客户数据库运行缓慢，告诉都在等待物理读事件，远程连接查看statspack报告
DB Name DB Id Instance Inst Num Release Cluster Host ------------ ----------- ------------ -------- ----------- ------- ------------ xxxx 259264653 xxxx 1 9.2.0.1.0 NO M Snap Id Snap TIME Sessions Curs/Sess Comment ------- ------------------ -------- --------- ------------------- BEGIN Snap: 18 04-12月-14 00:00:03 56 5.2 END Snap: 26 04-12月-14 08:00:06 69 9.1 Elapsed: 480.05 (mins) Cache Sizes (END) ~~~~~~~~~~~~~~~~~ Buffer Cache: 256M Std Block SIZE: 8K Shared Pool SIZE: 600M Log Buffer: 2,048K LOAD Profile ~~~~~~~~~~~~ Per SECOND Per TRANSACTION --------------- --------------- Redo SIZE: 33,077.28 157,241.29 Logical reads: 46,175.56 219,507.30 Block changes: 12,514.78 59,492.17 Physical reads: 1,582.20 7,521.40 Physical writes: 79.65 378.62 USER calls: 41.86 199.01 Parses: 6.11 29.04 Hard parses: 0.02 0.07 Sorts: 9.51 45.23 Logons: 0.19 0.92 Executes: 6,285.99 29,882.05 Transactions: 0.21 |
这里发现Block changes 高达12000以上，也就是每秒修改的12000块以上。
Top SQL
逻辑读
Buffer Gets Executions Gets per EXEC %Total TIME (s) TIME (s) Hash VALUE --------------- ------------ -------------- ------ -------- --------- ---------- 802,758,097 15,772 50,897.7 60.4 ######## ######### 827999392 UPDATE "xxx" "A1" SET "col1" = :B1 WHERE "A1" ."col2"=:B2 AND "A1"."col1" IS NULL OR "A1"."col1" <> :B1 |
物理读
Physical Reads Executions Reads per EXEC %Total TIME (s) TIME (s) Hash VALUE --------------- ------------ -------------- ------ -------- --------- ---------- 21,011,968 15,772 1,332.2 46.1 ######## ######### 827999392 UPDATE "xxx" "A1" SET "col1" = :B1 WHERE "A1" ."col2"=:B2 AND "A1"."col1" IS NULL OR "A1"."col1"<>:B1 |
表空间IO问题
Tablespace
------------------------------ Av Av Av Av Buffer Av Buf Reads Reads/s Rd(ms) Blks/Rd Writes Writes/s Waits Wt(ms) -------------- ------- ------ ------- ------------ -------- ---------- ------ DATA 11,721,822 407 1.6 3.2 196,380 7 281 8.7 UNDOTBS1 7,477,389 260 9.2 1.0 2,088,770 73 37,178 8.0 ------------------------------------------------------------- |
数据文件
Tablespace Filename ------------------------ ---------------------------------------------------- Av Av Av Av Buffer Av Buf Reads Reads/s Rd(ms) Blks/Rd Writes Writes/s Waits Wt(ms) -------------- ------- ------ ------- ------------ -------- ---------- ------ DATA D:\ORACLE\ORA92\xxx\xxxxxx.ORA 5,839,496 203 1.9 3.2 118,428 4 46 8.7 D:\ORACLE\ORA92\xxx\xxxxx.ORA 5,882,326 204 1.4 3.3 77,952 3 235 8.7 UNDOTBS1 D:\ORACLE\ORA92\xxxx\UNDOTBS01.DBF 7,477,389 260 9.2 1.0 2,088,770 73 37,178 8.0 ------------------------------------------------------------- |
从上面可以看出Tablespace data、undotbs1的IO使用比较严重，严重了上面的update语句导致。
查看下undo的使用
ROLLBACK Segment Storage FOR DB: FSJZ Instance: fsjz Snaps: 18 -26 ->Optimal SIZE should be larger than Avg Active RBS No Segment SIZE Avg Active Optimal SIZE Maximum SIZE ------ --------------- --------------- --------------- --------------- 0 385,024 6,553 385,024 1 35,774,464 1,003,908 129,097,728 2 10,608,640 996,987 75,620,352 3 13,819,904 772,807 18,997,248 4 5,365,760 1,355,300 14,802,944 5 8,511,488 5,457,853 3,885,096,960 6 4,183,941,120 2,771,161,530 4,252,098,560 7 5,365,760 619,248 10,608,640 8 14,802,944 7,010,756 21,094,400 9 3,268,608 2,908,152 22,142,976 10 4,317,184 567,499 4,317,184 ------------------------------------------------------------- |
上面的信息可以得出6号undo段已经使用了4G。
查看事物开始时间
会话的信息
SELECT --t.xid xid s.sid sid , t.inst_id , p.spid , s.STATUS , t.start_time , t.STATUS transaction_status , t.SPACE , t.recursive , t.noundo , t.ptx , ROUND((t.used_ublk * p.VALUE)/1024/1024) used_undo , t.log_io , t.phy_io , t.cr_get , s.username , s.osuser --, s.sql_id --, s.sql_child_number , s.program FROM gv$transaction t INNER JOIN gv$session s ON t.inst_id = s.inst_id AND t.ses_addr = s.saddr INNER JOIN gv$process p ON p.inst_id = s.inst_id AND p.addr = s.paddr ORDER BY t.start_time ; |
从上面来看这个事物的Undo使用达到了17G
查看Update 语句的执行计划
SQL> EXPLAIN plan FOR 2 UPDATE xxxxx."xxxxxxxxxxxxxxxxx" "A1" 3 SET "col2" = '1' 4 WHERE "A1" ."col2"=96 5 AND "A1"."col1" IS NULL 6 OR "A1"."col1"<>'0' 7 ; 已解释。 SQL> SELECT * FROM TABLE(dbms_xplan.display()); PLAN_TABLE_OUTPUT ------------------------------------------------------------------------------ --------------------------------------------------------------------------- | Id | Operation | Name | ROWS | Bytes | Cost | --------------------------------------------------------------------------- | 0 | UPDATE STATEMENT | | | | | | 1 | UPDATE | xxxxxxxxxxxxxxxxx | | | | |* 2 | TABLE ACCESS FULL | xxxxxxxxxxxxxxxxx | | | | --------------------------------------------------------------------------- Predicate Information (IDENTIFIED BY operation id): --------------------------------------------------- PLAN_TABLE_OUTPUT ------------------------------------------------------------------------------ 2 - FILTER("A1"."col2"=96 AND "A1"."col1" IS NULL OR "A 1"."col1"<>'0') Note: rule based optimization |
已选择16行。
这里得知SQL走的全表扫，仔细查看SQL 发现
《AND "A1"."col1" IS NULL OR "A1"."col1"<>'0'》 |
这两个条件应该增加括号，否则会oracle会找到col1不等于某个值的所有结果进行更新。
手工KILL掉这个进程后，数据库事物回滚异常缓慢
查看数据库参数
SQL> SHOW parameter cpu_count NAME TYPE VALUE ------------------------------------ ----------- ------------------------ cpu_count INTEGER 16 SQL> SHOW parameter fast_start NAME TYPE VALUE ------------------------------------ ----------- ------------------------ fast_start_parallel_rollback string LOW SQL> SHOW parameter parallel_max NAME TYPE VALUE ------------------------------------ ----------- ------------------------ parallel_max_servers INTEGER 5 |
fast_start_parallel_rollback参数是加快回滚速度，这个参数默认是low，即2倍cpu数的并发度，进行并发的回滚。这个参数可以设置成high，即为4倍cpu数的并发度进行回滚。
如果设置了此参数为high或low，也就是4倍或2倍的cpu数，也会受到另外一些参数的影响，如PARALLEL_MAX_SERVERS，这个才是真正最大的并发度设置。
PARALLEL_MAX_SERVERS这个参数的默认值为PARALLEL_THREADS_PER_CPU * CPU_COUNT * concurrent_parallel_users * 5。PARALLEL_THREADS_PER_CPU 和CPU_COUNT都是初始化参数。
在一般情况下，并发的回滚总是比串行的快，但是当并发的子进程之间存在资源冲突的情况。
在并发子进程之间需要的资源冲突时，往往此时smon的等待事件是长期处于
Wait for stopper event to be increased |
而子进程的等待事件是较多出现
Wait for a undo record。 |
此时，就是子进程冲突了。并发的回滚速度反而不如串行的回滚速度。
查询数据库得知现在最大并行进程设置5，远远不够并行恢复需要的进程。查看smon进程等待事件时，竟然没有记录。
通过关闭并行进程恢复，提高恢复速度
ALTER system SET fast_start_parallel_rollback=FALSE; |
查看恢复进度
SQL> DECLARE 2 l_start NUMBER; 3 l_end NUMBER; 4 BEGIN 5 SELECT ktuxesiz INTO l_start FROM x$ktuxe WHERE KTUXEUSN=6 AND KTUXESLT=46; 6 dbms_lock.sleep(60); 7 SELECT ktuxesiz INTO l_end FROM x$ktuxe WHERE KTUXEUSN=6 AND KTUXESLT=46; 8 dbms_output.put_line('time est Day:'|| round(l_end/(l_start -l_end)/60/24,2)); 9 END; 10 / TIME est DAY:.14 SQL> SELECT 0.14*24 FROM dual 2 ; 0.14*24 ---------- 3.36 |
大体评估事务回滚需要3.36个小时
