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[20190101]块内重整.txt
--//我不知道用什么术语表达这样的情况,我仅仅一次开会对方这么讲,我现在也照用这个术语.
--//当dml插入数据到数据块时,预留一定的空间(pctfree的百分比)不再插入.保留一些空间主要目的为了itl的增加,以及update时空间增长.
--//避免大量的行迁移情况出现.
--//当到达顶部时,会出现一次块内重整(也有叫块内重组).通过演示了解这个过程.
1.环境:
scott@test01p> @ ver1
port_string version banner con_id
-------------------- ---------- ---------------------------------------------------------------------------- ------
ibmpc/win_nt64-9.1.0 12.2.0.1.0 oracle database 12c enterprise edition release 12.2.0.1.0 - 64bit production 0
scott@test01p> create table t as select rownum id,to_char(rownum)||lpad('x',800,'x') name from dual connect by level<=8;
table created.
scott@test01p> @desc t
name null? type
---- -------- -------------
id number
name varchar2(840)
--//name 定义varchar2(840).
scott@test01p> select rowid ,id from t;
rowid id
------------------ --
aaaf61aalaaaacraaa 1
aaaf61aalaaaacraab 2
aaaf61aalaaaacraac 3
aaaf61aalaaaacraad 4
aaaf61aalaaaacraae 5
aaaf61aalaaaacraaf 6
aaaf61aalaaaacraag 7
aaaf61aalaaaacraah 8
8 rows selected.
--//有8条记录在一个块中.
scott@test01p> @rowid aaaf61aalaaaacraaa
object file block row rowid_dba dba text
---------- ---------- ---------- ---------- -------------------- -------------------- ----------------------------------------
24245 11 171 0 0x2c000ab 11,171 alter system dump datafile 11 block 171
scott@test01p> alter system checkpoint ;
system altered.
2.通过bbed观察:
bbed> set dba 11,172
dba 0x02c000ac (46137516 11,172)
--//windows版本block+1.
bbed> map
file: d:\app\oracle\oradata\test\test01p\users01.dbf (11)
block: 172 dba:0x02c000ac
------------------------------------------------------------
ktb data block (table/cluster)
struct kcbh, 20 bytes @0
struct ktbbh, 96 bytes @20
struct kdbh, 14 bytes @124
struct kdbt[1], 4 bytes @138
sb2 kdbr[8] @142
ub1 freespace[1550] @158
ub1 rowdata[6480] @1708
ub4 tailchk @8188
bbed> p kdbr
sb2 kdbr[0] @142 7254
sb2 kdbr[1] @144 6444
sb2 kdbr[2] @146 5634
sb2 kdbr[3] @148 4824
sb2 kdbr[4] @150 4014
sb2 kdbr[5] @152 3204
sb2 kdbr[6] @154 2394
sb2 kdbr[7] @156 1584
bbed> x /rnc *kdbr[0]
rowdata[5670] @7378
-------------
flag@7378: 0x2c (kdrhfl, kdrhff, kdrhfh)
lock@7379: 0x00
cols@7380: 2
col 0[2] @7381: 1
col 1[801] @7384: 1xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx ..trunc.
--//注这里的偏移是相对偏移 7254+124(kdbh地址) = 7378.
--//共占用3+1+2+3+801 = 810,注前面有1个字节保持长度指示器.另外注意字符串长度大于250,需要使用3个字节保存长度指示器.
--//可以从数据的偏移量可以看出数据是从块底开始插入的.当前空余空间是1550.
3.继续测试,删除部分数据:
--//先删除一部分数据,仅仅打上标识标识删除.
scott@test01p> delete from t where id in (1,3,6,7);
4 rows deleted.
scott@test01p> commit ;
commit complete.
scott@test01p> alter system checkpoint ;
system altered.
--//通过bbed观察:
bbed> map
file: d:\app\oracle\oradata\test\test01p\users01.dbf (11)
block: 172 dba:0x02c000ac
------------------------------------------------------------
ktb data block (table/cluster)
struct kcbh, 20 bytes @0
struct ktbbh, 96 bytes @20
struct kdbh, 14 bytes @124
struct kdbt[1], 4 bytes @138
sb2 kdbr[8] @142
ub1 freespace[1550] @158
ub1 rowdata[6480] @1708
ub4 tailchk @8188
--//删除部分记录,freespace空间保持不变.
bbed> p kdbr
sb2 kdbr[0] @142 7254
sb2 kdbr[1] @144 6444
sb2 kdbr[2] @146 5634
sb2 kdbr[3] @148 4824
sb2 kdbr[4] @150 4014
sb2 kdbr[5] @152 3204
sb2 kdbr[6] @154 2394
sb2 kdbr[7] @156 1584
bbed> x /rnc *kdbr[0]
rowdata[5670] @7378
-------------
flag@7378: 0x3c (kdrhfl, kdrhff, kdrhfd, kdrhfh)
lock@7379: 0x02
cols@7380: 0
--//仅仅flag从2c变成3c,加入了kdrhfd标识.
4.然后修改一部分数据看看:
scott@test01p> update t set name=lpad('a',811,'a') where id=2;
1 row updated.
scott@test01p> commit ;
commit complete.
scott@test01p> alter system checkpoint ;
system altered.
--//我修改的长度与原来不等,这样增加长度增加10个字节.这样还剩下1550-820 = 730字节.
--//验证看看:
bbed> map
file: d:\app\oracle\oradata\test\test01p\users01.dbf (11)
block: 172 dba:0x02c000ac
------------------------------------------------------------
ktb data block (table/cluster)
struct kcbh, 20 bytes @0
struct ktbbh, 96 bytes @20
struct kdbh, 14 bytes @124
struct kdbt[1], 4 bytes @138
sb2 kdbr[8] @142
ub1 freespace[730] @158
ub1 rowdata[7300] @888
ub4 tailchk @8188
--//ok正确!!
bbed> p kdbr
sb2 kdbr[0] @142 7254
sb2 kdbr[1] @144 764
sb2 kdbr[2] @146 5634
sb2 kdbr[3] @148 4824
sb2 kdbr[4] @150 4014
sb2 kdbr[5] @152 3204
sb2 kdbr[6] @154 2394
sb2 kdbr[7] @156 1584
--//kdbr[1] 指向新的位置.对应id=2的记录.其它不动.
bbed> x /rnc *kdbr[1]
rowdata[0] @888
----------
flag@888: 0x2c (kdrhfl, kdrhff, kdrhfh)
lock@889: 0x03
cols@890: 2
col 0[2] @891: 2
col 1[811] @894: aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa... trunc..
--//如果继续修改记录,长度消耗大于730并且长度与原来不一样,这样就会出现块内重整的情况:
scott@test01p> update t set name=lpad('b',811,'b') where id=4;
1 row updated.
scott@test01p> commit ;
commit complete.
scott@test01p> alter system checkpoint ;
system altered.
bbed> map
file: d:\app\oracle\oradata\test\test01p\users01.dbf (11)
block: 172 dba:0x02c000ac
------------------------------------------------------------
ktb data block (table/cluster)
struct kcbh, 20 bytes @0
struct ktbbh, 96 bytes @20
struct kdbh, 14 bytes @124
struct kdbt[1], 4 bytes @138
sb2 kdbr[8] @142
ub1 freespace[4762] @158
ub1 rowdata[3268] @4920
ub4 tailchk @8188
--//freespace变成了4762,出现一次块内重整,而且收回删除记录的空间.
bbed> p kdbr
sb2 kdbr[0] @142 8062
sb2 kdbr[1] @144 7242
sb2 kdbr[2] @146 7240
sb2 kdbr[3] @148 4796
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~当前偏移最小.
sb2 kdbr[4] @150 6430
sb2 kdbr[5] @152 6428
sb2 kdbr[6] @154 6426
sb2 kdbr[7] @156 5616
--//可以发现行目录发生变化对比前面的情况,做了整理,整体下移.我前面删除的记录是id in (1,3,6,7);
--//id=6,7的记录应该对应kdbr[5],kdbr[6],可以发现记录的偏移地址6428,6426,挨的很近.
--//可以大致猜测它的算法,移动除kdbr[3](对应id=4)的记录下移腾出空间,你可以发现kdbr[3]指向的偏移是当前最小的.
bbed> x /rnc *kdbr[5]
rowdata[1632] @6552
-------------
flag@6552: 0x3c (kdrhfl, kdrhff, kdrhfd, kdrhfh)
lock@6553: 0x02
cols@6554: 0
bbed> x /rnc *kdbr[6]
rowdata[1630] @6550
-------------
flag@6550: 0x3c (kdrhfl, kdrhff, kdrhfd, kdrhfh)
lock@6551: 0x02
cols@6552: 0
bbed> dump /v offset 6550 count 20
file: d:\app\oracle\oradata\test\test01p\users01.dbf (11)
block: 172 offsets: 6550 to 6569 dba:0x02c000ac
-----------------------------------------------------------------------------------------------------------
3c023c02 2c000202 c106fe21 03357878 78787878 l <.<.,...??.5xxxxxx
<32 bytes per line>
--//仅仅保留2个字节.
bbed> x /rnc *kdbr[0]
rowdata[3266] @8186
-------------
flag@8186: 0x3c (kdrhfl, kdrhff, kdrhfd, kdrhfh)
lock@8187: 0x02
cols@8188: 0
bbed> dump /v count 20
file: d:\app\oracle\oradata\test\test01p\users01.dbf (11)
block: 172 offsets: 8186 to 8191 dba:0x02c000ac
-----------------------------------------------------------------------------------------------------------
3c020106 ff18 l <.....
<32 bytes per line>
--//这个时候按照以前使用bbed修复删除记录的方法是无用的.因为对应的记录信息已经被覆盖了.
5.继续插入记录看看.
scott@test01p> insert into t values(9,to_char(9)||lpad('y',800,'y'));
1 row created.
scott@test01p> commit ;
commit complete.
scott@test01p> alter system checkpoint ;
system altered.
--//当前块已经空出许多空间,可以继续插入数据,看看这时的情况:
bbed> map
file: d:\app\oracle\oradata\test\test01p\users01.dbf (11)
block: 172 dba:0x02c000ac
------------------------------------------------------------
ktb data block (table/cluster)
struct kcbh, 20 bytes @0
struct ktbbh, 96 bytes @20
struct kdbh, 14 bytes @124
struct kdbt[1], 4 bytes @138
sb2 kdbr[9] @142
ub1 freespace[3950] @160
ub1 rowdata[4078] @4110
ub4 tailchk @8188
--//还有freespace=3950.
bbed> p kdbr
sb2 kdbr[0] @142 2
sb2 kdbr[1] @144 7242
sb2 kdbr[2] @146 5
sb2 kdbr[3] @148 4796
sb2 kdbr[4] @150 6430
sb2 kdbr[5] @152 6
sb2 kdbr[6] @154 -1
sb2 kdbr[7] @156 5616
sb2 kdbr[8] @158 3986
--//可以发现当前块又插入1条.
bbed> x /rnc *kdbr[8]
rowdata[0] @4110
----------
flag@4110: 0x2c (kdrhfl, kdrhff, kdrhfh)
lock@4111: 0x02
cols@4112: 2
col 0[2] @4113: 9
col 1[801] @4116: 9yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy..trunc..
--//注意看前面的kdbr[0],kdbr[2],kdbr[5],kdbr[6]对应的偏移量(2,5,6,-1),通过偏移量链接起来.
--//也就是这个这时的偏移量指向的地址根本不是rowdate区域.
--//我不知道是否可以得出这样结论:如果kdbr指向的值如果小于当前的行目录数量(9),这些行对应的记录应该是删除的.
--//理论讲这时对应记录很难恢复,已经覆盖了.
6.测试重用的情况:
--//是否会重用呢?理论讲应该会重用.打开新的会话看看:
scott@test01p> insert into t values(10,to_char(10)||lpad('x',800,'x'));
1 row created.
scott@test01p> commit ;
commit complete.
scott@test01p> alter system checkpoint;
system altered.
bbed> map
file: d:\app\oracle\oradata\test\test01p\users01.dbf (11)
block: 172 dba:0x02c000ac
------------------------------------------------------------
ktb data block (table/cluster)
struct kcbh, 20 bytes @0
struct ktbbh, 96 bytes @20
struct kdbh, 14 bytes @124
struct kdbt[1], 4 bytes @138
sb2 kdbr[9] @142
ub1 freespace[3139] @160
ub1 rowdata[4889] @3299
ub4 tailchk @8188
--//freespace=3139.
bbed> p kdbr
sb2 kdbr[0] @142 3175
sb2 kdbr[1] @144 7242
sb2 kdbr[2] @146 5
sb2 kdbr[3] @148 4796
sb2 kdbr[4] @150 6430
sb2 kdbr[5] @152 6
sb2 kdbr[6] @154 -1
sb2 kdbr[7] @156 5616
sb2 kdbr[8] @158 3986
--//理论讲不同的会话不会插入相同的块,因为我建表比较特殊,使用ctas建立的.
--//可以发现插入占用了kdbr[0].
bbed> x /rnc *kdbr[0]
rowdata[0] @3299
----------
flag@3299: 0x2c (kdrhfl, kdrhff, kdrhfh)
lock@3300: 0x03
cols@3301: 2
col 0[2] @3302: 10
col 1[802] @3305: 10xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx...trunc..
scott@test01p> select rowid,id from t;
rowid id
------------------ ----------
aaaf61aalaaaacraaa 10
aaaf61aalaaaacraab 2
aaaf61aalaaaacraad 4
aaaf61aalaaaacraae 5
aaaf61aalaaaacraah 8
aaaf61aalaaaacraai 9
6 rows selected.
--//oracle在插入时能快速定位到kdbr[0],估计在块内有相关记录.
bbed> p kdbh
struct kdbh, 14 bytes @124
ub1 kdbhflag @124 0x00 (none)
b1 kdbhntab @125 1
b2 kdbhnrow @126 9
sb2 kdbhfrre @128 2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
sb2 kdbhfsbo @130 36
sb2 kdbhfseo @132 3175
b2 kdbhavsp @134 3147
b2 kdbhtosp @136 3147
--//猜测记录在kdbh.kdbhfrre中,当前是2.如果继续插入,变成5就可以验证我的判断。
--//注意在新的会话插入(session 2)：
scott@test01p> insert into t values(11,to_char(11)||lpad('w',800,'w'));
1 row created.
scott@test01p> commit ;
commit complete.
scott@test01p> alter system checkpoint;
system altered.
bbed> map
file: d:\app\oracle\oradata\test\test01p\users01.dbf (11)
block: 172 dba:0x02c000ac
------------------------------------------------------------
ktb data block (table/cluster)
struct kcbh, 20 bytes @0
struct ktbbh, 96 bytes @20
struct kdbh, 14 bytes @124
struct kdbt[1], 4 bytes @138
sb2 kdbr[9] @142
ub1 freespace[2328] @160
ub1 rowdata[5700] @2488
ub4 tailchk @8188
--//freespace继续减少。
bbed> p kdbr
sb2 kdbr[0] @142 3175
sb2 kdbr[1] @144 7242
sb2 kdbr[2] @146 2364
sb2 kdbr[3] @148 4796
sb2 kdbr[4] @150 6430
sb2 kdbr[5] @152 6
sb2 kdbr[6] @154 -1
sb2 kdbr[7] @156 5616
sb2 kdbr[8] @158 3986
--//插入记录占用了kdbr[2]。
bbed> p kdbh
struct kdbh, 14 bytes @124
ub1 kdbhflag @124 0x00 (none)
b1 kdbhntab @125 1
b2 kdbhnrow @126 9
sb2 kdbhfrre @128 5
--//~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~变成5，验证的判断。
sb2 kdbhfsbo @130 36
sb2 kdbhfseo @132 2364
b2 kdbhavsp @134 2336
b2 kdbhtosp @136 2336
--//这样通过kdbr[5],里面记录的是6，继续插入可以很容易行目录，最后记录-1,表示已经没有了。
scott@test01p> select rowid,id from t;
rowid id
------------------ ----------
aaaf61aalaaaacraaa 10
aaaf61aalaaaacraab 2
aaaf61aalaaaacraac 11
aaaf61aalaaaacraad 4
aaaf61aalaaaacraae 5
aaaf61aalaaaacraah 8
aaaf61aalaaaacraai 9
7 rows selected.
总结:
--//仅仅通过一些简单的例子演示这个过程,也许许多情况下更加复杂.
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