分析一例 TX Enqueue contention案例

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标题:分析一例 TX Enqueue contention案例

作者:惜分飞©版权所有[未经本人同意,不得以任何形式转载,否则有进一步追究法律责任的权利.]

应用反馈某个业务比较慢,需要紧急处理
查询等待事件

SQL> select * from v$version;

BANNER
----------------------------------------------------------------
Oracle9i Enterprise Edition Release 9.2.0.8.0 - 64bit Production
PL/SQL Release 9.2.0.8.0 - Production
CORE    9.2.0.8.0       Production
TNS for IBM/AIX RISC System/6000: Version 9.2.0.8.0 - Production
NLSRTL Version 9.2.0.8.0 - Production

SQL> select A.INST_ID,count(*), event
  2    from Gv$session_wait a
  3  where event not in ('SQL*Net more data to client',
  4  'rdbms ipc message',
  5               'smon timer',
  6               'pmon timer',
  7               'SQL*Net message from client',
  8               'lock manager wait for remote message',
  9               'ges remote message',
 10               'gcs remote message',
 11               'gcs for action',
 12               'client message',
 13               'pipe get',
 14               'null event',
 15               'PX Idle Wait',
 16               'single-task message',
 17               'PX Deq: Execution Msg',
 18               'KXFQ: kxfqdeq - normal deqeue',
 19               'listen endpoint status',
 20               'slave wait',
 21               'wakeup time manager','jobq slave wait')
 22  group by INST_ID,event
 23  order by 1 desc,2 desc;

   INST_ID   COUNT(*) EVENT
---------- ---------- --------------------------------------
         2          8 enqueue
         2          1 async disk IO
         2          1 db file sequential read
         2          1 SQL*Net message to client
         2          1 PX Deq: reap credit
         1          2 global cache cr request
         1          1 async disk IO
         1          1 PX Deq: reap credit
         1          1 PX Deq: Execute Reply

9 rows selected.

发现enqueue等待有些多,怀疑是TX enquenue,查询阻塞者

SQL> set linesize 100
SQL> set pagesize 66
SQL> col c1 for a15
SQL> col c1 heading "Program Name "
SQL> select l.inst_id,l.SID,program c1,l.TYPE,l.ID1,l.ID2,l.LMODE,l.REQUEST
  2  from gv$lock l,gv$session s
  3  where l.type like 'TX' and l.REQUEST =6
  4  and l.inst_id=s.inst_id and l.sid=s.sid
  5  order by id1
  6  / 

   INST_ID        SID Program Name    TY        ID1        ID2      LMODE    REQUEST
---------- ---------- --------------- -- ---------- ---------- ---------- ----------
         2        295 rtStopMain@zwq_ TX    1441805    2391806          0          6
                      bill_2 (TNS V1-
                      V3)

         2        992 rtStopMain@zwq_ TX    1441805    2391806          0          6
                      bill_2 (TNS V1-
                      V3)

         2       1238 rtStopMain@zwq_ TX    6946827    2546365          0          6
                      bill_2 (TNS V1-
                      V3)

         2       1298 rtStopMain@zwq_ TX    6946827    2546365          0          6
                      bill_2 (TNS V1-
                      V3)

         2       1684 rtStopMain@zwq_ TX    6946827    2546365          0          6
                      bill_2 (TNS V1-
                      V3)

         2       1553 rtStopMain@zwq_ TX    6946827    2546365          0          6
                      bill_2 (TNS V1-
                      V3)

         2         75 rtStopMain@zwq_ TX   12451856     199146          0          6
                      bill_2 (TNS V1-
                      V3)

         2       1125 rtStopMain@zwq_ TX   14352404      63837          0          6
                      bill_2 (TNS V1-
                      V3)

查询持有者

SQL> set linesize 100
SQL> set pagesize 66
SQL> col c1 for a15
SQL> col c1 heading "Program Name "
SQL> select l.inst_id,l.SID,program c1,l.TYPE,l.ID1,l.ID2,l.LMODE,l.REQUEST
  2  from gv$lock l,gv$session s
  3  where l.type like 'TX' and l.LMODE =6 and (l.ID1,l.ID2) in
  4  (select id1,id2 from gv$lock where type like 'TX' and REQUEST =6)
  5  and l.inst_id=s.inst_id and l.sid=s.sid
  6  order by id1
  7  / 

   INST_ID        SID Program Name    TY        ID1        ID2      LMODE    REQUEST
---------- ---------- --------------- -- ---------- ---------- ---------- ----------
         2         75 rtStopMain@zwq_ TX    1441805    2391806          6          0
                      bill_2 (TNS V1-
                      V3)

         2        992 rtStopMain@zwq_ TX    6946827    2546365          6          0
                      bill_2 (TNS V1-
                      V3)

         2        295 rtStopMain@zwq_ TX   12451856     199146          6          0
                      bill_2 (TNS V1-
                      V3)

         2       1553 rtStopMain@zwq_ TX   14352404      63837          6          0
                      bill_2 (TNS V1-
                      V3)

通过持有者和阻塞者可以得出:
1.持有者和阻塞者都是在2号实例上
2.持有者75阻塞了295/992的会话
3.持有者992阻塞了1238/1298/1684/1553的会话
4.持有者295阻塞了75的会话
5.持有者1553阻塞了1125的会话
6.同时分析发现,所有的持有者sid也在阻塞者中,也就是持有者阻塞了某个sid,而自身又被其他sid给阻塞,形成了多级阻塞或者环.如:75阻塞了295,而295有阻塞了75;992阻塞了1553,而1553阻塞了1125

查询阻塞和持有者对象

SQL> set linesize 110
SQL> col c0 for 999
SQL> col c0 heading "INS"
SQL> col c1 for a15
SQL> col c1 heading "Program Name "
SQL> select inst_id c0,sid,program c1,ROW_WAIT_OBJ# object_no, ROW_WAIT_FILE# Rfile_no,
  2  ROW_WAIT_BLOCK# Block_no ,ROW_WAIT_ROW# Row_no
  3  from gv$session
  4  where (inst_id,sid) in (select inst_id,sid from gv$session_wait where p1='1415053318')
  5  / 

 INS        SID Program Name     OBJECT_NO   RFILE_NO   BLOCK_NO     ROW_NO
---- ---------- --------------- ---------- ---------- ---------- ----------
   2         75 rtStopMain@zwq_    1323132         13     122601        111
                bill_2 (TNS V1-
                V3)

   2        295 rtStopMain@zwq_    1323132         13     122601        100
                bill_2 (TNS V1-
                V3)

   2        992 rtStopMain@zwq_    1323132         13     122601        101
                bill_2 (TNS V1-
                V3)

   2       1125 rtStopMain@zwq_    1323132         84      38445         70
                bill_2 (TNS V1-
                V3)

   2       1238 rtStopMain@zwq_    1323132         15     255066         41
                bill_2 (TNS V1-
                V3)

   2       1298 rtStopMain@zwq_    1323132         14     118411          8
                bill_2 (TNS V1-
                V3)

   2       1553 rtStopMain@zwq_    1323132         15     255066         19
                bill_2 (TNS V1-
                V3)

   2       1684 rtStopMain@zwq_    1323132         14     118411         21
                bill_2 (TNS V1-
                V3)

8 rows selected.

SQL> set linesize 100
SQL> set pagesize 100
SQL> col owner for a10
SQL> col object_name for a20
SQL> col object_type for a10
SQL> select owner,object_name,object_id,object_type
  2  from dba_objects
  3  where
  4  object_id in (select ROW_WAIT_OBJ# from gv$session
  5  where (inst_id, sid) in (select inst_id,sid from gv$session_wait where p1='1415053318'))
  6  / 
 

OWNER      OBJECT_NAME           OBJECT_ID OBJECT_TYP
---------- -------------------- ---------- ----------
DBACCADM   DCUSTCREDITBALANCE      1323132 TABLE

通过查询的出来,所有操作的聚焦点都是在DBACCADM.DCUSTCREDITBALANCE表上面

查询相关sql语句

SQL> SQL> set linesize 120
SQL> set pagesize 66
SQL> col c0 for 999
SQL> col c0 heading "INS"
SQL> col c1 for a9
SQL> col c1 heading "OS User"
SQL> col c2 for a9
SQL> col c2 heading "Oracle User"
SQL> col c3 for a15
SQL> col c3 heading "Program Name"
SQL> col b1 for a9
SQL> col b1 heading "Unix PID"
SQL> col b2 for 9999 justify left
SQL> col b2 heading "ORA SID"
SQL> col b3 for 999999 justify left
SQL> col b3 heading "SERIAL#"
SQL> col sql_text for a45
SQL> set space 1
SQL> break on b1 nodup on c0 nodup on c3 nodup on c1 nodup on c2 nodup on b2 nodup on b3 skip 2
SQL> select a.inst_id c0,b.sid b2,c.spid b1, b.program c3, b.username c2,b.serial# b3, a.sql_text
  2    from gv$sql a, gv$session b, gv$process c
  3   where
  4     a.address = b.sql_address
  5     and b.paddr = c.addr
  6     and a.hash_value = b.sql_hash_value
  7     and a.inst_id=b.inst_id and a.inst_id=c.inst_id
  8     and a.inst_id like '&inst_id' and b.sid like '&sid'
  9   order by c.spid,a.hash_value
 10  / 
Enter value for inst_id: 2
Enter value for sid: 75
old   8:    and a.inst_id like '&inst_id' and b.sid like '&sid'
new   8:    and a.inst_id like '2' and b.sid like '75'

 INS ORA SID Unix PID  Program Name    Oracle Us SERIAL# SQL_TEXT
---- ------- --------- --------------- --------- ------- ---------------------------------------------
   2      75 1167392   rtStopMain@zwq_ DBCUSTOPR   42815 update dcustcreditbalance  set limit_owe=:b0,
                       bill_2 (TNS V1-                   unbill_fee=:b1,prepay_fee=:b2,owe_fee=:b3,op_
                       V3)                               time=sysdate where id_no=:b4




SQL> /
Enter value for inst_id: 2
Enter value for sid: 992
old   8:    and a.inst_id like '&inst_id' and b.sid like '&sid'
new   8:    and a.inst_id like '2' and b.sid like '992'

 INS ORA SID Unix PID  Program Name    Oracle Us SERIAL# SQL_TEXT
---- ------- --------- --------------- --------- ------- ---------------------------------------------
   2     992 2760870   rtStopMain@zwq_ DBCUSTOPR   56282 update dcustcreditbalance  set limit_owe=:b0,
                       bill_2 (TNS V1-                   unbill_fee=:b1,prepay_fee=:b2,owe_fee=:b3,op_
                       V3)                               time=sysdate where id_no=:b4




SQL> /
Enter value for inst_id: 2
Enter value for sid: 295
old   8:    and a.inst_id like '&inst_id' and b.sid like '&sid'
new   8:    and a.inst_id like '2' and b.sid like '295'

 INS ORA SID Unix PID  Program Name    Oracle Us SERIAL# SQL_TEXT
---- ------- --------- --------------- --------- ------- ---------------------------------------------
   2     295 1639008   rtStopMain@zwq_ DBCUSTOPR   35740 update dcustcreditbalance  set limit_owe=:b0,
                       bill_2 (TNS V1-                   unbill_fee=:b1,prepay_fee=:b2,owe_fee=:b3,op_
                       V3)                               time=sysdate where id_no=:b4

其他阻塞者和持有者执行sql语句均和该语句相同,省略其他查询.通过这些查询可以确定是因为对dcustcreditbalance表的更新操作导致了这样的现象发生.

处理方案
1.临时处理方案:kill掉持有者
2.永久处理方案:修改这部分程序业务逻辑

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