码迷,mamicode.com
首页 > 其他好文 > 详细

【故障处理】队列等待之TX - allocate ITL entry引起的死锁处理

时间:2016-09-10 13:02:49      阅读:545      评论:0      收藏:0      [点我收藏+]

标签:

【故障处理】队列等待之TX - allocate ITL entry引起的死锁处理

1  BLOG文档结构图

技术分享 

 

 

2  前言部分

2.1  导读和注意事项

各位技术爱好者,看完本文后,你可以掌握如下的技能,也可以学到一些其它你所不知道的知识,~O(∩_∩)O~

① enq: TX - allocate ITL entry等待事件的解决

② 一般等待事件的解决办法

③ 队列等待的基本知识

ITL死锁解决

ITL死锁模拟

Merge语句的非关联形式的查询优化

  Tips:

① 本文在itpubhttp://blog.itpub.net/26736162)、博客园(http://www.cnblogs.com/lhrbest)和微信公众号(xiaomaimiaolhr)有同步更新

② 文章中用到的所有代码,相关软件,相关资料请前往小麦苗的云盘下载(http://blog.itpub.net/26736162/viewspace-1624453/

网页文章代码格式有错乱,推荐使用360浏览器,也可以下载pdf格式的文档来查看,pdf文档下载地址:http://blog.itpub.net/26736162/viewspace-1624453/,另外itpub格式显示有问题,也可以去博客园地址阅读

本篇BLOG中命令的输出部分需要特别关注的地方我都用灰色背景和粉红色字体来表示,比如下边的例子中,thread 1的最大归档日志号为33thread 2的最大归档日志号为43是需要特别关注的地方;而命令一般使用黄色背景和红色字体注;对代码或代码输出部分的注释一般采用蓝色字体表示

  List of Archived Logs in backup set 11

  Thrd Seq     Low SCN    Low Time            Next SCN   Next Time

  ---- ------- ---------- ------------------- ---------- ---------

  1    32      1621589    2015-05-29 11:09:52 1625242    2015-05-29 11:15:48

  1    33      1625242    2015-05-29 11:15:48 1625293    2015-05-29 11:15:58

  2    42      1613951    2015-05-29 10:41:18 1625245    2015-05-29 11:15:49

  2    43      1625245    2015-05-29 11:15:49 1625253    2015-05-29 11:15:53

 

[ZHLHRDB1:root]:/>lsvg -o

T_XLHRD_APP1_vg

rootvg

[ZHLHRDB1:root]:/>

00:27:22 SQL> alter tablespace idxtbs read write;

 

====》2097152*512/1024/1024/1024=1G

 

本文如有错误或不完善的地方请大家多多指正,ITPUB留言或QQ皆可,您的批评指正是我写作的最大动力。

 

 

3  故障分析及解决过程

 

3.1  故障环境介绍

 

项目

source db

db 类型

RAC

db version

11.2.0.3.0

db 存储

ASM

OS版本及kernel版本

AIX 64位 7.1.0.0

 

3.2  故障发生现象及故障分析解决

早上刚来上班,同事就发了一个SQL过来,说是有锁,然后我就查了查系统里的锁,结果一个锁都没得。好吧,还是得干点事的,先看看SQL语句:

MERGE INTO TLHR.TLHRBOKBAL S

USING (SELECT A.BOOKACCOUNT AS BOOKACCOUNT,

              (A.CURRBALANCE + NVL(B.BAL, 0.00)) AS BANKAMT

         FROM TLHR.TLHRBOKBAL_TMP A,

              (SELECT T1.BOOKACCOUNT AS BOOKACCOUNT,

                      SUM(DECODE(T1.DCFLAG, ‘D‘, -T1.AMT, ‘C‘, T1.AMT, 0)) AS BAL

                 FROM TLHR.TLHRBOKBALJN T1

                WHERE T1.BOOKACCOUNT LIKE ‘13500000%‘

                  AND T1.TRANDATE = ‘20150901‘

                  AND (T1.REASON IN (‘1‘, ‘2‘) OR

                      (T1.REASON = ‘0‘ AND T1.ONLINEFLAG = ‘1‘))

                GROUP BY T1.BOOKACCOUNT) B

        WHERE A.BOOKACCOUNT = B.BOOKACCOUNT(+)

          AND A.BOOKACCOUNT LIKE ‘13500000%‘) T

ON (S.BOOKACCOUNT = T.BOOKACCOUNT)

WHEN MATCHED THEN

  UPDATE

     SET S.LASTBALANCE = T.BANKAMT,

         S.CURRBALANCE = T.BANKAMT,

         S.DEBITAMT    = 0.00,

         S.CREDITAMT   = 0.00;

看起来是一个MERGE语句,按照小麦苗以前的经验,这一类的SQL最好是修改为MERGE的非关联形式比较好,我们先看看执行计划有没有问题:

先找到SQL_ID53qv858pwwwwb

SELECT a.ELAPSED_TIME,a.EXECUTIONS,a.* FROM v$sql a WHERE a.SQL_TEXT LIKE ‘%MERGE INTO TLHRBOKBAL S%‘ AND A.SQL_TEXT LIKE ‘%13500000%‘  ;

查询历史执行计划:

SELECT * FROM TABLE(DBMS_XPLAN.DISPLAY_AWR(SQL_ID => ‘53qv858pwwwwb‘ )) ;

Plan hash value: 2695089823

 

------------------------------------------------------------------------------------------------------------

| Id  | Operation                        | Name            | Rows  | Bytes |TempSpc| Cost (%CPU)| Time     |

------------------------------------------------------------------------------------------------------------

|   0 | MERGE STATEMENT                  |                 |       |       |       |   155K(100)|          |

|   1 |  MERGE                           | TLHRBOKBAL      |       |       |       |            |          |

|   2 |   VIEW                           |                 |       |       |       |            |          |

|   3 |    HASH JOIN RIGHT OUTER         |                 |   153K|    15M|       |   155K  (2)| 00:31:04 |

|   4 |     VIEW                         |                 |     1 |    31 |       |     6   (0)| 00:00:01 |

|   5 |      SORT GROUP BY               |                 |     1 |    41 |       |     6   (0)| 00:00:01 |

|   6 |       TABLE ACCESS BY INDEX ROWID| TLHRBOKBALJN    |     1 |    41 |       |     6   (0)| 00:00:01 |

|   7 |        INDEX RANGE SCAN          | PK_TLHRBOKBALJN |     2 |       |       |     4   (0)| 00:00:01 |

|   8 |     HASH JOIN                    |                 |   153K|    10M|  5848K|   155K  (2)| 00:31:04 |

|   9 |      TABLE ACCESS FULL           | TLHRBOKBAL_TMP  |   153K|  4048K|       | 85415   (2)| 00:17:05 |

|  10 |      TABLE ACCESS FULL           | TLHRBOKBAL      |   305K|    13M|       | 68755   (3)| 00:13:46 |

------------------------------------------------------------------------------------------------------------

可以看到,该执行计划的顺序为【7-->6-->5-->4-->9-->10-->8-->3-->2-->1-->0】,而耗费性能的地方在91083个步骤上,走的是全表扫描,我们先看看2个大表的数据量:

SELECT COUNT(*)  FROM TLHR.TLHRBOKBAL_TMP A WHERE  A.BOOKACCOUNT LIKE ‘13500000%‘;   --306043/38998765

SELECT COUNT(*) FROM TLHR.TLHRBOKBAL A WHERE  A.BOOKACCOUNT LIKE ‘13500000%‘;    --306043/38826275

3000万的数据里边取出30万的数据,还是比较少的,所以应该去走索引的,看了一下统计信息,也是最新收集的,好吧,算了,先修改一下SQL让其走索引扫描看看,

MERGE  INTO TLHR.TLHRBOKBAL S

USING (SELECT S.ROWID ROWIDS,

              A.BOOKACCOUNT AS BOOKACCOUNT,

              (A.CURRBALANCE + NVL(B.BAL, 0.00)) AS BANKAMT

         FROM (SELECT /*+index(NB,PK_TLHRBOKBAL_TMP)*/NB.CURRBALANCE,NB.BOOKACCOUNT

                 FROM TLHR.TLHRBOKBAL_TMP NB

                WHERE NB.BOOKACCOUNT LIKE ‘13500000%‘) A,

              TLHR.TLHRBOKBAL S,

              (SELECT T1.BOOKACCOUNT AS BOOKACCOUNT,

                      SUM(DECODE(T1.DCFLAG, ‘D‘, -T1.AMT, ‘C‘, T1.AMT, 0)) AS BAL

                 FROM TLHR.TLHRBOKBALJN T1

                WHERE T1.BOOKACCOUNT LIKE ‘13500000%‘

                  AND T1.TRANDATE = ‘20150901‘

                  AND (T1.REASON IN (‘1‘, ‘2‘) OR

                      (T1.REASON = ‘0‘ AND T1.ONLINEFLAG = ‘1‘))

                GROUP BY T1.BOOKACCOUNT) B

        WHERE A.BOOKACCOUNT = B.BOOKACCOUNT(+)

          AND S.BOOKACCOUNT = A.BOOKACCOUNT

          AND S.BOOKACCOUNT LIKE ‘13500000%‘) T

ON (T.ROWIDS = S.ROWID)

WHEN MATCHED THEN

  UPDATE

     SET S.LASTBALANCE = T.BANKAMT,

         S.CURRBALANCE = T.BANKAMT,

         S.DEBITAMT    = 0.00,

         S.CREDITAMT   = 0.00

Plan Hash Value  : 273017430

----------------------------------------------------------------------------------------------------------

| Id   | Operation                           | Name              | Rows   | Bytes    | Cost   | Time     |

----------------------------------------------------------------------------------------------------------

|    0 | MERGE STATEMENT                     |                   | 152885 |  4280780 | 283362 | 00:56:41 |

|    1 |   MERGE                             | TLHRBOKBAL        |        |          |        |          |

|    2 |    VIEW                             |                   |        |          |        |          |

|    3 |     NESTED LOOPS                    |                   | 152885 | 20945245 | 283362 | 00:56:41 |

|  * 4 |      HASH JOIN RIGHT OUTER          |                   | 152885 | 14065420 | 130342 | 00:26:05 |

|    5 |       VIEW                          |                   |    124 |     3844 |  15668 | 00:03:09 |

|    6 |        SORT GROUP BY                |                   |    124 |     5084 |  15668 | 00:03:09 |

|  * 7 |         TABLE ACCESS BY INDEX ROWID | TLHRBOKBALJN      |    124 |     5084 |  15668 | 00:03:09 |

|  * 8 |          INDEX RANGE SCAN           | PK_TLHRBOKBALJN   |    165 |          |  15501 | 00:03:07 |

|  * 9 |       HASH JOIN                     |                   | 152885 |  9325985 | 114671 | 00:22:57 |

|   10 |        TABLE ACCESS BY INDEX ROWID  | TLHRBOKBAL_TMP    | 153563 |  4146201 | 112930 | 00:22:36 |

| * 11 |         INDEX RANGE SCAN            | PK_TLHRBOKBAL_TMP | 153563 |          |   1159 | 00:00:14 |

| * 12 |        INDEX RANGE SCAN             | PK_TLHRBOKBAL     | 152884 |  5198056 |   1117 | 00:00:14 |

|   13 |      TABLE ACCESS BY USER ROWID     | TLHRBOKBAL        |      1 |       45 |      1 | 00:00:01 |

----------------------------------------------------------------------------------------------------------

 

Predicate Information (identified by operation id):

------------------------------------------

* 4 - access("NB"."BOOKACCOUNT"="B"."BOOKACCOUNT"(+))

* 7 - filter("T1"."REASON"=‘0‘ AND "T1"."ONLINEFLAG"=‘1‘ OR "T1"."REASON"=‘1‘ OR "T1"."REASON"=‘2‘)

* 8 - access("T1"."BOOKACCOUNT" LIKE ‘13500000%‘ AND "T1"."TRANDATE"=‘20150901‘)

* 8 - filter("T1"."BOOKACCOUNT" LIKE ‘13500000%‘ AND "T1"."TRANDATE"=‘20150901‘)

* 9 - access("S"."BOOKACCOUNT"="NB"."BOOKACCOUNT")

* 11 - access("NB"."BOOKACCOUNT" LIKE ‘13500000%‘)

* 11 - filter("NB"."BOOKACCOUNT" LIKE ‘13500000%‘)

* 12 - access("S"."BOOKACCOUNT" LIKE ‘13500000%‘)

* 12 - filter("S"."BOOKACCOUNT" LIKE ‘13500000%‘)

 

执行计划中,基本都走了索引了,跑了一下,大约1分种多,但是里边有个HINTS,分析了一下表TLHRBOKBAL_TMP上的索引情况,发现是个主键索引,且有2个列(BOOKACCOUNT,CURRENCY),但是不包含列CURRBALANCE,可能是Oracle觉得回表读的耗费比较大吧,那这里可以使用虚拟索引测试一下索引的性能:

CREATE INDEX IX_VI01_ID ON TLHR.TLHRBOKBAL_TMP(CURRBALANCE, CURRENCY,BOOKACCOUNT) NOSEGMENT;

ALTER SESSION SET "_USE_NOSEGMENT_INDEXES"=TRUE;

EXPLAIN PLAN FOR  MERGE INTO TLHR.TLHRBOKBAL S

USING (SELECT S.ROWID ROWIDS,

              A.BOOKACCOUNT AS BOOKACCOUNT,

              (A.CURRBALANCE + NVL(B.BAL, 0.00)) AS BANKAMT

         FROM (SELECT NB.CURRBALANCE,NB.BOOKACCOUNT

                 FROM TLHR.TLHRBOKBAL_TMP NB

                WHERE NB.BOOKACCOUNT LIKE ‘13500000%‘) A,

              TLHR.TLHRBOKBAL S,

              (SELECT T1.BOOKACCOUNT AS BOOKACCOUNT,

                      SUM(DECODE(T1.DCFLAG, ‘D‘, -T1.AMT, ‘C‘, T1.AMT, 0)) AS BAL

                 FROM TLHR.TLHRBOKBALJN T1

                WHERE T1.BOOKACCOUNT LIKE ‘13500000%‘

                  AND T1.TRANDATE = ‘20150901‘

                  AND (T1.REASON IN (‘1‘, ‘2‘) OR

                      (T1.REASON = ‘0‘ AND T1.ONLINEFLAG = ‘1‘))

                GROUP BY T1.BOOKACCOUNT) B

        WHERE A.BOOKACCOUNT = B.BOOKACCOUNT(+)

          AND S.BOOKACCOUNT = A.BOOKACCOUNT

          AND S.BOOKACCOUNT LIKE ‘13500000%‘) T

ON (T.ROWIDS = S.ROWID)

WHEN MATCHED THEN

  UPDATE

     SET S.LASTBALANCE = T.BANKAMT,

         S.CURRBALANCE = T.BANKAMT,

         S.DEBITAMT    = 0.00,

         S.CREDITAMT   = 0.00;

SELECT * FROM TABLE(DBMS_XPLAN.display);

Plan hash value: 983878991

 

-------------------------------------------------------------------------------------------------------------

| Id  | Operation                         | Name            | Rows  | Bytes |TempSpc| Cost (%CPU)| Time     |

-------------------------------------------------------------------------------------------------------------

|   0 | MERGE STATEMENT                   |                 |   152K|  4180K|       |   170K  (1)| 00:34:06 |

|   1 |  MERGE                            | TLHRBOKBAL      |       |       |       |            |          |

|   2 |   VIEW                            |                 |       |       |       |            |          |

|   3 |    NESTED LOOPS                   |                 |   152K|    19M|       |   170K  (1)| 00:34:06 |

|*  4 |     HASH JOIN RIGHT OUTER         |                 |   152K|    13M|       | 17421   (1)| 00:03:30 |

|   5 |      VIEW                         |                 |   124 |  3844 |       | 15668   (1)| 00:03:09 |

|   6 |       SORT GROUP BY               |                 |   124 |  5084 |       | 15668   (1)| 00:03:09 |

|*  7 |        TABLE ACCESS BY INDEX ROWID| TLHRBOKBALJN    |   124 |  5084 |       | 15668   (1)| 00:03:09 |

|*  8 |         INDEX RANGE SCAN          | PK_TLHRBOKBALJN |   165 |       |       | 15501   (1)| 00:03:07 |

|*  9 |      HASH JOIN                    |                 |   152K|  9107K|  5856K|  1750   (1)| 00:00:22 |

|* 10 |       INDEX FAST FULL SCAN        | IX_VI01_ID      |   153K|  4049K|       |     9   (0)| 00:00:01 |

|* 11 |       INDEX RANGE SCAN            | PK_TLHRBOKBAL   |   152K|  5076K|       |  1117   (1)| 00:00:14 |

|  12 |     TABLE ACCESS BY USER ROWID    | TLHRBOKBAL      |     1 |    45 |       |     1   (0)| 00:00:01 |

-------------------------------------------------------------------------------------------------------------

 

Predicate Information (identified by operation id):

---------------------------------------------------

 

   4 - access("NB"."BOOKACCOUNT"="B"."BOOKACCOUNT"(+))

   7 - filter("T1"."REASON"=‘0‘ AND "T1"."ONLINEFLAG"=‘1‘ OR "T1"."REASON"=‘1‘ OR "T1"."REASON"=‘2‘)

   8 - access("T1"."BOOKACCOUNT" LIKE ‘13500000%‘ AND "T1"."TRANDATE"=‘20150901‘)

       filter("T1"."BOOKACCOUNT" LIKE ‘13500000%‘ AND "T1"."TRANDATE"=‘20150901‘)

   9 - access("S"."BOOKACCOUNT"="NB"."BOOKACCOUNT")

  10 - filter("NB"."BOOKACCOUNT" LIKE ‘13500000%‘)

  11 - access("S"."BOOKACCOUNT" LIKE ‘13500000%‘)

       filter("S"."BOOKACCOUNT" LIKE ‘13500000%‘)

 

说明创建3个列的索引是可以的。我们先将该虚拟索引删除DROP INDEX IX_VI01_ID;

3.2.1  ITL死锁问题解决

另外一个问题,是开发说上边的SQL语句产生了死锁,起初我还半信半疑,先去告警日志中用命令(more alert* | grep Deadlock)搜了一下:

技术分享 

结果发现很多的死锁,拿到相关的文件,看到如下一段:

user session for deadlock lock 0x7000008094d14e0

  sid: 332 ser: 47221 audsid: 991000 user: 84/TLHR

    flags: (0x41) USR/- flags_idl: (0x1) BSY/-/-/-/-/-

    flags2: (0x40009) -/-/INC

  pid: 101 O/S info: user: grid, term: UNKNOWN, ospid: 6489034

    image: oracle@ZFTLHRDB1

  client details:

    O/S info: user: TLHR, term: , ospid: 34406578

    machine: ZFTLHRAP1 program: bat_CheckBookBal@ZFTLHRAP1 (TNS V1-V3)

    application name: bat_CheckBookBal@ZFTLHRAP1 (TNS V1-V3), hash value=446537749

  current SQL:

   MERGE INTO TLHRBOKBAL S USING  (SELECT A.BOOKACCOUNT AS BOOKACCOUNT,  (A.CURRBALANCE + nvl(B.BAL, 0.00)) AS BANKAMT  FROM  TLHRBOKBAL_TMP A, (SELECT T1.BOOKACCOUNT AS BOOKACCOUNT,  SUM(DECODE(T1.DCFLAG, ‘D‘, -T1.AMT, ‘C‘, T1.AMT, 0)) AS BAL  FROM TLHRBOKBALJN T1  WHERE T1.BOOKACCOUNT LIKE ‘13450000‘||‘%‘  AND T1.TRANDATE=‘20160901‘  AND (T1.REASON = ‘2‘ OR T1.REASON = ‘1‘ OR  (T1.REASON = ‘0‘ AND T1.ONLINEFLAG = ‘1‘))  GROUP BY T1.BOOKACCOUNT) B  WHERE A.BOOKACCOUNT = B.BOOKACCOUNT(+)  AND A.BOOKACCOUNT LIKE ‘13450000‘||‘%‘) T  ON (S.BOOKACCOUNT = T.BOOKACCOUNT)  WHEN MATCHED THEN UPDATE  SET S.LASTBALANCE=T.BANKAMT,S.CURRBALANCE=T.BANKAMT,S.DEBITAMT=0.00,S.CREDITAMT=0.00

DUMP LOCAL BLOCKER: initiate state dump for DEADLOCK

  possible owner[101.6489034] on resource TX-00EE0009-00005EA6

 

*** 2016-09-01 18:30:38.014

Submitting asynchronized dump request [28]. summary=[ges process stack dump (kjdglblkrdm1)].

Global blockers dump end:-----------------------------------

Global Wait-For-Graph(WFG) at ddTS[0.2fe0] :

BLOCKED 0x7000008e9c8bc28 3 wq 2 cvtops x1 TX 0x159001e.0x2379(ext 0x5,0x0)[1006-0065-0019365C] inst 1

BLOCKER 0x700000809ab4b28 3 wq 1 cvtops x28 TX 0x159001e.0x2379(ext 0x5,0x0)[2005-005E-00185E15] inst 2

BLOCKED 0x700000891b48708 3 wq 2 cvtops x1 TX 0x1c2001d.0x4b82(ext 0x2,0x0)[2005-005E-00185E15] inst 2

BLOCKER 0x7000008e9c8b148 3 wq 1 cvtops x28 TX 0x1c2001d.0x4b82(ext 0x2,0x0)[1004-004D-0000C03E] inst 1

BLOCKED 0x70000089a636970 3 wq 2 cvtops x1 TX 0x1c0000b.0x18f6(ext 0x2,0x0)[1004-004D-0000C03E] inst 1

BLOCKER 0x7000008e9c8b4e8 3 wq 1 cvtops x28 TX 0x1c0000b.0x18f6(ext 0x2,0x0)[1005-0058-000DD3D9] inst 1

BLOCKED 0x700000891d5fc50 3 wq 2 cvtops x1 TX 0xee0009.0x5ea6(ext 0x2,0x0)[1005-0058-000DD3D9] inst 1

BLOCKER 0x7000008094d14e0 3 wq 1 cvtops x28 TX 0xee0009.0x5ea6(ext 0x2,0x0)[1006-0065-0019365C] inst 1

* Cancel deadlock victim lockp 0x7000008e9c8bc28

 

*** 2016-09-01 18:30:43.001

kjddt2vb: valblk  [0.2fe1] > local ts [0.2fe0]

 

*** 2016-09-01 18:30:47.000

kjddt2vb: valblk  [0.2fe5] > local ts [0.2fe4]

 

*** 2016-09-01 18:40:38.062

kjddt2vb: valblk  [0.2ff1] > local ts [0.2ff0]

 

*** 2016-09-01 18:42:01.084

kjddt2vb: valblk  [0.2ff4] > local ts [0.2ff3]

2016-09-01 22:33:52.213848 : Setting 3-way CR grants to 0 global-lru off? 0

 

*** 2016-09-01 22:34:23.163

2016-09-01 22:34:23.163681 : Setting 3-way CR grants to 1 global-lru off? 0

2016-09-01 22:50:00.603305 : Setting 3-way CR grants to 0 global-lru off? 0

 

*** 2016-09-01 22:51:33.104

2016-09-01 22:51:33.104615 : Setting 3-way CR grants to 1 global-lru off? 0

2016-09-02 05:30:18.751891 : Setting 3-way CR grants to 0 global-lru off? 0

2016-09-02 05:49:01.360730 : Setting 3-way CR grants to 1 global-lru off? 0

2016-09-02 10:28:55.429293 : Setting 3-way CR grants to 0 global-lru off? 0

果然,产生死锁的SQL还是上边分析优化的SQL,其中会话信息为:(33247221),我们去DBA_HIST_ACTIVE_SESS_HISTORY视图里查询:

SELECT D.SQL_ID, D.CURRENT_OBJ#, D.EVENT, COUNT(1)

  FROM DBA_HIST_ACTIVE_SESS_HISTORY D

 WHERE D.SAMPLE_TIME BETWEEN

       TO_DATE(‘2016-09-01 18:25:00‘, ‘YYYY-MM-DD HH24:MI:SS‘) AND

       TO_DATE(‘2016-09-01 18:45:00‘, ‘YYYY-MM-DD HH24:MI:SS‘)

   AND D.BLOCKING_SESSION_STATUS = ‘VALID‘

   AND D.SESSION_ID = 332

   AND D.SESSION_SERIAL# = 47221

 GROUP BY D.SQL_ID, D.CURRENT_OBJ#, D.EVENT;

技术分享 

可以看到该会话的等待事件是enq: TX - allocate ITL entry。可以猜测是由于ITL事务槽引起的问题。

SELECT DISTINCT D.BLOCKING_SESSION, D.BLOCKING_SESSION_SERIAL#, D.SQL_ID

  FROM DBA_HIST_ACTIVE_SESS_HISTORY D

 WHERE D.SAMPLE_TIME BETWEEN

       TO_DATE(‘2016-09-01 18:25:00‘, ‘YYYY-MM-DD HH24:MI:SS‘) AND

       TO_DATE(‘2016-09-01 18:45:00‘, ‘YYYY-MM-DD HH24:MI:SS‘)

   AND D.EVENT = ‘enq: TX - allocate ITL entry‘

   AND D.BLOCKING_SESSION_STATUS = ‘VALID‘

   AND D.SESSION_ID = 332

   AND D.SESSION_SERIAL# = 47221;

技术分享 

可以看出会话(332,47221)共阻塞了3个会话,由于有死锁,那么我们看看上边查询出来的3个会话阻塞了哪些会话:

SELECT DISTINCT D.INSTANCE_NUMBER,

                D.SESSION_ID,

                D.SESSION_SERIAL#,

                D.BLOCKING_INST_ID,

                D.BLOCKING_SESSION,

                D.BLOCKING_SESSION_SERIAL#,

                D.SQL_ID

  FROM DBA_HIST_ACTIVE_SESS_HISTORY D

 WHERE D.SAMPLE_TIME BETWEEN

       TO_DATE(‘2016-09-01 18:25:00‘, ‘YYYY-MM-DD HH24:MI:SS‘) AND

       TO_DATE(‘2016-09-01 18:45:00‘, ‘YYYY-MM-DD HH24:MI:SS‘)

   AND D.EVENT = ‘enq: TX - allocate ITL entry‘

   AND D.BLOCKING_SESSION_STATUS = ‘VALID‘

   AND ((D.SESSION_ID = 332 AND D.SESSION_SERIAL# = 47221) OR

       (D.SESSION_ID = 2602 AND D.SESSION_SERIAL# = 4343) OR

       (D.SESSION_ID = 2995 AND D.SESSION_SERIAL# = 46891) OR

       (D.SESSION_ID = 1894 AND D.SESSION_SERIAL# = 30761));

技术分享 

可以看到,1894和2602相互阻塞(绿色表示),3322602相互阻塞(红色表示),2995332相互阻塞(粉色表示),这么多的相互阻塞就产生了死锁,这里由于SQL_ID不同,而且产生的等待事件是enq: TX - allocate ITL entry,所以推测出生成的是ITL死锁。

解决这类问题就是增大ini_trans和PCT_FREE值。

SELECT * FROM DBA_TABLES D WHERE D.TABLE_NAME = TLHRBOKBAL‘;

技术分享 

可以看到,ini_trans和PCT_FREE值都是默认的,太小了,根据MOSTroubleshooting waits for ‘enq: TX - allocate ITL entry‘ (Doc ID 1472175.1)  地址:http://blog.itpub.net/26736162/viewspace-2124531/)我们需要修改该参数,SQL如下:

ALTER TABLE TLHR.TLHRBOKBAL PCTFREE 20  INITRANS 16 ;

ALTER TABLE TLHR.TLHRBOKBAL MOVE NOLOGGING PARALLEL 12;

ALTER TABLE TLHR.TLHRBOKBAL LOGGING NOPARALLEL;

ALTER INDEX TLHR.PK_TLHRBOKBAL  REBUILD PCTFREE 20 INITRANS 16 NOLOGGING PARALLEL 12;

ALTER INDEX TLHR.PK_TLHRBOKBAL LOGGING NOPARALLEL;

 

由于表里有3000W的数据量,开了并行,本来我预估的是5分钟,结果move表的时候10秒都不到还是比较快的。

调整之后的值:

SELECT * FROM DBA_TABLES D WHERE D.TABLE_NAME = TLHRBOKBAL‘;

技术分享 

 

SELECT * FROM dba_indexes d WHERE d.index_name=‘PK_TLHRBOKBAL‘;

技术分享 

修改已经生效,接下来就看开发那边是否还报死锁的错误,这个等待需要明天看了。

终于等到第2天了,看来没有报错了:

技术分享 

4  这里我们模拟一个ITL死锁

有人的地方就有江湖,有资源阻塞的地方就可能有死锁。所谓死锁: 是指两个或两个以上的进程在执行过程中,因争夺资源而造成的一种互相等待的现象,若无外力作用,它们都将无法推进下去。此时称系统处于死锁状态或系统产生了死锁,这些永远在互相等待的进程称为死锁进程。其最常见的锁的类型为:行级锁(row-level locks)和块级锁(block-level locks,这里的行级锁其实就是指的ITL死锁。有关死锁的问题,有许多需要介绍的,这篇blog主要是故障处理,所以这里我们模拟一个ITL死锁产生的过程即可,后边我会系统的发一次有关死锁的内容,还有ITL的内容,希望大家持续关注小麦苗的微信公众号(xiaomaimiaolhr)

实验部分:

实验的设计过程来源于网络!

我们首先创建一张表T_ITL_LHR,这里指定PCTFREE0INITRANS1,就是为了观察到ITL的真实等待情况,然后我们给这些块内插入数据,把块填满,让它不能有空间分配。

SYS@lhrdb21> SELECT * FROM V$VERSION;

BANNER

--------------------------------------------------------------------------------

Oracle Database 11g Enterprise Edition Release 11.2.0.4.0 - 64bit Production

PL/SQL Release 11.2.0.4.0 - Production

CORE    11.2.0.4.0      Production

TNS for IBM/AIX RISC System/6000: Version 11.2.0.4.0 - Production

NLSRTL Version 11.2.0.4.0 - Production

SYS@lhrdb21> SHOW PARAMETER CLUSTER

 

NAME                                 TYPE        VALUE

------------------------------------ ----------- ------------------------------

cluster_database                     boolean     TRUE

cluster_database_instances           integer     2

cluster_interconnects                string

 

SYS@lhrdb21> CREATE TABLE T_ITL_LHR(A INT) PCTFREE 0 INITRANS 1;

Table created.

SYS@lhrdb21> BEGIN

  2    FOR I IN 1 .. 2000 LOOP

  3      INSERT INTO T_ITL_LHR VALUES (I);

  4    END LOOP;

  5  END;

  6  /

 

PL/SQL procedure successfully completed.

 

SYS@lhrdb21> COMMIT;

 

Commit complete.

 

 

 

我们检查数据填充的情况:

SYS@lhrdb21> SELECT F, B, COUNT(*)

  2    FROM (SELECT DBMS_ROWID.ROWID_RELATIVE_FNO(ROWID) F,

  3                 DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID) B

  4            FROM T_ITL_LHR)

  5   GROUP BY F, B

  6   ORDER BY F,B;

 

         F          B      COUNT(*)

---------- ---------- ----------

         1      94953        734

         1      94954        734

         1      94955        532

 

可以发现,这2000条数据分布在3个块内部,其中有2个块(9495394954)填满了,一个块(94955)是半满的。因为有2ITL槽位,我们需要拿2个满的数据块,4个进程来模拟ITL死锁:

实验步骤

会话

SID

要更新的块号

要更新的行号

是否有阻塞

步骤一

1

19

94953

94953

1

N

2

79

2

N

3

78

94954

94954

1

N

4

139

2

N

 

会话1

SYS@lhrdb21> SELECT USERENV(‘SID‘) FROM DUAL;

USERENV(‘SID‘)

--------------

            19

 

SYS@lhrdb21> UPDATE T_ITL_LHR SET A=A

  2   WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94953 

  3  AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=1;

 

1 row updated.

 

会话2

SYS@lhrdb21> SELECT USERENV(‘SID‘) FROM DUAL;

USERENV(‘SID‘)

--------------

            79

 

SYS@lhrdb21> UPDATE T_ITL_LHR SET A=A

  2   WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94953

  3  AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=2;

 

1 row updated.

 

会话3

SYS@lhrdb21> SELECT USERENV(‘SID‘) FROM DUAL;

USERENV(‘SID‘)

--------------

            78

 

SYS@lhrdb21> UPDATE T_ITL_LHR SET A=A

  2   WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94954

  3  AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=1;

 

1 row updated.

会话4

SYS@lhrdb21> SELECT USERENV(‘SID‘) FROM DUAL;

 

USERENV(‘SID‘)

--------------

           139

 

SYS@lhrdb21> UPDATE T_ITL_LHR SET A=A

  2   WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94954

  3  AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=2;

 

1 row updated.

这个时候系统不存在阻塞,

SELECT NVL(A.SQL_ID, A.PREV_SQL_ID) SQL_ID,

       A.BLOCKING_SESSION,

       A.SID,

       A.SERIAL#,

       A.LOGON_TIME,

       A.EVENT

  FROM GV$SESSION A

 WHERE A.SID IN (19, 79,78,139)

 ORDER BY A.LOGON_TIME;

技术分享 

 

以上4个进程把2个不同块的4ITL槽位给消耗光了,现在的情况,就是让他们互相锁住,达成死锁条件,回到会话1,更新块94954,注意,以上4个操作,包括以下的操作,更新的根本不是同一行数据,主要是为了防止出现的是行锁等待。

实验步骤

会话

SID

要更新的块号

要更新的行号

是否有阻塞

步骤一

1

19

94953

94953

1

N

2

79

2

N

3

78

94954

94954

1

N

4

139

2

N

步骤二

1

19

94954

3

Y

3

78

94953

3

Y

 

会话1

UPDATE T_ITL_LHR SET A=A

WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94954

AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=3;

技术分享 

会话1出现了等待。

 

会话3

UPDATE T_ITL_LHR SET A=A

WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94953

AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=3;

技术分享 

会话3发现出现了等待。

 

我们查询阻塞的具体情况:

SELECT NVL(A.SQL_ID, A.PREV_SQL_ID) SQL_ID,

       A.BLOCKING_SESSION,

       A.SID,

       A.SERIAL#,

       A.LOGON_TIME,

       A.EVENT

  FROM GV$SESSION A

 WHERE A.SID IN (19, 79,78,139)

 ORDER BY A.LOGON_TIME;

技术分享 

可以看到,会话1被会话4阻塞了,会话3被会话2阻塞了。

注意,如果是9i,在这里就报死锁了,但是在10g里面,这个时候,死锁是不会发生的,因为这里的会话1还可以等待会话4释放资源,会话3还可以等待会话2释放资源,只要会话2与会话4释放了资源,整个环境又活了,那么我们需要把这两个进程也塞住。

实验步骤

会话

SID

要更新的块号

要更新的行号

是否有阻塞

步骤一

1

19

94953

94953

1

N

2

79

2

N

3

78

94954

94954

1

N

4

139

2

N

步骤二

1

19

94954

3

Y

3

78

94953

3

Y

步骤三

2

79

94954

4

Y

4

139

94953

4

Y

 

会话2,注意,我们也不是更新的同一行数据

UPDATE T_ITL_LHR SET A=A

WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94954

AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=4;

技术分享 

会话2出现了等待,具体阻塞情况:

技术分享 

我做了几次实验,会话2执行完SQL后,会话3到这里就报出了死锁,但有的时候并没有产生死锁,应该跟系统的阻塞顺序有关,若没有产生死锁,我们可以继续会话4的操作。

SYS@lhrdb21> UPDATE T_ITL_LHR SET A=A

  2   WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94953

  3  AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=3;

 

 

UPDATE T_ITL_LHR SET A=A

       *

ERROR at line 1:

ORA-00060: deadlock detected while waiting for resource

 

会话4,注意,我们也不是更新的同一行数据

UPDATE T_ITL_LHR SET A=A

WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94953 

AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=4;

技术分享 

会话4发现出现了等待。

技术分享 

虽然,以上的每个更新语句,更新的都不是同一个数据行,但是,的确,所有的进程都被阻塞住了,那么,死锁的条件也达到了,等待一会(这个时间有个隐含参数来控制的:_lm_dd_interval),我们可以看到,会话2出现提示,死锁:

SYS@lhrdb21> UPDATE T_ITL_LHR SET A=A

  2   WHERE DBMS_ROWID.ROWID_BLOCK_NUMBER(ROWID)=94954

  3  AND DBMS_ROWID.ROWID_ROW_NUMBER(ROWID)=4;

 

UPDATE T_ITL_LHR SET A=A

       *

ERROR at line 1:

ORA-00060: deadlock detected while waiting for resource

 

报出死锁之后的阻塞情况:

技术分享 

我们可以在会话2上继续执行步骤三中的SQL,依然会产生死锁。生成死锁后,在告警日志中有下边的语句:

Fri Sep 09 17:56:55 2016

Global Enqueue Services Deadlock detected. More info in file

/oracle/app/oracle/diag/rdbms/lhrdb2/lhrdb21/trace/lhrdb21_lmd0_17039368.trc.

 

其中的内容有非常经典的一段Global Wait-For-Graph(WFG)

*** 2016-09-09 17:48:22.216

Submitting asynchronized dump request [1c]. summary=[ges process stack dump (kjdglblkrdm1)].

Global blockers dump end:-----------------------------------

Global Wait-For-Graph(WFG) at ddTS[0.395] :

BLOCKED 0x700010063d59b90 3 wq 2 cvtops x1001 TX 0x7000b.0xa67(ext 0x2,0x0)[1002-0029-00008387] inst 1

BLOCKER 0x700010063c6d268 3 wq 1 cvtops x28 TX 0x7000b.0xa67(ext 0x2,0x0)[1002-002D-00003742] inst 1

BLOCKED 0x700010063d5adc8 3 wq 2 cvtops x1 TX 0x30021.0x848(ext 0x2,0x0)[1002-002D-00003742] inst 1

BLOCKER 0x700010063d5a4b8 3 wq 1 cvtops x28 TX 0x30021.0x848(ext 0x2,0x0)[1002-0029-00008387] inst 1

 

至于每个参数到底是什么意思,目前还没有去研究,等待大神可以无偿解释一下。至于如何解决ITL产生的死锁,无非就是增大表和索引的initransPCT_FREE的值,可以参考本BLOG中的ITL死锁问题解决。

该实验过程可能有点复杂,小麦苗画了个图来说明整个实验过程:

技术分享

 

5  与文章有关的相关连接

【推荐】 update修改为merge(max+decode)

http://blog.itpub.net/26736162/viewspace-1244055/

【推荐】 采用merge语句的非关联形式再次显神能

http://blog.itpub.net/26736162/viewspace-1222423/

【推荐】 采用MERGE语句的非关联形式提升性能 ---后传

http://blog.itpub.net/26736162/viewspace-1222417/

【推荐】 采用MERGE 语句的非关联形式提升性能

http://blog.itpub.net/26736162/viewspace-1218671/

自相矛盾:一个进程可以自成死锁么

http://blog.itpub.net/26736162/viewspace-2080712/

oracle死锁类型和原因分析

http://blog.itpub.net/26736162/viewspace-1744719/

【DEADLOCK】Oracle“死锁”模拟

http://blog.itpub.net/26736162/viewspace-1744705/

[转]:深入研究ITL阻塞与ITL死锁 

http://blog.itpub.net/26736162/viewspace-2124539/

 

 

  About Me

..........................................................................................................................................................................................................................................................................................................

● 本文作者:小麦苗,只专注于数据库的技术,更注重技术的运用

● 本文在itpub(http://blog.itpub.net/26736162)、博客园(http://www.cnblogs.com/lhrbest)和个人微信公众号(xiaomaimiaolhr)上有同步更新

● QQ群:230161599  微信群:私聊

● 本文itpub地址:http://blog.itpub.net/26736162/viewspace-2124771/  本文博客园地址:http://www.cnblogs.com/lhrbest/p/5859095.html

● 本文pdf版:http://yunpan.cn/cdEQedhCs2kFz (提取码:ed9b)

● 小麦苗云盘地址http://blog.itpub.net/26736162/viewspace-1624453/

● 联系我请加QQ好友(642808185),注明添加缘由

● 于 2016-08-01 15:00~ 2016-08-01 19:00 在中行完成

● 【版权所有,文章允许转载,但须以链接方式注明源地址,否则追究法律责任】

..........................................................................................................................................................................................................................................................................................................

手机长按下图识别二维码或微信客户端扫描下边的二维码来关注小麦苗的微信公众号:xiaomaimiaolhr,免费学习最实用的数据库技术。

技术分享

 

【故障处理】队列等待之TX - allocate ITL entry引起的死锁处理

标签:

原文地址:http://www.cnblogs.com/lhrbest/p/5859095.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!