SQLServer BIT字段存储
Author:zfive5
Email:[email protected]
引子
和同事探讨BIT怎么存储，发生了分歧
create
table A1
(
a
CHAR(5),
b
bit,
c
CHAR(5),
d
BIT
)
由于之前成天再看《存储引擎》，深知这样定义表字段也会把两个BIT压缩成1个BYTE
而是同事说会分别存2个BYTE
开始证明
正文
在证明前，列出一些工具SQL:
|
CREATE FUNCTION convert_page_nums(@page_num binary(6)) RETURNS AS BEGIN RETURN( CONVERT(VARCHAR(2),( CONVERT(INT,SUBSTRING(@page_num,6,1))* ) (CONVERT(INT,SUBSTRING(@page_num,5,1))) ) CONVERT(VARCHAR(11), (CONVERT(INT,SUBSTRING(@page_num,4,1)) (CONVERT(INT,SUBSTRING(@page_num,3,1)) (CONVERT(INT,SUBSTRING(@page_num,2,1)) (CONVERT(INT,SUBSTRING(@page_num,1,1))) ) ) END select allocation_unit_id from sys.partitions on p.partition_id where select i.name partition_id,partition_number allocation_unit_id from sys.indexes join sys.allocation_units where i.object_id |
执行如下SQL:
select
convert(char(8),object_name(i.object_id))
as table_name,
indexproperty(i.object_id,i.name,‘minlen‘)
as minlen,
i.name
as index_name,i.index_id,i.type_desc
as index_type,
partition_id,partition_number
as pnum,first_page,rows,
a.allocation_unit_id
as au_id,a.type_desc
as page_type_desc,a.total_pages
as pages
from sys.indexes
i join
sys.partitions
p on
i.object_id
= p.object_id
and i.index_id
= p.index_id
join sys.system_internals_allocation_units
a on
p.partition_id
= a.container_id
where i.object_id
= object_id(‘dbo.a1‘)
SELECT DBO.convert_page_nums(0xEC2D00000100)
执行如下：
insert A1(a,b,c,d)
values(‘AAAAA‘,1,‘BBBBB‘,1)
insert A1(a,b,c,d)
values(‘BBBBB‘,0,‘CCCCC‘,0)
insert A1(a,b,c,d)
values(‘CCCCC‘,0,‘DDDDD‘,1)
insert A1(a,b,c,d)
values(‘DDDDD‘,1,‘FFFFF‘,0)
dbcc TRACEon(3604)
DBCC page
(A,1,121,3)
得到如下信息：
|
DBCC 执行完毕。如果 DBCC 输出了错误信息，请与系统管理员联系。 PAGE: (1:121) BUFFER: BUF @0x000000046E165B80 bpage = 0x000000045DDDA000 bhash = 0x0000000000000000 bpageno = (1:121) bdbid = 9 breferences = 0 bcputicks = 0 bsampleCount = 0 bUse1 = 3353 bstat = 0x10b blog = 0x15acc bnext = 0x0000000000000000 PAGE HEADER: Page @0x000000045DDDA000 m_pageId = (1:121) m_headerVersion = 1 m_type = 1 m_typeFlagBits = 0x0 m_level = 0 m_flagBits = 0x8000 m_objId (AllocUnitId.idObj) = 85 m_indexId (AllocUnitId.idInd) = 256 Metadata: AllocUnitId = 72057594043498496 Metadata: PartitionId = 72057594039107584 Metadata: IndexId = 0 Metadata: ObjectId = 581577110 m_prevPage = (0:0) m_nextPage = (0:0) pminlen = 15 m_slotCnt = 4 m_freeCnt = 8016 m_freeData = 168 m_reservedCnt = 0 m_lsn = (34:25:2) m_xactReserved = 0 m_xdesId = (0:0) m_ghostRecCnt = 0 m_tornBits = 0 DB Frag ID = 1 Allocation Status GAM (1:2) = ALLOCATED SGAM (1:3) = ALLOCATED PFS (1:1) = 0x61 MIXED_EXT ALLOCATED 50_PCT_FULL DIFF (1:6) = CHANGED ML (1:7) = NOT MIN_LOGGED Slot 0 Offset 0x60 Length 18 Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP Record Size = 18 Memory Dump @0x0000000013D6A060 0000000000000000: 10000f00 41414141 41034242 42424204 0000 ....AAAAA.BBBBB... Slot 0 Column 1 Offset 0x4 Length 5 Length (physical) 5 a = AAAAA Slot 0 Column 2 Offset 0x9 Length 1 (Bit position 0) b = 1 Slot 0 Column 3 Offset 0xa Length 5 Length (physical) 5 c = BBBBB Slot 0 Column 4 Offset 0x9 Length 1 (Bit position 1) d = 1 Slot 1 Offset 0x72 Length 18 Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP Record Size = 18 Memory Dump @0x0000000013D6A072 0000000000000000: 10000f00 42424242 42004343 43434304 0000 ....BBBBB.CCCCC... Slot 1 Column 1 Offset 0x4 Length 5 Length (physical) 5 a = BBBBB Slot 1 Column 2 Offset 0x9 Length 1 (Bit position 0) b = 0 Slot 1 Column 3 Offset 0xa Length 5 Length (physical) 5 c = CCCCC Slot 1 Column 4 Offset 0x9 Length 1 (Bit position 1) d = 0 Slot 2 Offset 0x84 Length 18 Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP Record Size = 18 Memory Dump @0x0000000013D6A084 0000000000000000: 10000f00 43434343 43024444 44444404 0000 ....CCCCC.DDDDD... Slot 2 Column 1 Offset 0x4 Length 5 Length (physical) 5 a = CCCCC Slot 2 Column 2 Offset 0x9 Length 1 (Bit position 0) b = 0 Slot 2 Column 3 Offset 0xa Length 5 Length (physical) 5 c = DDDDD Slot 2 Column 4 Offset 0x9 Length 1 (Bit position 1) d = 1 Slot 3 Offset 0x96 Length 18 Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP Record Size = 18 Memory Dump @0x0000000013D6A096 0000000000000000: 10000f00 44444444 44014646 46464604 0000 ....DDDDD.FFFFF... Slot 3 Column 1 Offset 0x4 Length 5 Length (physical) 5 a = DDDDD Slot 3 Column 2 Offset 0x9 Length 1 (Bit position 0) b = 1 Slot 3 Column 3 Offset 0xa Length 5 Length (physical) 5 c = FFFFF Slot 3 Column 4 Offset 0x9 Length 1 (Bit position 1) d = 0 DBCC 执行完毕。如果 DBCC 输出了错误信息，请与系统管理员联系。 |
更直观的比较
1、10000f00 41414141 41034242 42424204 0000
2、10000f00 42424242 42004343 43434304 0000
3、10000f00 44444444 44014646 46464604 0000
4、10000f00 44444444 44014646 46464604 0000
二进制的中间分别为 03 00 02 01
insert A1(a,b,c,d)
values(‘AAAAA‘,1,‘BBBBB‘,1)
03
insert A1(a,b,c,d)
values(‘BBBBB‘,0,‘CCCCC‘,0)
00
insert A1(a,b,c,d)
values(‘CCCCC‘,0,‘DDDDD‘,1)
02
insert A1(a,b,c,d)
values(‘DDDDD‘,1,‘FFFFF‘,0)
01
足以证明SQLServer 不管创建的顺序，都是试图压缩到一个BYTE上去，当然超过8位，会再一次申请一个新BYTE。
SQLServer BIT字段存储,布布扣,bubuko.com