Travel大神-记一次故障诊断:事物回滚

本文是转载自Travel大神的文章:

记一次故障诊断:事物回滚

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 04-12-14 00:00:03      56       5.2
 
  END Snap:                 26 04-12-14 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
;

Travel大神-记一次故障诊断:事物回滚_第1张图片

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

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