Oracle中number数据类型存储的是整型,碰巧看到这篇文章讲解了通过分析索引了解0和1的存储机制,值得学习一下。
P.S. https://www.modb.pro/db/605566
create table t1 as select * from dba_objects;insert into t1 select * from t1; --执行5次commit;create table t2 as select * from t1;update t1 set object_name=null where object_id<7000;commit;update t2 set object_name=null where object_id<7000;commit;create index ind_t1_name on t1(object_name,0);create index ind_t2_name on t2(object_name,1);exec dbms_stats.gather_table_stats('MYTEST','T1',degree => 4,cascade => true,method_opt=>'for all columns size auto',estimate_percent=>100);exec dbms_stats.gather_table_stats('MYTEST','T2',degree => 4,cascade => true,method_opt=>'for all columns size auto',estimate_percent=>100);
通过检查索引大小及占用块数,我们发现,常数0的复合索引占用空间更小,
--查看表和索引大小col segment_name for a20select segment_name,segment_type,bytes/1024/1024 from user_segments where segment_name in('T1','T2','IND_T1_NAME','IND_T2_NAME');SEGMENT_NAME SEGMENT_TYPE BYTES/1024/1024-------------------- ------------------ ---------------IND_T1_NAME INDEX 120IND_T2_NAME INDEX 128T1 TABLE 352T2 TABLE 352col index_name for a20col table_name for a10col LEAF_BLOCKS for 99999999col NUM_ROWS for 99999999select index_name,TABLE_NAME,LEAF_BLOCKS,NUM_ROWS from user_indexes where index_name in ('IND_T1_NAME','IND_T2_NAME');INDEX_NAME TABLE_NAME LEAF_BLOCKS NUM_ROWS-------------------- ---------- ----------- ---------IND_T1_NAME T1 15004 2310176IND_T2_NAME T2 15328 2310176
IND_T1_NAME内容,
col object_name for a16select object_id,object_name,object_type from user_objects where object_name in ('IND_T1_NAME','IND_T2_NAME');OBJECT_ID OBJECT_NAME OBJECT_TYPE---------- ---------------- -----------------------242122 IND_T1_NAME INDEX242124 IND_T2_NAME INDEXALTER SESSION SET EVENTS 'immediate trace name treedump level 242122';----- begin tree dumpbranch: 0x2816183 42033539 (0: nrow: 98, level: 2)branch: 0x281aa74 42052212 (-1: nrow: 230, level: 1)leaf: 0x2816184 42033540 (-1: row:254.254 avs:820)leaf: 0x2816185 42033541 (0: row:199.199 avs:819)......leaf: 0x241dd88 37870984 (229: row:512.512 avs:828)leaf: 0x241dd89 37870985 (230: row:512.512 avs:828)select DBMS_UTILITY.data_block_address_file(37870984) RELATIVE_FNO,DBMS_UTILITY.data_block_address_block(37870984) blk_id from dual;--注意,上述转换是相对文件号,需要转换成绝对文件号file_id.select RELATIVE_FNO,FILE_ID from dba_data_files where RELATIVE_FNO=9;alter system dump datafile <file_id> block <block_id>;--IND_T1_NAME ,我们可以看到,索引有列,复合索引两列,第一列是空值,第二列是80,也就是我们写的0,第三列是rowid。Block header dump: 0x0241dd88Object id on Block? Yseg/obj: 0x3b1ca csc: 0x000000000c43aa16 itc: 2 flg: E typ: 2 - INDEXbrn: 0 bdba: 0x241da03 ver: 0x01 opc: 0inc: 0 exflg: 0Itl Xid Uba Flag Lck Scn/Fsc0x01 0x0000.000.00000000 0x00000000.0000.00 ---- 0 fsc 0x0000.000000000x02 0xffff.000.00000000 0x00000000.0000.00 C--- 0 scn 0x000000000c43aa16Leaf block dump===============header address 3595444324=0xd64e2064kdxcolev 0KDXCOLEV Flags = - - -kdxcolok 0kdxcoopc 0x80: opcode=0: iot flags=--- is converted=Ykdxconco 3kdxcosdc 0kdxconro 512kdxcofbo 1060=0x424kdxcofeo 1888=0x760kdxcoavs 828kdxlespl 0kdxlende 0kdxlenxt 37870985=0x241dd89kdxleprv 37870983=0x241dd87kdxledsz 0kdxlebksz 8032row#0[8020] flag: -------, lock: 0, len=12col 0; NULLcol 1; len 1; (1): 80col 2; len 6; (6): 02 81 3e ed 00 0b
IND_T2_NAME内容,
--通过上述方式,我们dump 索引IND_T2_NAMEselect DBMS_UTILITY.data_block_address_file(42066656) RELATIVE_FNO,(select file_id from dba_data_files where relative_fno=DBMS_UTILITY.data_block_address_file(42066656)) file_id,DBMS_UTILITY.data_block_address_block(42066656) blk_id from dual;RELATIVE_FNO FILE_ID BLK_ID------------ ---------- ----------10 15 123616alter system dump datafile 15 block 123616;--如下所示,最有一行,可看出,IND_T2_NAME 也有三列,空值列,常数1,rowid。其中常数1占用了两个字节Block header dump: 0x0281e2e0Object id on Block? Yseg/obj: 0x3b1cc csc: 0x000000000c43b071 itc: 2 flg: E typ: 2 - INDEXbrn: 0 bdba: 0x281e200 ver: 0x01 opc: 0inc: 0 exflg: 0Itl Xid Uba Flag Lck Scn/Fsc0x01 0x0000.000.00000000 0x00000000.0000.00 ---- 0 fsc 0x0000.000000000x02 0xffff.000.00000000 0x00000000.0000.00 C--- 0 scn 0x000000000c43b071Leaf block dump===============header address 139662300258404=0x7f05a9cba064kdxcolev 0KDXCOLEV Flags = - - -kdxcolok 0kdxcoopc 0x80: opcode=0: iot flags=--- is converted=Ykdxconco 3kdxcosdc 0kdxconro 478kdxcofbo 992=0x3e0kdxcofeo 1818=0x71akdxcoavs 826kdxlespl 0kdxlende 0kdxlenxt 42066657=0x281e2e1kdxleprv 42066655=0x281e2dfkdxledsz 0kdxlebksz 8032row#0[8019] flag: -------, lock: 0, len=13col 0; NULLcol 1; len 2; (2): c1 02col 2; len 6; (6): 02 81 a3 32 00 27
因此,我们能知道,常数0存储占用一个字节,常数1占用两个字节,这跟Oracle数据库存储number机制有关系。
因为有负数、小数点等,Oracle采用了如下方式表示,
Oracle中存储的number类型包含3个部分:HEAD(标记占用了几位),DATA,符号位。对正数来说,符号位省略,对0来说,只有80。
Oracle是以十六进制00-FF来表示所有的number,所以为了编码的对称,首先将number分为正负,所以以00-FF的中间位置80,即十进制的128来表示0,HEAD部分小于80,即为负数,大于80即为正数。
00-3E表示 x <= -1
3F-7F 表示 -1< x <0
81-C0 表示 0< x < 1
C1-FF 表示 1<= x
参考资料,
http://www.itpub.net/forum.php?mod=viewthread&tid=308317

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