so unique non-clustered index created on unique clustered index use smallest space, while non-unique non-clustered index created on non-unique clustered index use largest space.
Showing posts with label SQL Internal. Show all posts
Showing posts with label SQL Internal. Show all posts
Friday, January 4, 2013
Row of nonclustered index page
Below is the formula of calculating the row size of non-clustered index page
so unique non-clustered index created on unique clustered index use smallest space, while non-unique non-clustered index created on non-unique clustered index use largest space.
so unique non-clustered index created on unique clustered index use smallest space, while non-unique non-clustered index created on non-unique clustered index use largest space.
Thursday, January 3, 2013
Heap might take more space than clustered Index when inserting row
Sometimes when you insert row on heap, even if the page has enough free space, the new row can not be inserted into that page, and a new page will be created for the new row. Here is interesting example:
1. Create testing DB
use master
go
CREATE DATABASE [test] ON PRIMARY
( NAME = N'test', FILENAME = N'C:\Program Files\Microsoft SQL Server\MSSQL10_50.MSSQLSERVER\MSSQL\DATA\test.mdf' , SIZE = 2048KB , FILEGROWTH = 1024KB )
LOG ON
( NAME = N'test_log', FILENAME = N'C:\Program Files\Microsoft SQL Server\MSSQL10_50.MSSQLSERVER\MSSQL\DATA\test_log.ldf' , SIZE = 1024KB , FILEGROWTH = 1024KB)
GO
2. Create table
CREATE TABLE myHeap
(
a char(500) NOT NULL,
b char(100) NOT NULL,
c char(153) NOT NULL,
) ON [PRIMARY]
GO
1. Create testing DB
use master
go
CREATE DATABASE [test] ON PRIMARY
( NAME = N'test', FILENAME = N'C:\Program Files\Microsoft SQL Server\MSSQL10_50.MSSQLSERVER\MSSQL\DATA\test.mdf' , SIZE = 2048KB , FILEGROWTH = 1024KB )
LOG ON
( NAME = N'test_log', FILENAME = N'C:\Program Files\Microsoft SQL Server\MSSQL10_50.MSSQLSERVER\MSSQL\DATA\test_log.ldf' , SIZE = 1024KB , FILEGROWTH = 1024KB)
GO
2. Create table
CREATE TABLE myHeap
(
a char(500) NOT NULL,
b char(100) NOT NULL,
c char(153) NOT NULL,
) ON [PRIMARY]
GO
so the row length will be 753+7=760 bytes, in theory, the every page can hold 10 rows(8060/760=10)
3, insert 10 rows
Declare @int int
set @int=1
while (@int <= 10)
begin
insert into myHeap(a,b,c)
values(right('00000'+CONVERT(varchar(5),@int),5),'bbbbb','ccccc')
set @int=@int+1
end
4. check the table page status
SELECT %%physloc%% rowid, sys.fn_PhysLocFormatter (%%physloc%%) AS [Physical RID],DATALENGTH(a) a_length, DATALENGTH(b) b_length,DATALENGTH(c) d_length, *
FROM test.dbo.myHeap
| rowid | Physical RID | a_length | b_length | d_length |
| 0x9900000001000000 | (1:153:0) | 500 | 100 | 153 |
| 0x9900000001000100 | (1:153:1) | 500 | 100 | 153 |
| 0x9900000001000200 | (1:153:2) | 500 | 100 | 153 |
| 0x9900000001000300 | (1:153:3) | 500 | 100 | 153 |
| 0x9900000001000400 | (1:153:4) | 500 | 100 | 153 |
| 0x9900000001000500 | (1:153:5) | 500 | 100 | 153 |
| 0x9900000001000600 | (1:153:6) | 500 | 100 | 153 |
| 0x9900000001000700 | (1:153:7) | 500 | 100 | 153 |
| 0x9900000001000800 | (1:153:8) | 500 | 100 | 153 |
| 0x9B00000001000000 | (1:155:0) | 500 | 100 | 153 |
in fact, the first 9 rows are in page 153, however, the 10th row are in the other page 155.
4. let's check how much free space is in page 153
DBCC TRACEON(3604)
GO
dbcc page(test,1,153,3)
GO
m_pageId = (1:153) m_headerVersion = 1 m_type = 1
m_typeFlagBits = 0x4 m_level = 0 m_flagBits = 0x8000
m_objId (AllocUnitId.idObj) = 27 m_indexId (AllocUnitId.idInd) = 256
Metadata: AllocUnitId = 72057594039697408
Metadata: PartitionId = 72057594038779904 Metadata: IndexId = 0
Metadata: ObjectId = 2105058535 m_prevPage = (0:0) m_nextPage = (0:0)
pminlen = 757 m_slotCnt = 9 m_freeCnt = 1238
m_freeData = 6936 m_reservedCnt = 0 m_lsn = (32:86:2)
m_xactReserved = 0 m_xdesId = (0:0) m_ghostRecCnt = 0
m_tornBits = 0
Allocation Status
GAM (1:2) = ALLOCATED SGAM (1:3) = ALLOCATED
PFS (1:1) = 0x63 MIXED_EXT ALLOCATED 95_PCT_FULL DIFF (1:6) = CHANGED
ML (1:7) = NOT MIN_LOGGED
Slot 0 Offset 0x60 Length 760
Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP Record Size = 760
well, there is 1238 bytes free space, the row size is 760 byte, so there is enough free space to hold the 10th rows. why sql server create a new page to keep 10th row??? here we found a interesting info in the page head : PFS indicts the page is 95% full, so there is only 8060*5%=403 bytes free space.
let's review how sql server insert rows on Heap, because there is no clustered index, a new row is always inserted wherever room is available in the heap table. sql server will use IAMs and the PFS pages to keep track of which extents in ta file already belong to a table and which of the pages in those extens have space available.
5. let's check the PFS page:
DBCC TRACEON(3604)
GO
dbcc page(test,1,1,3)
GO
(1:153) - = ALLOCATED 95_PCT_FULL Mixed Ext
The PFS page contains 1 byte for each page in a 8088-age range of a file, and the last three bits are used to indicate the age fullness.
The bits in each byte has some meaning as illustrated below
- bits 0-2: how much free space is on the page
- 000 is empty
- 001 is 1 to 50% full
- 010 is 51 to 80% full
- 011 is 81 to 95% full
- 100 is 96 to 100% full
- bit 3 : 1 if page has ghost records else 0
- bit 4 : 1 if an IAM page else 0
- bit 5 : 1 if mixed extent, 0 if uniform
- bit 6 : 1 if page allocated, 0 if unallocated
we have 9 rows in page 153, 9*760/8060 =84.9% full, so PFS indicts it is 95% full. when sql server check page 153, it think the free space on page 153 is 8060*5%=403 bytes < row size 760bytes, so a new page 156 was created for 10th row.
6. In order to save space and keep the 10 rows in the same page, we can create a clustered index,
ALTER TABLE dbo.myheap ADD CONSTRAINT
PK_mytest PRIMARY KEY CLUSTERED
(
a
) WITH( STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
GO
then we can find all 10 rows are in the same page
SELECT %%physloc%% rowid, sys.fn_PhysLocFormatter (%%physloc%%) AS [Physical RID],DATALENGTH(a) a_length, DATALENGTH(b) b_length,DATALENGTH(c) d_length, *
FROM test.dbo.myHeap
| rowid | Physical RID | a_length | b_length | d_length |
| 0x9C00000001000000 | (1:156:0) | 500 | 100 | 153 |
| 0x9C00000001000100 | (1:156:1) | 500 | 100 | 153 |
| 0x9C00000001000200 | (1:156:2) | 500 | 100 | 153 |
| 0x9C00000001000300 | (1:156:3) | 500 | 100 | 153 |
| 0x9C00000001000400 | (1:156:4) | 500 | 100 | 153 |
| 0x9C00000001000500 | (1:156:5) | 500 | 100 | 153 |
| 0x9C00000001000600 | (1:156:6) | 500 | 100 | 153 |
| 0x9C00000001000700 | (1:156:7) | 500 | 100 | 153 |
| 0x9C00000001000800 | (1:156:8) | 500 | 100 | 153 |
| 0x9C00000001000900 | (1:156:9) | 500 | 100 | 153 |
DBCC TRACEON(3604)
GO
dbcc page(test,1,156,3)
GO
m_pageId = (1:156) m_headerVersion = 1 m_type = 1
m_typeFlagBits = 0x0 m_level = 0 m_flagBits = 0x4
m_objId (AllocUnitId.idObj) = 28 m_indexId (AllocUnitId.idInd) = 256
Metadata: AllocUnitId = 72057594039762944
Metadata: PartitionId = 72057594038845440 Metadata: IndexId = 1
Metadata: ObjectId = 2105058535 m_prevPage = (0:0) m_nextPage = (0:0)
pminlen = 757 m_slotCnt = 10 m_freeCnt = 476
m_freeData = 7696 m_reservedCnt = 0 m_lsn = (32:93:21)
m_xactReserved = 0 m_xdesId = (0:0) m_ghostRecCnt = 0
m_tornBits = 0
Allocation Status
GAM (1:2) = ALLOCATED SGAM (1:3) = ALLOCATED
PFS (1:1) = 0x60 MIXED_EXT ALLOCATED 0_PCT_FULL DIFF (1:6) = CHANGED
this example is one of the reasons why clustered index is recommended on big table
reference
http://sqlsimplified.blogspot.com/2012/01/page-free-space.html
Sunday, November 25, 2012
Explore file physical structure - IAM
1. Create sample db
use master
go
CREATE DATABASE [test] ON PRIMARY
( NAME = N'test', FILENAME = N'C:\Program Files\Microsoft SQL Server\MSSQL10_50.MSSQLSERVER\MSSQL\DATA\master\test.mdf' , SIZE = 2048KB , FILEGROWTH = 1024KB )
LOG ON
( NAME = N'test_log', FILENAME = N'C:\Program Files\Microsoft SQL Server\MSSQL10_50.MSSQLSERVER\MSSQL\DATA\master\test_log.ldf' , SIZE = 1024KB , FILEGROWTH = 1024KB)
GO
2. Create table
use test
GO
BEGIN TRANSACTION
GO
CREATE TABLE dbo.mytest
(
a int NOT NULL,
b varchar(500) NOT NULL,
d varchar(400) NULL,
) ON [PRIMARY]
GO
ALTER TABLE dbo.mytest ADD CONSTRAINT
PK_mytest PRIMARY KEY CLUSTERED
(
a
) WITH( STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
GO
COMMIT
3. Insert test data
use test
GO
Declare @int int
set @int=1
while (@int <= 300)
begin
insert into mytest(a,b,d)
values(@int,CONVERT(char(500),@int),CONVERT(char(400),@int))
set @int=@int+1
end
4. check IAM chain
DBCC TRACEON(3604)
GO
dbcc
ind ('test','mytest',1)
GO
output of the first 4 rows :
| PageFID | PagePID | IAMFID | IAMPID | ObjectID | IndexID | PartitionNumber | PartitionID | iam_chain_type | PageType |
| 1 | 154 | NULL | NULL | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 10 |
| 1 | 153 | 1 | 154 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 1 |
| 1 | 155 | 1 | 154 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 2 |
| 1 | 156 | 1 | 154 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 1 |
here the IAM page is pagePID 154type =10).
5. check the page 153
DBCC TRACEON(3604)
GO
dbcc page(test,1,154,3)
GO
IAM: Single Page Allocations @0x00000000109EA08E
Slot 0 = (1:153) Slot 1 = (1:155) Slot 2 = (1:156)
Slot 3 = (1:157) Slot 4 = (1:158) Slot 5 = (1:159)
Slot 6 = (1:176) Slot 7 = (1:177)
IAM: Extent Alloc Status Slot 1 @0x00000000109EA0C2
(1:0) - (1:176) = NOT ALLOCATED
(1:184) - (1:208) = ALLOCATED
(1:216) - (1:248) = NOT ALLOCATED
so the first 8 page are single-page allocation page, after first 8 page, sql use uniform extents.
6. you can verify this by DBCC Extentinfo
dbcc extentinfo ( 'test','mytest',1 )
go
| file_id | page_id | pg_alloc | ext_size | object_id | index_id | partition_number | partition_id | iam_chain_type | pfs_bytes |
| 1 | 153 | 1 | 1 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x6000000000000000 |
| 1 | 155 | 1 | 1 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x6000000000000000 |
| 1 | 156 | 1 | 1 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x6000000000000000 |
| 1 | 157 | 1 | 1 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x6000000000000000 |
| 1 | 158 | 1 | 1 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x6000000000000000 |
| 1 | 159 | 1 | 1 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x6000000000000000 |
| 1 | 176 | 1 | 1 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x6000000000000000 |
| 1 | 177 | 1 | 1 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x6000000000000000 |
| 1 | 184 | 8 | 8 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x4040404040404040 |
| 1 | 192 | 8 | 8 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x4040404040404040 |
| 1 | 200 | 8 | 8 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x4040404040404040 |
| 1 | 208 | 7 | 8 | 2105058535 | 1 | 1 | 72057594038845440 | In-row data | 0x4040404040404000 |
so the first 8 pages are single-page allocations, then from page 184, the ext_size is 8(pages), they are uniform extent.
7. -T1118
Trace flag -T1118 can force uniform extent allocation, if you start sql server service with -T1118. then rebuild the index.
alter index PK_mytest on mytest rebuild
go
dbcc extentinfo ( 'test','mytest',1 )
go
| file_id | page_id | pg_alloc | ext_size | object_id | index_id | partition_number | partition_id | iam_chain_type | pfs_bytes |
| 1 | 184 | 1 | 8 | 2105058535 | 1 | 1 | 72057594038976512 | In-row data | 0x4000000000000000 |
| 1 | 192 | 8 | 8 | 2105058535 | 1 | 1 | 72057594038976512 | In-row data | 0x4040404040404040 |
| 1 | 200 | 8 | 8 | 2105058535 | 1 | 1 | 72057594038976512 | In-row data | 0x4040404040404040 |
| 1 | 208 | 8 | 8 | 2105058535 | 1 | 1 | 72057594038976512 | In-row data | 0x4040404040404040 |
| 1 | 256 | 8 | 8 | 2105058535 | 1 | 1 | 72057594038976512 | In-row data | 0x4040404040404040 |
| 1 | 264 | 1 | 8 | 2105058535 | 1 | 1 | 72057594038976512 | In-row data | 0x4000000000000000 |
| 1 | 272 | 5 | 8 | 2105058535 | 1 | 1 | 72057594038976512 | In-row data | 0x4040404040000000 |
now, all pages are in uniform extents(ext_size=8)
you can also get the IAM page number by query below:
select sys.fn_PhysLocFormatter(first_iam_page) from sys.system_internals_allocation_units sau
join sys.partitions pt on sau.container_id = pt.partition_id and pt.object_id = object_id('mytest')
reference:
http://sqlsimplified.blogspot.com/2012/01/as-bol-says-index-allocation-map-iam.html
http://www.sqlskills.com/blogs/paul/post/Inside-the-Storage-Engine-IAM-pages-IAM-chains-and-allocation-units.aspx
Wednesday, November 21, 2012
Explore file physical structure - Delete
1. Create sample db
use master
go
CREATE DATABASE [test] ON PRIMARY
( NAME = N'test', FILENAME = N'C:\Program Files\Microsoft SQL Server\MSSQL10_50.MSSQLSERVER\MSSQL\DATA\master\test.mdf' , SIZE = 2048KB , FILEGROWTH = 1024KB )
LOG ON
( NAME = N'test_log', FILENAME = N'C:\Program Files\Microsoft SQL Server\MSSQL10_50.MSSQLSERVER\MSSQL\DATA\master\test_log.ldf' , SIZE = 1024KB , FILEGROWTH = 1024KB)
GO
2. Create table
use test
GO
BEGIN TRANSACTION
GO
CREATE TABLE dbo.mytest
(
a int NOT NULL,
b varchar(500) NOT NULL,
d varchar(400) NULL,
) ON [PRIMARY]
GO
ALTER TABLE dbo.mytest ADD CONSTRAINT
PK_mytest PRIMARY KEY CLUSTERED
(
a
) WITH( STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
GO
COMMIT
3. Insert test data
use test
go
Declare @int int
set @int=1
while (@int <= 3000)
begin
insert into mytest(a,b,d)
values(@int,CONVERT(char(500),@int),CONVERT(char(400),@int))
set @int=@int+1
end
4. select data and check page number
use master
GO
SELECT %%physloc%% rowid, sys.fn_PhysLocFormatter (%%physloc%%) AS [Physical RID],DATALENGTH(a) a_length, DATALENGTH(b) b_length,DATALENGTH(d) d_length, *
FROM test.dbo.mytest where a<9
GO
output
| rowid | Physical RID | a_length | b_length | d_length | a | b | d |
| 0x9900000001000000 | (1:153:0) | 4 | 500 | 400 | 1 | 1 | 1 |
| 0x9900000001000100 | (1:153:1) | 4 | 500 | 400 | 2 | 2 | 2 |
| 0x9900000001000200 | (1:153:2) | 4 | 500 | 400 | 3 | 3 | 3 |
| 0x9900000001000300 | (1:153:3) | 4 | 500 | 400 | 4 | 4 | 4 |
| 0x9900000001000400 | (1:153:4) | 4 | 500 | 400 | 5 | 5 | 5 |
| 0x9900000001000500 | (1:153:5) | 4 | 500 | 400 | 6 | 6 | 6 |
| 0x9900000001000600 | (1:153:6) | 4 | 500 | 400 | 7 | 7 | 7 |
| 0x9900000001000700 | (1:153:7) | 4 | 500 | 400 | 8 | 8 | 8 |
so 8 rows in pages: (1:153)
5. delete row and print page
use test
go
delete from mytest where a=3
go
DBCC TRACEON(3604)
GO
dbcc page(test,1,153,2)
GO
6. check page 153.
Here I only print the content which is changed after data deleting
a) in page header
m_xactReserved = 0 m_xdesId = (0:4099) m_ghostRecCnt = 1
#m_ghostRecCnt=1: there is 1 ghost record
b) in the data section:
000000001485A780: 20202020 20202020 20203c00 08000300 ? <.....
000000001485A790: 00000300 00020005 02950333 20202020 ?.........?.3
000000001485A7A0: 20202020 20202020 20202020 20202020 ?
#the first 2 bytes of row(a=3) is changed from "30" to "3c"= 0011 1100
0011=Bit 4 means the record has a NULL bitmap and bit 5 means the record has variable length columns
1100 : Bits 1-3 of byte 0 give the record type, 110=6=ghost record
c) there is no change in the offset table, still 8 records in it.
Row - Offset
7 (0x7) - 6515 (0x1973)
6 (0x6) - 5598 (0x15de)
5 (0x5) - 4681 (0x1249)
4 (0x4) - 3764 (0xeb4)
3 (0x3) - 2847 (0xb1f)
2 (0x2) - 1930 (0x78a)
1 (0x1) - 1013 (0x3f5)
0 (0x0) - 96 (0x60)
7. wait for while and check the page again
DBCC TRACEON(3604)
GO
dbcc page(test,1,153,2)
GO
a) page header
m_xactReserved = 0 m_xdesId = (0:4099) m_ghostRecCnt = 0
#ghost record is gone now
b) row data
00000000140DA780: 20202020 20202020 20203c00 08000300 ? <.....
00000000140DA790: 00000300 00020005 02950333 20202020 ?.........?.3
00000000140DA7A0: 20202020 20202020 20202020 20202020 ?
#still no change in the row data
c) offset table
Row - Offset
6 (0x6) - 6515 (0x1973)
5 (0x5) - 5598 (0x15de)
4 (0x4) - 4681 (0x1249)
3 (0x3) - 3764 (0xeb4)
2 (0x2) - 2847 (0xb1f)
1 (0x1) - 1013 (0x3f5)
0 (0x0) - 96 (0x60)
#only 7 rows now, the row (a=3) is gone from the offset table
Here is a question:
if we insert the same data back, is it inserted into the same place? or reorganize the page?
8) insert data back again.
use test
go
insert into mytest(a,b,d)
values(3,CONVERT(char(500),3),CONVERT(char(400),3))
GO
use master
go
SELECT %%physloc%% rowid, sys.fn_PhysLocFormatter (%%physloc%%) AS [Physical RID],DATALENGTH(a) a_length, DATALENGTH(b) b_length,DATALENGTH(d) d_length, *
FROM test.dbo.mytest where a<10
GO
output:
| rowid | Physical RID | a_length | b_length | d_length | a | b | d |
| 0x9900000001000000 | (1:153:0) | 4 | 500 | 400 | 1 | 1 | 1 |
| 0x9900000001000100 | (1:153:1) | 4 | 500 | 400 | 2 | 2 | 2 |
| 0x9900000001000200 | (1:153:2) | 4 | 500 | 400 | 3 | 3 | 3 |
| 0x9900000001000300 | (1:153:3) | 4 | 500 | 400 | 4 | 4 | 4 |
| 0x9900000001000400 | (1:153:4) | 4 | 500 | 400 | 5 | 5 | 5 |
| 0x9900000001000500 | (1:153:5) | 4 | 500 | 400 | 6 | 6 | 6 |
| 0x9900000001000600 | (1:153:6) | 4 | 500 | 400 | 7 | 7 | 7 |
| 0x9900000001000700 | (1:153:7) | 4 | 500 | 400 | 8 | 8 | 8 |
so confirmed the original row(a=3) is inserted into page 153, and rowid is still 1:153:2.
9) let's check the page
DBCC TRACEON(3604)
GO
dbcc page(test,1,153,2)
GO
the Most important change is:
the page is reorganized, rows (4,5,6,7) have been moved up, and the deleted slot is overwrote. then the new row is added at the tail of the original content. here is the page, I marked the row header with red
PAGE: (1:153)
DATA:
Memory Dump @0x00000000140DA000
00000000140DA000: 01010400 00000001 00000000 00000800 ?................
00000000140DA010: 9c000000 01000800 1c000000 e802081d ??...........è...
00000000140DA020: 99000000 01000000 30000000 8d010000 ?........0.......
00000000140DA030: 04000000 03100000 00000000 6f0c6a1c ?............o.j.
00000000140DA040: 01000000 00000000 00000000 00000000 ?................
00000000140DA050: 00000000 00000000 00000000 00000000 ?................
00000000140DA060: 30000800 01000000 03000002 00050295 ?0..............?
00000000140DA070: 03312020 20202020 20202020 20202020 ?.1
00000000140DA080: 20202020 20202020 20202020 20202020 ?
00000000140DA090: 20202020 20202020 20202020 20202020 ?
00000000140DA0A0: 20202020 20202020 20202020 20202020 ?
00000000140DA0B0: 20202020 20202020 20202020 20202020 ?
00000000140DA0C0: 20202020 20202020 20202020 20202020 ?
00000000140DA0D0: 20202020 20202020 20202020 20202020 ?
00000000140DA0E0: 20202020 20202020 20202020 20202020 ?
00000000140DA0F0: 20202020 20202020 20202020 20202020 ?
00000000140DA100: 20202020 20202020 20202020 20202020 ?
00000000140DA110: 20202020 20202020 20202020 20202020 ?
00000000140DA120: 20202020 20202020 20202020 20202020 ?
00000000140DA130: 20202020 20202020 20202020 20202020 ?
00000000140DA140: 20202020 20202020 20202020 20202020 ?
00000000140DA150: 20202020 20202020 20202020 20202020 ?
00000000140DA160: 20202020 20202020 20202020 20202020 ?
00000000140DA170: 20202020 20202020 20202020 20202020 ?
00000000140DA180: 20202020 20202020 20202020 20202020 ?
00000000140DA190: 20202020 20202020 20202020 20202020 ?
00000000140DA1A0: 20202020 20202020 20202020 20202020 ?
00000000140DA1B0: 20202020 20202020 20202020 20202020 ?
00000000140DA1C0: 20202020 20202020 20202020 20202020 ?
00000000140DA1D0: 20202020 20202020 20202020 20202020 ?
00000000140DA1E0: 20202020 20202020 20202020 20202020 ?
00000000140DA1F0: 20202020 20202020 20202020 20202020 ?
00000000140DA200: 20202020 20202020 20202020 20202020 ?
00000000140DA210: 20202020 20202020 20202020 20202020 ?
00000000140DA220: 20202020 20202020 20202020 20202020 ?
00000000140DA230: 20202020 20202020 20202020 20202020 ?
00000000140DA240: 20202020 20202020 20202020 20202020 ?
00000000140DA250: 20202020 20202020 20202020 20202020 ?
00000000140DA260: 20202020 20312020 20202020 20202020 ? 1
00000000140DA270: 20202020 20202020 20202020 20202020 ?
00000000140DA280: 20202020 20202020 20202020 20202020 ?
00000000140DA290: 20202020 20202020 20202020 20202020 ?
00000000140DA2A0: 20202020 20202020 20202020 20202020 ?
00000000140DA2B0: 20202020 20202020 20202020 20202020 ?
00000000140DA2C0: 20202020 20202020 20202020 20202020 ?
00000000140DA2D0: 20202020 20202020 20202020 20202020 ?
00000000140DA2E0: 20202020 20202020 20202020 20202020 ?
00000000140DA2F0: 20202020 20202020 20202020 20202020 ?
00000000140DA300: 20202020 20202020 20202020 20202020 ?
00000000140DA310: 20202020 20202020 20202020 20202020 ?
00000000140DA320: 20202020 20202020 20202020 20202020 ?
00000000140DA330: 20202020 20202020 20202020 20202020 ?
00000000140DA340: 20202020 20202020 20202020 20202020 ?
00000000140DA350: 20202020 20202020 20202020 20202020 ?
00000000140DA360: 20202020 20202020 20202020 20202020 ?
00000000140DA370: 20202020 20202020 20202020 20202020 ?
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00000000140DBFB0: af0a6a0a 210ad609 8f094c09 0509c008 ?ˉ.j.!.? . L . à.
00000000140DBFC0: 73082a08 df079807 57071007 d1068c06 ?s.*.?...W...?.?.
00000000140DBFD0: 47060806 c7058405 43050205 bb047a04 ?G...?.?.C...?.z.
00000000140DBFE0: 3504f203 ab036a03 2903e402 a3025a02 ?5.ò.?.j.).?.￡.Z.
00000000140DBFF0: de154912 b40e1f0b 8a077319 f5036000 ?T.I.′...?.s.?.`.
OFFSET TABLE:
Row - Offset
7 (0x7) - 5598 (0x15de)
6 (0x6) - 4681 (0x1249)
5 (0x5) - 3764 (0xeb4)
4 (0x4) - 2847 (0xb1f)
3 (0x3) - 1930 (0x78a)
2 (0x2) - 6515 (0x1973)
1 (0x1) - 1013 (0x3f5)
0 (0x0) - 96 (0x60)
DBCC execution completed. If DBCC printed error messages, contact your system administrator.
so SQL Server did a huge work for the insert operation, it re-organize the page, then insert the data.
10) last time we inserted a same length row, what happen if I insert a small row? does the page re-organize happen again? let's test it
11) delete row(a=4)
delete from mytest where a=4
12) wait for a while, and make sure the ghost row is gone,
13) insert a small row with same cluster key(4)
insert into mytest(a,b,d)
values(4,CONVERT(char(50),4),CONVERT(char(50),4))
14) check the page
you can find the deleted row is not overwrote, there is no page re-organize, the new row is just inserted at the tail of the existing Data. that's because there is still enough free space at the page tail, while last time the free space at the page tail is not enough to save the new row, so the page need to be re-organized.
In a summary:
1. When delete row, the row is just marked a ghost row or delete row with flag. then ghost clean up thread will remove the ghost row by updating the offset table.
2. when insert row.
a)if there is enough free space in the page,
i) if there is enough free space from the end of last row to beginning of the offset table, insert it direct after the last row.
ii) or else, re-organize the page, then insert the new row after the last row.
b) if there is no enough free space in the page, then create a new page, move half data to the new page, then insert the new row in the old page.
15) what happen if we delete all the rows in the page?let's run
delete from mytest where a<9
when checking page, we can see all rows are marked as ghost row in the row header. however, in the offset table, ghost clean up thread doesn't remove all the slot, it still keep one row in the offset table.
Row - Offset
0 (0x0) - 5598 (0x15de)
let's check if the page has been removed from page chain.
select sys.fn_PhysLocFormatter(first_page) from sys.system_internals_allocation_units where allocation_unit_id=72057594039762944
output is (1:153:0)
so page 153 is still in the page chain. what happen if we delete other page? let's run
delete from mytest where a<=24 and a>=17
we deleted all 8 rows from page 157, check the page:
m_pageId = (1:157) m_headerVersion = 1 m_type = 1
m_typeFlagBits = 0x4 m_level = 0 m_flagBits = 0x0
m_objId (AllocUnitId.idObj) = 28 m_indexId (AllocUnitId.idInd) = 256
Metadata: AllocUnitId = 72057594039762944
Metadata: PartitionId = 72057594038845440 Metadata: IndexId = 1
Metadata: ObjectId = 2105058535 m_prevPage = (1:156) m_nextPage = (0:0)
pminlen = 8 m_slotCnt = 1 m_freeCnt = 7177
m_freeData = 7432 m_reservedCnt = 0 m_lsn = (48:473:12)
m_xactReserved = 0 m_xdesId = (0:4106) m_ghostRecCnt = 1
m_tornBits = 1214600220
Row - Offset
0 (0x0) - 6515 (0x1973)
Although the ghost cleanup thread still keeps 1 ghost row, the m_nextpage is 0 now. let's check its original previous page 156:
DBCC TRACEON(3604)
GO
dbcc page(test,1,156,2)
GO
output:
m_pageId = (1:156) m_headerVersion = 1 m_type = 1
m_typeFlagBits = 0x4 m_level = 0 m_flagBits = 0x0
m_objId (AllocUnitId.idObj) = 28 m_indexId (AllocUnitId.idInd) = 256
Metadata: AllocUnitId = 72057594039762944
Metadata: PartitionId = 72057594038845440 Metadata: IndexId = 1
Metadata: ObjectId = 2105058535 m_prevPage = (1:153) m_nextPage = (1:158)
pminlen = 8 m_slotCnt = 8 m_freeCnt = 744
m_freeData = 7432 m_reservedCnt = 0 m_lsn = (48:473:10)
m_xactReserved = 0 m_xdesId = (0:0) m_ghostRecCnt = 0
m_tornBits = -677617447
the next page of page 156 is 158 now. let's check page 158:
output:
m_pageId = (1:158) m_headerVersion = 1 m_type = 1
m_typeFlagBits = 0x4 m_level = 0 m_flagBits = 0x0
m_objId (AllocUnitId.idObj) = 28 m_indexId (AllocUnitId.idInd) = 256
Metadata: AllocUnitId = 72057594039762944
Metadata: PartitionId = 72057594038845440 Metadata: IndexId = 1
Metadata: ObjectId = 2105058535 m_prevPage = (1:156) m_nextPage = (1:159)
pminlen = 8 m_slotCnt = 8 m_freeCnt = 744
m_freeData = 7432 m_reservedCnt = 0 m_lsn = (48:473:11)
m_xactReserved = 0 m_xdesId = (0:0) m_ghostRecCnt = 0
m_tornBits = 2036857631
the m_prevPage of page 158 is 156 now. so page 157 has been removed from page chain, if you check table space by
sp_spaceused 'mytest',@updateusage = N'TRUE';
you can see 8kb reduced. so 157 has been deallocated.
reference:
Inside the Storage Engine: Ghost cleanup in depth
http://www.sqlskills.com/blogs/paul/post/inside-the-storage-engine-ghost-cleanup-in-depth.aspx
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