[20150728]关于block record flag2.txt
--昨天被别人问及这个block record flag，google找到如下链接：
http://www.hcdba.com/?p=18
--做一个记录。
#define KDRHFK 0x80 Cluster Key
#define KDRHFC 0x40 Clustered table member
#define KDRHFH 0x20 Head piece of row
#define KDRHFD 0x10 Deleted row
#define KDRHFF 0x08 First data piece
#define KDRHFL 0x04 Last data piece
#define KDRHFP 0x02 First column continues from Previous piece
#define KDRHFN 0x01 Last column continues in Next piece
---------------------------------------------------
AC====>0x80+0x20+0x08+0x04 ===>dump中的(K-H-FL--)
6C====>0x40+0x20+0x08+0x04 ===>dump中的(-CH-FL--)
2C====>0x20+0x08+0x04 ===>dump中的(--H-FL--)
3C====>0x20+0x10+0x08+0x04 ===>dump中的(--HDFL--)
--为了加强记忆，通过一些例子讲解这8个bit。
--还剩下2个
#define KDRHFK 0x80 Cluster Key
#define KDRHFC 0x40 Clustered table member
--应该与cluster有关
1.建立测试例子：
SCOTT@test> @ver1
PORT_STRING VERSION BANNER
------------------------------ -------------- --------------------------------------------------------------------------------
x86_64/Linux 2.4.xx 11.2.0.3.0 Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production
--关于cluser 可以参考：
http://blog.itpub.net/267265/viewspace-1266411/
http://blog.itpub.net/267265/viewspace-1276613/
create cluster cluster_dept (deptno NUMBER(2)) ;
create index i_cluster_deptno on cluster cluster_dept;
create table dept1 cluster cluster_dept(deptno) as select * from dept;
create table emp1 cluster cluster_dept(deptno) as select * from emp;
SCOTT@test> select owner,object_name,object_id,data_object_id,object_type from dba_objects where owner=user and object_name in ('DEPT1','EMP1','CLUSTER_DEPT');
OWNER OBJECT_NAME OBJECT_ID DATA_OBJECT_ID OBJECT_TYPE
------ -------------------- ---------- -------------- -------------------
SCOTT CLUSTER_DEPT 291016 291016 CLUSTER
SCOTT DEPT1 291018 291016 TABLE
SCOTT EMP1 291019 291016 TABLE
--可以发现这些对象的DATA_OBJECT_ID=291016.
SCOTT@test> select owner,segment_name,segment_type,header_file,header_block,bytes from dba_segments where owner=user and segment_name in ('DEPT1','EMP1','CLUSTER_DEPT','I_CLUSTER_DEPTNO');
OWNER SEGMENT_NAME SEGMENT_TYPE HEADER_FILE HEADER_BLOCK BYTES
------ -------------------- ------------------ ----------- ------------ ----------
SCOTT I_CLUSTER_DEPTNO INDEX 4 1986 65536
SCOTT CLUSTER_DEPT CLUSTER 4 1978 131072
--注意1些细节，仅仅存在CLUSTER_DEPT,I_CLUSTER_DEPTNO段.
2.继续测试:
SCOTT@test> select rowid,dept1.* from dept1;
ROWID DEPTNO DNAME LOC
------------------ ---------- -------------- -------------
AABHDIAAEAAAAe7AAA 20 RESEARCH DALLAS1
AABHDIAAEAAAAe8AAA 30 SALES CHICAGO
AABHDIAAEAAAAe9AAA 40 OPERATIONS BOSTON
AABHDIAAEAAAAe+AAA 50 aaa bbb
AABHDIAAEAAAAe/AAA 10 ACCOUNTING NEW YORK
AABHDIAAEAAAAIgAAA 70 aaaa BBBB
AABHDIAAEAAAAInAAA 60 cc AAA
7 rows selected.
SCOTT@test> @ lookup_rowid AABHDIAAEAAAAe7AAA
OBJECT FILE BLOCK ROW DBA TEXT
---------- ---------- ---------- ---------- -------------------- ----------------------------------------
291016 4 1979 0 4,1979 alter system dump datafile 4 block 1979
--bbed观察：
BBED> set dba 4,1979
DBA 0x010007bb (16779195 4,1979)
BBED> p *kdbr[0]
rowdata[202]
------------
ub1 rowdata[202] @8166 0xac
BBED> x /rccn
rowdata[202] @8166
------------
flag@8166: 0xac (KDRHFL, KDRHFF, KDRHFH, KDRHFK)
lock@8167: 0x00
cols@8168: 1
kref@8169: 6
mref@8171: 6
hrid@8173:0x010007bb.0
nrid@8179:0x010007bb.0
col 0[2] @8185: ..
--flag =0xac，很奇怪col显示不正常！
SCOTT@test> @dfb16 0x010007bb
RFILE# BLOCK# TEXT
---------- ---------- ------------------------------------------------------------
4 1979 alter system dump datafile 4 block 1979 ;
SCOTT@test> alter system dump datafile 4 block 1979 ;
System altered.
--查看转储:
Block header dump: 0x010007bb
Object id on Block? Y
seg/obj: 0x470c8 csc: 0x02.a63f479c itc: 2 flg: E typ: 1 - DATA
brn: 0 bdba: 0x10007b8 ver: 0x01 opc: 0
inc: 0 exflg: 0
Itl Xid Uba Flag Lck Scn/Fsc
0x01 0x0001.00a.000044bd 0x00c005b5.1c5b.0f C--- 0 scn 0x0002.a63f4789
0x02 0x0005.008.0000725e 0x00c007a0.2518.1a --U- 5 fsc 0x0000.a63f47a3
bdba: 0x010007bb
data_block_dump,data header at 0x2a974de264
===============
tsiz: 0x1f98
hsiz: 0x28
pbl: 0x2a974de264
76543210
flag=--------
ntab=3
-- 从这里看出有3个表。
nrow=7
frre=-1
fsbo=0x28
fseo=0x1eb8
avsp=0x1e90
tosp=0x1e90
0xe:pti[0] nrow=1 offs=0
0x12:pti[1] nrow=1 offs=1
0x16:pti[2] nrow=5 offs=2
0x1a:pri[0] offs=0x1f82
0x1c:pri[1] offs=0x1f6d
0x1e:pri[2] offs=0x1f4a
0x20:pri[3] offs=0x1f24
0x22:pri[4] offs=0x1eff
0x24:pri[5] offs=0x1edc
0x26:pri[6] offs=0x1eb8
block_row_dump:
tab 0, row 0, @0x1f82
tl: 22 fb: K-H-FL-- lb: 0x0 cc: 1
curc: 6 comc: 6 pk: 0x010007bb.0 nk: 0x010007bb.0
col 0: [ 2] c1 15 => 对应数字20.
--flag=fb: K-H-FL--, 对应0xac
--组合起来：
#define KDRHFK 0x80 Cluster Key
#define KDRHFH 0x20 Head piece of row
#define KDRHFF 0x08 First data piece
#define KDRHFL 0x04 Last data piece
---------------------------------------
tab 1, row 0, @0x1f6d
tl: 21 fb: -CH-FL-- lb: 0x0 cc: 2 cki: 0
col 0: [ 8] 52 45 53 45 41 52 43 48
col 1: [ 7] 44 41 4c 4c 41 53 31
-------------------------------------
SCOTT@test> @conv_c 5245534541524348
C60
------------------------------------------------------------
RESEARCH
SCOTT@test> @conv_c 44414c4c415331
C60
------------------------------------------------------------
DALLAS1
SCOTT@test> select * from dept1 where deptno=20;
DEPTNO DNAME LOC
---------- -------------- -------------
20 RESEARCH DALLAS1
--可以发现是对上的。
--表1dept1，flag=fb: -CH-FL--，对应0x6c.
--组合起来：
#define KDRHFC 0x40 Clustered table member
#define KDRHFH 0x20 Head piece of row
#define KDRHFF 0x08 First data piece
#define KDRHFL 0x04 Last data piece
-----------------------------------------------------------------------
tab 2, row 0, @0x1f4a
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
SCOTT@test> @conv_n c24a46
N20
----------
7369
SCOTT@test> select * from emp1 where empno=7369;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- ------------------- ---------- ---------- ----------
7369 SMITH CLERK 7902 1980-12-17 00:00:00 800 20
--转换都是对上的。
总结：
--总结：
#define KDRHFK 0x80 Cluster Key =>使用K表示
#define KDRHFC 0x40 Clustered table member =>使用C表示
#define KDRHFH 0x20 Head piece of row =>使用H表示
#define KDRHFD 0x10 Deleted row =>使用D表示
#define KDRHFF 0x08 First data piece =>使用F表示
#define KDRHFL 0x04 Last data piece =>使用L表示
#define KDRHFP 0x02 First column continues from Previous piece =>使用P表示
#define KDRHFN 0x01 Last column continues in Next piece =>使用N表示