEN_INSTANCE"."PARTIAL_ID"))
INDEX(@"SEL$0E991E55" "S"@"SEL$1" ("F_SCREEN_INSTANCE_BUF"."DAY_ID"
"F_SCREEN_INSTANCE_BUF"."PARTIAL_ID"))
OUTLINE(@"SEL$9EC647DD")
OUTLINE(@"SEL$2")
MERGE(@"SEL$9EC647DD")
OUTLINE_LEAF(@"SEL$0E991E55")
ALL_ROWS
DB_VERSION('12.1.0.1')
OPTIMIZER_FEATURES_ENABLE('12.1.0.1')
IGNORE_OPTIM_EMBEDDED_HINTS
END_OUTLINE_DATA
*/
Run Code Online (Sandbox Code Playgroud)
可能解决方案
--#3: Join the large table to the smaller result set. This uses the largest table twice,
--but the plan can use the ROWID for a very quick join.
explain plan for
merge into F_SCREEN_INSTANCE t
using
(
--#2: Now get the missing rows with an outer join. Since the _BUF table is
--small I assume it does not make a big difference exactly how it it joind
--to the 100K result set.
--The hints NO_MERGE and NO_PUSH_PRED are required to keep the INNER_JOIN
--inline view intact.
select /*+ no_merge(inner_join) no_push_pred(inner_join) */ inner_join.*
from f_screen_instance_buf s
left join
(
--#1: Get 100K rows efficiently with an inner join.
--Note that the ROWID is retrieved here.
select /*+ use_merge(t s) leading(t s) */ s.*, s.rowid s_rowid
from f_screen_instance_buf s
join f_screen_instance t
on (s.DAY_ID = t.DAY_ID and s.PARTIAL_ID = t.PARTIAL_ID)
) inner_join
on (s.DAY_ID = inner_join.DAY_ID and s.PARTIAL_ID = inner_join.PARTIAL_ID)
) s
on (s.s_rowid = t.rowid)
when matched then update set
t.ACTIVE_TIME_SUM = t.ACTIVE_TIME_SUM + s.ACTIVE_TIME_SUM,
t.IDLE_TIME_SUM = t.IDLE_TIME_SUM + s.IDLE_TIME_SUM
when not matched then insert values (
s.DAY_ID, s.PARTIAL_ID, s.ID, s.AGENT_USER_ID, s.COMPUTER_ID, s.RAW_APPLICATION_ID, s.APP_USER_ID, s.APPLICATION_ID, s.USER_ID, s.RAW_MODULE_ID, s.MODULE_ID, s.START_TIME, s.RAW_SCREEN_NAME, s.SCREEN_ID, s.SCREEN_TYPE, s.ACTIVE_TIME_SUM, s.IDLE_TIME_SUM);
Run Code Online (Sandbox Code Playgroud)
它并不漂亮,但至少它会在排序合并连接中首先生成一个包含大表的计划.
select * from table(dbms_xplan.display);
Plan hash value: 1086560566
-------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes |TempSpc| Cost (%CPU)| Time |
-------------------------------------------------------------------------------------------------------------
| 0 | MERGE STATEMENT | | 500G| 173T| | 5355K (43)| 00:03:30 |
| 1 | MERGE | F_SCREEN_INSTANCE | | | | | |
| 2 | VIEW | | | | | | |
|* 3 | HASH JOIN OUTER | | 500G| 179T| 29M| 5355K (43)| 00:03:30 |
|* 4 | HASH JOIN OUTER | | 100K| 28M| 3712K| 8663 (53)| 00:00:01 |
| 5 | INDEX FAST FULL SCAN| F_SCREEN_INSTANCE_BUF_PK | 100K| 2539K| | 9 (0)| 00:00:01 |
| 6 | VIEW | | 100K| 25M| | 6898 (66)| 00:00:01 |
| 7 | MERGE JOIN | | 100K| 12M| | 6898 (66)| 00:00:01 |
| 8 | INDEX FULL SCAN | F_SCREEN_INSTANCE_PK | 500M| 12G| | 4504 (100)| 00:00:01 |
|* 9 | SORT JOIN | | 100K| 9765K| 26M| 2393 (1)| 00:00:01 |
| 10 | TABLE ACCESS FULL| F_SCREEN_INSTANCE_BUF | 100K| 9765K| | 34 (6)| 00:00:01 |
| 11 | INDEX FAST FULL SCAN | F_SCREEN_INSTANCE_PK | 500M| 46G| | 2703 (100)| 00:00:01 |
-------------------------------------------------------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
3 - access("INNER_JOIN"."S_ROWID"=("T".ROWID(+)))
4 - access("S"."PARTIAL_ID"="INNER_JOIN"."PARTIAL_ID"(+) AND
"S"."DAY_ID"="INNER_JOIN"."DAY_ID"(+))
9 - access("S"."DAY_ID"="T"."DAY_ID" AND "S"."PARTIAL_ID"="T"."PARTIAL_ID")
filter("S"."PARTIAL_ID"="T"."PARTIAL_ID" AND "S"."DAY_ID"="T"."DAY_ID")
Run Code Online (Sandbox Code Playgroud)