[20181226]简单探究cluster table.txt
–//简单探究cluster table.以前也做过,有点生疏了.
1.环境:
scott@book> @ ver1
port_string version banner
——————- ———- —————————————————————————-
x86_64/linux 2.4.xx 11.2.0.4.0 oracle database 11g enterprise edition release 11.2.0.4.0 – 64bit production
2.建立测试环境:
create cluster deptx_cluster (deptno number(2));
create table deptx
(
deptno number(2) ,
dname varchar2(14 byte),
loc varchar2(13 byte)
) cluster deptx_cluster (deptno);
alter table deptx add constraint pk_deptx primary key (deptno);
create table empx
(
empno number(4) ,
ename varchar2(10 byte),
job varchar2(9 byte),
mgr number(4),
hiredate date,
sal number(7,2),
comm number(7,2),
deptno number(2) references deptx
) cluster deptx_cluster (deptno);
alter table empx add constraint constraint_name primary key (empno);
create index i_deptx_cluster_deptno on cluster deptx_cluster;
–//注这里不能使用unique,否则报ora-01715: unique may not be used with a cluster index
insert into deptx select * from dept where deptno=10;
insert into empx select * from emp where deptno=10;
insert into deptx select * from dept where deptno=20;
insert into empx select * from emp where deptno=20;
insert into deptx select * from dept where deptno=30;
insert into empx select * from emp where deptno=30;
insert into deptx select * from dept where deptno=40;
insert into empx select * from emp where deptno=40;
commit;
3.查看数据:
scott@book> select rowid,deptx.* from deptx;
rowid deptno dname loc
—————— ———- ————– ————-
aaawefaaeaaaairaaa 20 research dallas
~~~~~~~~~~~~~~~~~~
aaawefaaeaaaaitaaa 30 sales chicago
aaawefaaeaaaaiuaaa 40 operations boston
aaawefaaeaaaaivaaa 10 accounting new york
scott@book> select rowid,empx.* from empx where deptno=20;
rowid empno ename job mgr hiredate sal comm deptno
—————— ———- ———- ——— ———- ——————- ———- ———- ———-
aaawefaaeaaaairaaa 7369 smith clerk 7902 1980-12-17 00:00:00 800 20
~~~~~~~~~~~~~~~~~~
aaawefaaeaaaairaab 7566 jones manager 7839 1981-04-02 00:00:00 2975 20
aaawefaaeaaaairaac 7788 scott analyst 7566 1987-04-19 00:00:00 3000 20
aaawefaaeaaaairaad 7876 adams clerk 7788 1987-05-23 00:00:00 1100 20
aaawefaaeaaaairaae 7902 ford analyst 7566 1981-12-03 00:00:00 3000 20
scott@book> @ dpc ” ”
plan_table_output
————————————-
sql_id fw7tmp1r74rf4, child number 0
————————————-
select rowid,empx.* from empx where deptno=20
plan hash value: 1833007843
————————————————————————————————
| id | operation | name | e-rows |e-bytes| cost (%cpu)| e-time |
————————————————————————————————
| 0 | select statement | | | | 2 (100)| |
| 1 | table access cluster| empx | 5 | 190 | 2 (0)| 00:00:01 |
|* 2 | index unique scan | i_deptx_cluster_deptno | 1 | | 1 (0)| 00:00:01 |
————————————————————————————————
query block name / object alias (identified by operation id):
————————————————————-
1 – sel$1 / empx@sel$1
2 – sel$1 / empx@sel$1
predicate information (identified by operation id):
—————————————————
2 – access(“deptno”=20)
–//可以发现查询empx表的deptno=20可以利用cluster table的索引.另外你可以注意一个特点cluster table里面的表rowid可以相同.比如下划线的内容.
–//实际上这样设计相关表的查询都保存在相同块中,连接访问会快许多.
scott@book> @ rowid aaawefaaeaaaairaaa
object file block row rowid_dba dba text
———- ———- ———- ———- ——————– ——————– —————————————-
90399 4 555 0 0x100022b 4,555 alter system dump datafile 4 block 555 ;
–//转储dba=4,555看看.
3.转储分析:
scott@book> alter system flush buffer_cache;
system altered.
scott@book> alter system dump datafile 4 block 555 ;
system altered.
block header dump: 0x0100022b
object id on block? y
seg/obj: 0x1611f csc: 0x03.175f42b6 itc: 2 flg: e typ: 1 – data
brn: 0 bdba: 0x1000228 ver: 0x01 opc: 0
inc: 0 exflg: 0
itl xid uba flag lck scn/fsc
0x01 0x0009.01b.00000a44 0x00c00356.0394.31 c— 0 scn 0x0003.175f42b6
0x02 0x000a.001.00004cfa 0x00c02e94.0ef9.27 –u- 6 fsc 0x0000.175f42cc
bdba: 0x0100022b
data_block_dump,data header at 0x7f286b5f1064
===============
tsiz: 0x1f98
hsiz: 0x28
pbl: 0x7f286b5f1064
76543210
flag=——-k
ntab=3
nrow=7
frre=-1
fsbo=0x28
fseo=0x1eb9
avsp=0x1e91
tosp=0x1e91
0xe:pti[0] nrow=1 offs=0
0x12:pti[1] nrow=1 offs=1
0x16:pti[2] nrow=5 offs=2
–//说明有3个表,cluster表deptx_cluster以及表deptx,empx.行数分别是1,1,5(下面可以看到对于7条记录).后面offs表示偏移量.
0x1a:pri[0] offs=0x1f82
0x1c:pri[1] offs=0x1f6e
0x1e:pri[2] offs=0x1f4b
0x20:pri[3] offs=0x1f25
0x22:pri[4] offs=0x1f00
0x24:pri[5] offs=0x1edd
0x26:pri[6] offs=0x1eb9
block_row_dump:
tab 0, row 0, @0x1f82 –//表0 cluster table deptx_cluster
tl: 22 fb: k-h-fl– lb: 0x0 cc: 1
–//fb: k-h-fl–, k meaning cluster key.
–//the remaining rows have the second high order bit set (fb: -ch-fl–),
–//c meaning cluster table member.
curc: 6 comc: 6 pk: 0x0100022b.0 nk: 0x0100022b.0
–//curc: 6 current row count for this key in this block
–//comc: 6 committed row count for this key in this block
–//pk: pk: 0x0100022b.0 rowid of previous block for this cluster key
–//nk: 0x0100022b.0 rowid of next block for this cluster key
col 0: [ 2] c1 15 –//数字20.
tab 1, row 0, @0x1f6e
tl: 20 fb: -ch-fl– lb: 0x2 cc: 2 cki: 0 –//表1 deptx
–//c meaning cluster table member.
col 0: [ 8] 52 45 53 45 41 52 43 48 –//对应内容’research’
col 1: [ 6] 44 41 4c 4c 41 53 –//对应内容’dallas’
tab 2, row 0, @0x1f4b –//表2 empx
tl: 35 fb: -ch-fl– lb: 0x2 cc: 6 cki: 0
col 0: [ 3] c2 4a 46
col 1: [ 5] 53 4d 49 54 48
col 2: [ 5] 43 4c 45 52 4b
col 3: [ 3] c2 50 03
col 4: [ 7] 77 b4 0c 11 01 01 01
col 5: [ 2] c2 09
tab 2, row 1, @0x1f25
tl: 38 fb: -ch-fl– lb: 0x2 cc: 6 cki: 0
col 0: [ 3] c2 4c 43
col 1: [ 5] 4a 4f 4e 45 53
col 2: [ 7] 4d 41 4e 41 47 45 52
col 3: [ 3] c2 4f 28
col 4: [ 7] 77 b5 04 02 01 01 01
col 5: [ 3] c2 1e 4c
tab 2, row 2, @0x1f00
tl: 37 fb: -ch-fl– lb: 0x2 cc: 6 cki: 0
col 0: [ 3] c2 4e 59
col 1: [ 5] 53 43 4f 54 54
col 2: [ 7] 41 4e 41 4c 59 53 54
col 3: [ 3] c2 4c 43
col 4: [ 7] 77 bb 04 13 01 01 01
col 5: [ 2] c2 1f
tab 2, row 3, @0x1edd
tl: 35 fb: -ch-fl– lb: 0x2 cc: 6 cki: 0
col 0: [ 3] c2 4f 4d
col 1: [ 5] 41 44 41 4d 53
col 2: [ 5] 43 4c 45 52 4b
col 3: [ 3] c2 4e 59
col 4: [ 7] 77 bb 05 17 01 01 01
col 5: [ 2] c2 0c
tab 2, row 4, @0x1eb9
tl: 36 fb: -ch-fl– lb: 0x2 cc: 6 cki: 0
col 0: [ 3] c2 50 03
col 1: [ 4] 46 4f 52 44
col 2: [ 7] 41 4e 41 4c 59 53 54
col 3: [ 3] c2 4c 43
col 4: [ 7] 77 b5 0c 03 01 01 01
col 5: [ 2] c2 1f
end_of_block_dump
end dump data blocks tsn: 4 file#: 4 minblk 555 maxblk 555
–//通过bbed观察:
bbed> p dba 4,555 kdbt
struct kdbt[0], 4 bytes @114
sb2 kdbtoffs @114 0
sb2 kdbtnrow @116 1
struct kdbt[1], 4 bytes @118
sb2 kdbtoffs @118 1
sb2 kdbtnrow @120 1
struct kdbt[2], 4 bytes @122
sb2 kdbtoffs @122 2
sb2 kdbtnrow @124 5
–//与如下对应:
0xe:pti[0] nrow=1 offs=0
0x12:pti[1] nrow=1 offs=1
0x16:pti[2] nrow=5 offs=2
–//比如可以看出对于当前块表empx位于*kdbr[2]开始.
bbed> x /rn *kdbr[0]
rowdata[201] @8166
————
flag@8166: 0xac (kdrhfl, kdrhff, kdrhfh, kdrhfk) –//cluster key.行头为ac 普通表为2c
lock@8167: 0x00
cols@8168: 1
kref@8169: 6
mref@8171: 6
hrid@8173:0x0100022b.0
nrid@8179:0x0100022b.0
col 0[2] @8185: 20
bbed> x /rcc *kdbr[1]
rowdata[181] @8146
————
flag@8146: 0x6c (kdrhfl, kdrhff, kdrhfh, kdrhfc) –//cluser table member .行头为6c 删除为7c,实际上加入kdrhfd标志.
lock@8147: 0x02
cols@8148: 2
col 0[8] @8150: research
col 1[6] @8159: dallas
–//注:不包括cluster table key.
bbed> x /rnccntnnn *kdbr[2]
rowdata[146] @8111
————
flag@8111: 0x6c (kdrhfl, kdrhff, kdrhfh, kdrhfc)
lock@8112: 0x02
cols@8113: 6
col 0[3] @8115: 7369
col 1[5] @8119: smith
col 2[5] @8125: clerk
col 3[3] @8131: 7902
col 4[7] @8135: 1980-12-17 00:00:00
col 5[2] @8143: 800
4.做一个删除的手工恢复看看.
scott@book> delete from empx where empno=7566;
1 row deleted.
scott@book> commit ;
commit complete.
scott@book> alter system flush buffer_cache;
system altered.
scott@book> alter system dump datafile 4 block 555 ;
system altered.
–//仅仅贴出改动部分:
block_row_dump:
tab 0, row 0, @0x1f82
tl: 22 fb: k-h-fl– lb: 0x0 cc: 1
curc: 6 comc: 5 pk: 0x0100022b.0 nk: 0x0100022b.0
col 0: [ 2] c1 15
–//comc 5 少1条记录.
…
tl: 35 fb: -ch-fl– lb: 0x0 cc: 6 cki: 0
col 0: [ 3] c2 4a 46
col 1: [ 5] 53 4d 49 54 48
col 2: [ 5] 43 4c 45 52 4b
col 3: [ 3] c2 50 03
col 4: [ 7] 77 b4 0c 11 01 01 01
col 5: [ 2] c2 09
tab 2, row 1, @0x1f25
tl: 4 fb: -chdfl– lb: 0x2 cc: 0 cki: 0 –//加入d标志.
tab 2, row 2, @0x1f00
–//如果手工修复,需要修改2处.
bbed> x /rn *kdbr[0]
rowdata[201] @8166
————
flag@8166: 0xac (kdrhfl, kdrhff, kdrhfh, kdrhfk)
lock@8167: 0x00
cols@8168: 1
kref@8169: 6
mref@8171: 5
~~~~~~~~~~~~~~~
hrid@8173:0x0100022b.0
nrid@8179:0x0100022b.0
col 0[2] @8185: 20
bbed> x /rnccntnnn *kdbr[3]
rowdata[108] @8073
————
flag@8073: 0x7c (kdrhfl, kdrhff, kdrhfd, kdrhfh, kdrhfc)
lock@8074: 0x02
cols@8075: 0
–//执行如下:
assign /d offset 8171=6;
assign /x offset 8073=6c;
bbed> assign /d offset 8171=6;
warning: contents of previous bifile will be lost. proceed? (y/n) y
ub1 rowdata[0] @8171 0x06
bbed> assign /x offset 8073=6c;
ub1 rowdata[0] @8073 0x6c
–//检查:
bbed> x /rn *kdbr[0]
rowdata[201] @8166
————
flag@8166: 0xac (kdrhfl, kdrhff, kdrhfh, kdrhfk)
lock@8167: 0x00
cols@8168: 1
kref@8169: 6
mref@8171: 6
hrid@8173:0x0100022b.0
nrid@8179:0x0100022b.0
col 0[2] @8185: 20
bbed> x /rnccntnnn *kdbr[2]
rowdata[146] @8111
————
flag@8111: 0x6c (kdrhfl, kdrhff, kdrhfh, kdrhfc)
lock@8112: 0x00
cols@8113: 6
col 0[3] @8115: 7369
col 1[5] @8119: smith
col 2[5] @8125: clerk
col 3[3] @8131: 7902
col 4[7] @8135: 1980-12-17 00:00:00
col 5[2] @8143: 800
bbed> sum apply
check value for file 4, block 555:
current = 0xf209, required = 0xf209
bbed> verify
dbverify – verification starting
file = /mnt/ramdisk/book/users01.dbf
block = 555
block checking: dba = 16777771, block type = ktb-managed data block
data header at 0x7fd6f1fd9264
kdbchk: the amount of space used is not equal to block size
used=263 fsc=34 avsp=7825 dtl=8088
block 555 failed with check code 6110
–//空间问题暂时不理会.
–//验证修改是否有效.
scott@book> select rowid,empx.* from empx where deptno=20;
rowid empno ename job mgr hiredate sal comm deptno
—————— ———- ———- ——— ———- ——————- ———- ———- ———-
aaawefaaeaaaairaaa 7369 smith clerk 7902 1980-12-17 00:00:00 800 20
aaawefaaeaaaairaab 7566 jones manager 7839 1981-04-02 00:00:00 2975 20
aaawefaaeaaaairaac 7788 scott analyst 7566 1987-04-19 00:00:00 3000 20
aaawefaaeaaaairaad 7876 adams clerk 7788 1987-05-23 00:00:00 1100 20
aaawefaaeaaaairaae 7902 ford analyst 7566 1981-12-03 00:00:00 3000 20
–//empno=7566可以查询到.如果通过索引是无法查询到信息.
scott@book> select rowid,empx.* from empx where empno=7566;
no rows selected
–//如果不修复comc的偏移,还原:
bbed> assign /d offset 8171=5;
ub1 rowdata[0] @8171 0x05
bbed> sum apply
check value for file 4, block 555:
current = 0xf109, required = 0xf109
bbed> verify
dbverify – verification starting
file = /mnt/ramdisk/book/users01.dbf
block = 555
block checking: dba = 16777771, block type = ktb-managed data block
data header at 0x105cc64
kdbchk: key comref count wrong
keyslot=0
block 555 failed with check code 6121
–//会报如上错误.不过查询没有问题.
scott@book> alter system flush buffer_cache;
system altered.
scott@book> select rowid,empx.* from empx where deptno=20;
rowid empno ename job mgr hiredate sal comm deptno
—————— ———- ———- ——— ———- ——————- ———- ———- ———-
aaawefaaeaaaairaaa 7369 smith clerk 7902 1980-12-17 00:00:00 800 20
aaawefaaeaaaairaab 7566 jones manager 7839 1981-04-02 00:00:00 2975 20
aaawefaaeaaaairaac 7788 scott analyst 7566 1987-04-19 00:00:00 3000 20
aaawefaaeaaaairaad 7876 adams clerk 7788 1987-05-23 00:00:00 1100 20
aaawefaaeaaaairaae 7902 ford analyst 7566 1981-12-03 00:00:00 3000 20
5.最后修复bbed verify的错误.
–//这个错误我建议不熟悉bbed,不再修复,比较繁琐.
bbed> assign /d offset 8171=6;
ub1 rowdata[0] @8171 0x06
bbed> sum apply
check value for file 4, block 555:
current = 0xf209, required = 0xf209
bbed> verify
dbverify – verification starting
file = /mnt/ramdisk/book/users01.dbf
block = 555
block checking: dba = 16777771, block type = ktb-managed data block
data header at 0x105cc64
kdbchk: the amount of space used is not equal to block size
used=263 fsc=34 avsp=7825 dtl=8088
block 555 failed with check code 6110
–//理论used+fsc+avsp=dtl
–//dtl-used-fsc= 8088-263-34 = 7791
–//然后assign kdbh.kdbhavsp=7791,如果修改事务itl槽信息,步骤也不少.
bbed> assign kdbh.kdbhavsp=7791
sb2 kdbhavsp @110 7791
bbed> sum apply
check value for file 4, block 555:
current = 0xf2f7, required = 0xf2f7
bbed> verify
dbverify – verification starting
file = /mnt/ramdisk/book/users01.dbf
block = 555
block checking: dba = 16777771, block type = ktb-managed data block
data header at 0x107ec64
kdbchk: space available on commit is incorrect
tosp=7863 fsc=34 stb=0 avsp=7791
block 555 failed with check code 6111
–//tosp – fsc – stb = avsp.
–//avsp+fsstb= 7791+34+0 = 7825.
bbed> assign kdbh.kdbhtosp=7825
sb2 kdbhtosp @112 7825
bbed> sum apply
check value for file 4, block 555:
current = 0xf2d1, required = 0xf2d1
bbed> verify
dbverify – verification starting
file = /mnt/ramdisk/book/users01.dbf
block = 555
dbverify – verification complete
total blocks examined : 1
total blocks processed (data) : 1
total blocks failing (data) : 0
total blocks processed (index): 0
total blocks failing (index): 0
total blocks empty : 0
total blocks marked corrupt : 0
total blocks influx : 0
message 531 not found; product=rdbms; facility=bbed
6.修复索引:
scott@book> alter index pk_empx rebuild online;
index altered.
–//注意一定要加online,不然不回表,无法修复错误索引错误.
scott@book> select rowid,empx.* from empx where empno=7566;
rowid empno ename job mgr hiredate sal comm deptno
—————— ———- ———- ——— ———- ——————- ———- ———- ———-
aaawefaaeaaaairaab 7566 jones manager 7839 1981-04-02 00:00:00 2975 20
总结:
–//修复cluster table要比普通表有难度.