Segments, extents and data blocks

I'm somewhat green when it comes to Oracle and databases in general, so bare with me. I bought the OCP Exam Guide to try and learn as much as I can. I'm at the chapter about segments, extents and data blocks and I'm having a bit of trouble grasping the conecpts of these. Are these logical? Does anyone have time for quick overview?
Thank you!

Ignoring the O/S, the basic storage unit in Oracle is a block (db_block). You define the size of the block with the db_block_size parameter, and it is effectively fixed at database creation. You can have tablespaces with a different block size from the database default, but we won't go there for now. A db_block is not the sme as an O/S block, so in that sense it is a logical unit.
In Oracle, a segment is anything that takes disc space, generally speaking, a table or an index. Every segment is stored in a tablespace. A segment can only be in one tablespace, but a tablespace can hold many segments. A tablespace gets its space from a data file (an O/S file). Each data file belongs to one tablespace, but a tablespace can have multiple data files.
Each segment is made up a a number of extents. The are the units of space that are allocated to the segment as it needs space. The size of each extent is controlled by either the INITIAL and NEXT parameters in CREATE TABLESPACE or in one of the CREATE <segment> commands. If you are using locally managed tablespaces, the tablespace parameters define the extent sizes. Each extent belongs to only one segment, but a segment can have many extents. Although the extent sizes are defined in bytes, they must be a multiple of the block size.So, if you have an 8K block size, you cannot have a 12K extent size.
So:
An extent is made up of one or more (contiginous) blocks within a data file.
A segment is made up of one or more (possibly non-contiginous) extents possibly in multiple data files.
Note that the blocks in an extent must be contiginous, but that is a logical concept within Oracle, since Oracle has no way of knowing how and where the O/S has actually allocated the physical space, it just has an internal map of the file space allocated by the O/S.
HTH
John

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Hi,
Look at this (might be more readable if you paste it somewhere and use a font with fixed width):
SQL>CONNECT SYSTEM/MANAGER
ConnectÚ.
SQL> CREATE TABLE MyTable
2 (
3 CH1 VARCHAR2(10),
4 CH2 VARCHAR2(10)
5 ) TABLESPACE TOOLS;
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SYSTEM MYTABLE TABLE TOOLS 0 7 1553 65536 16 7
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SQL>
SQL> -- Let's dump the first block and see what's in:
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SystÞme modifiÚ.
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/* Content:
Start dump data blocks tsn: 7 file#: 7 minblk 1553 maxblk 1553
buffer tsn: 7 rdba: 0x01c00611 (7/1553)
scn: 0x0000.0007b1fa seq: 0x02 flg: 0x00 tail: 0xb1fa2002
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unformatted: 13 total: 16 first useful block: 3
owning instance : 1
instance ownership changed at
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Freeness Status: nf1 0 nf2 0 nf3 0 nf4 0
Extent Map Block Offset: 4294967295
First free datablock : 3
Bitmap block lock opcode 0
Locker xid: : 0x0000.000.00000000
Highwater:: 0x01c00614 ext#: 0 blk#: 3 ext size: 16
#blocks in seg. hdr's freelists: 0
#blocks below: 0
mapblk 0x00000000 offset: 0
HWM Flag: HWM Set
DBA Ranges :
0x01c00611 Length: 16 Offset: 0
0:Metadata 1:Metadata 2:Metadata 3:unformatted
4:unformatted 5:unformatted 6:unformatted 7:unformatted
8:unformatted 9:unformatted 10:unformatted 11:unformatted
12:unformatted 13:unformatted 14:unformatted 15:unformatted
End dump data blocks tsn: 7 file#: 7 minblk 1553 maxblk 1553
SQL> -- Now add 2 lines
SQL> INSERT INTO MYTABLE VALUES ('ABC','123');
1 ligne crÚÚe.
SQL> COMMIT;
Validation effectuÚe.
SQL> INSERT INTO MYTABLE VALUES ('DEF','456');
1 ligne crÚÚe.
SQL> COMMIT;
Validation effectuÚe.
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SQL> SELECT ROWID FROM MYTABLE;
ROWID
AAAHotAAHAAAAYfAAA
AAAHotAAHAAAAYfAAB
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7
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SQL>
/* Content:
Start dump data blocks tsn: 7 file#: 7 minblk 1567 maxblk 1567
buffer tsn: 7 rdba: 0x01c0061f (7/1567)
scn: 0x0000.0007b2fd seq: 0x01 flg: 0x02 tail: 0xb2fd0601
frmt: 0x02 chkval: 0x0000 type: 0x06=trans data
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brn: 0 bdba: 0x1c00611 ver: 0x01
inc: 0 exflg: 0
Itl Xid Uba Flag Lck Scn/Fsc
0x01 0x0006.003.000002f8 0x008000b7.0081.06 --U-    1 fsc 0x0000.0007b2f9
0x02 0x0005.000.000002fd 0x00800099.008f.16 --U-    1 fsc 0x0000.0007b2fd
data_block_dump,data header at 0x2ff1064
===============
tsiz: 0xf98
hsiz: 0x16
pbl: 0x02ff1064
bdba: 0x01c0061f
76543210
flag=--------
ntab=1
nrow=2                                                                                                                                                                                              <--------- 2 Lines in the block
frre=-1
fsbo=0x16
fseo=0xf82
avsp=0xf6c
tosp=0xf6c
0xe:pti[0]     nrow=2     offs=0
0x12:pri[0]     offs=0xf8d
0x14:pri[1]     offs=0xf82
block_row_dump:
tab 0, row 0, @0xf8d
tl: 11 fb: H-FL lb: 0x1 cc: 2                                                                                                                        <--------- First line header
col 0: [ 3] 41 42 43                                                                                                                                                      <--------- Here is 'ABC'
col 1: [ 3] 31 32 33                                                                                                                                                      <--------- Here is '123'
tab 0, row 1, @0xf82
tl: 11 fb: H-FL lb: 0x2 cc: 2                                                                                                                        <--------- Second line header
col 0: [ 3] 44 45 46                                                                                                                                                      <--------- Here is 'DEF'
col 1: [ 3] 34 35 36                                                                                                                                                      <--------- Here is '456'
end_of_block_dump
End dump data blocks tsn: 7 file#: 7 minblk 1567 maxblk 1567
SQL> SELECT OBJECT_ID FROM DBA_OBJECTS WHERE OBJECT_NAME='MYTABLE';
OBJECT_ID
31277
Sincerely hope this'll help you understand.
Regards,
Yoann.

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I understand what are tablespaes,Segments,Extents, data blocks.But i want to know why do we have all these logical seggregations.Does DBA create all these while creating tablespaces?
Could some one plz explain me the exact concept behind having all these?

user11365275 wrote:
I understand what are tablespaes,Segments,Extents, data blocks.But i want to know why do we have all these logical seggregations.Does DBA create all these while creating tablespaces?
Could some one plz explain me the exact concept behind having all these?You should check and read the link that Helios has given. The DBA is only going to create tablespaces with datafiles added to them. The rest of the things are going to come into as and when the objects are going to get created, either by the DBA or by the end users. To answer why, this is the why oracle lets the dba manage the granularity within the datafile. These are all logical constructs which means, their existence is only within the oracle's own views . In real, its just bunch of files allocating space on the physical hard drive.
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TRACE FILE INFORMATION
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Hi,
This is from ADDM Report.
FINDING 8: 2% impact (159 seconds)
A hot data block with concurrent read and write activity was found. The block
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RECOMMENDATION 1: Application Analysis, 2% benefit (159 seconds)
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SYMPTOMS THAT LED TO THE FINDING:
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what does it mean by hot block with concurrent read and write??
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Hi,
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Hello Najm,
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Go to Runtime Repository Editor there.
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I will list out what i know so you guys can gauge me on where i am..
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Hi Aman
Hmm agreed.So when commit is issued , what happens at that time?Simply put:
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I lost here.In the ITL , the scn is transaction SCN or commit scn?As I just wrote, the ITL contains (if the cleanout occured) the commit SCN.
This sounds like high RBA information?What is RBA?
Commit SCNThis is the SCN associated with a committed transaction.
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BBED> print ktbbhitl[0]
struct ktbbhitl[0], 24 bytes     @44
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         ub2 kxidusn              @44       0x0009
         ub2 kxidslt              @46       0x002e
         ub4 kxidsqn              @48       0x0000fe77
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         ub4 kubadba              @52       0x00800249
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         ub1 kubarec              @58       0x4e
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         b2 _ktbitfsc             @62       0
         ub2 _ktbitwrp            @62       0x0000
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HTH
Chris

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Hi
I have a data block where I need that the first two items are a push button.
Now when user want to insert a new record, if you have a text item, since the cursor is in the text item, you can enter a new record.
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How to have a push button which is able to create a new record like a text item does?

In fact the original problem is :
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Space utilization and optimal row size in Data Blocks in Oracle 11g

Hi,
My main concern till now is to find the Optimal no. of rows/tuples per one oracle data block of size 8192. For that purpose i'm doing different kind of testing.
I created one table:
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2 (x char(2000));
Table created.
Inserted 4 rows in it.
Queried to check in which block no's these 4 rows exist.
SQL> SELECT x,DBMS_ROWID.ROWID_BLOCK_NUMBER(rowid) "Block No."
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x Block No.
A 422
B 422
C 422
D 423
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In this scenerio 3 rows (each of around 2006 bytes) are inserted into into one BLOCK of 8192 i.e. 8192-6006=2186. 2186 is the PCTFREE or what?
How can I see this CHAIN_CNT column value other than 0?
How can actually I find the optimal no. of rows/tuples in 1 data block that will not increase OVERHEAD on the system.
I'll highly appreciate the genuine help and solid suggestions.
Thanks in Advance.
Best Regards,
Kam

kamy555 wrote:
If you want to suggest something it'll be nice of yo.I can't disclose my main concern :)If you can't disclose what you mean by "optimal" and you can't disclose what "overhead" you are concerned about, I'm not sure that anyone could answer your question.
In this scenerio 3 rows (each of around 2006 bytes) are inserted into into one BLOCK of 8192 i.e. 8192-6006=2186. 2186 is the PCTFREE or
what?PCTFREE is a percentage. You haven't specified what you set it to, but that decreases the space in a block available for inserts to 8192 * (1 - PCTFREE/100).
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How can actually I find the optimal no. of rows/tuples in 1 data block that will not increase OVERHEAD on the system.Since you can't disclose what "optimal" means or what "overhead" you're concerned with, I don't see how this question could be answered.
Justin

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Hi,
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T2 = I created in a Tablespace with Blocksize 16k.
T3 = I created in a Tablespace with Blocksize 4k. In the same Instance.
Each table has approx. 500 rows (So, table sizes are same in all the cases to test Query execution time ). As these 3 tables are created under different data block sizes so the ALLOCATED no. of data blocks are different in all cases.
T1 = 8k = 256 Blocks = 00:00:04.76 (query execution time/elapsed time)
T2 = 16k=121 Blocks = 00:00:04.64
T3 = 4k = 490 Blocks = 00:00:04.91
Table Access is FULL i.e. I have used select * from table_name; in all 3 cases. No Index nothing.
My Question is why query execution time is nearly the same in all 3 cases because Oracle has to read all the data blocks in each case to fetch the records and there is a much difference in the allocated no. of blocks ???
In 4k block size example, Oracle has to read just 121 blocks and it's taking nearly the same time as it's taking to read 490 blocks???
This is just 1 example of different data blocks. I have around 40 tables in each block size tablespace and the result are nearly the same. It's very strange for me because there is a much difference in the no. of allocated blocks but execution time is almost the same, only difference in milliseconds.
I'll highly appreciate the expert opinions.
Bundle of thanks in advance.
Best Regards,

Hi Chris,
No I'm not using separate databases, it's 8k database with non-standard blocksizes of 16k and 4k.
Actually I wanted to test the Elapsed time of these 3 tables, so for that I tried to create the same size
tables.
And how I equalize these is like I have created one column table with char(2000).
555 MB is the figure I wanted to use for these 3 tables ( no special figure, just to make it bigger than the
RAM used for my db at the db startup to be sure of not retrieving the records from cache).
so row size with overhead is 2006 * 290,000 rows = 581740000(bytes) / 1024 = 568105KB / 1024 = 555MB.
Through this math calculation I thought It will be the total table size. So I Created the same no. of rows in 3 blocksizes.
If it's wrong then what a mes because I was calculating tables sizes in the same way from the last few months.
Can you please explain a little how you found out the tables sizes in different block sizes.Though I understood how you
calculated size in MB from these 3 block sizes
T8K =97177 BLOCKS=759MB *( 97177*8 = 777416KB / 1024 = 759MB )*
T16K=41639 BLOCKS=650MB
BT4K=293656 BLOCKS=1147MB
For me it's new to calculate the size of a table. Can you please tell me then how many rows I can create in each of
these 3 tables to make them equal in MB to test for elapsed time.
Then I'll again run my test and put the results here. Because If I've wrongly calculated table sizes then there is no need to talk about elapsed time. First I must equalize the table sizes properly.
SQL> select sum(bytes)/1024/1024 "Size in MB" from dba_segments> 2 where segment_name = 'T16K';
Size in MB
655
Is above SQL is correct to calculate the size or is it the correct alternative way to your method of calculating the size??
I created the same table again with everything same and the result is :
SQL> select num_rows,blocks from user_tables where table_name = 'T16K';NUM_ROWS BLOCKS
290000 41703
64 more blocks are allocated this time so may be that's y it's showing total size of 655 instead of 650.
Thanks alot for your help.
Best Regards,
KAm
Edited by: kam555 on Nov 20, 2009 5:57 PM

Data Blocks and the Elapsed Time

Hi,
I have created 3 tables with one column only. As an example Table 1 below:
SQL> create table T8k( x char(2000));
So 3 tables are created in this way i.e. T8k,T16K and T4K
T8 = in the default database tablespace of 8k (11g v11.1.0.6.0 - Production) (O.S=Windows).
T16 = I created in a Tablespace with Blocksize 16k.
T4K = I created in a Tablespace with Blocksize 4k. In the same Instance.
Each table has 290,000 rows and all the 3 tables have equal size of 555MB (2006(rowsize with overhead) * 290,000/1024/1024 = 555MB) to test Elapsed Time (set timing on).
As these 3 tables are created under different block sizes so the allocated no. of data blocks are different as below:
T8K = 97177 BLOCKS= 00:41:20.21 (Elapsed Time)
T16K=41639 BLOCKS= 00:44:11.59
BT4K=293656 BLOCKS=00:37:29.06
Please note the difference. First table i.e. 8k block size, allocated blocks are 97177 and taking around 41 mins. Third table i.e. BT4K (in a 4k block size tablespace), allocated blocks(293656 ) are almost 3 times bigger, taking around 37 mins to execute the query. I mean the difference is only 4 mins hardly and blocks difference is 3 times.
In case of any doubt, I've created these tables bigger than the memory used for my db(memory_max_size 408M) i.e. 555 MB that If Blocks are already in cache then reading the blocks and counting the rows will nearly be the same regardless of the block sizes or the number of blocks.
Need solid suggestions and if possilble, links also which has some serious discussions regarding my issue.
Bundle of thanks.
Best Regards,

Because I was not completely satisfied with the things last time. It doesn't seem so simple to me as people said
I thought may some guru have a look today, not just to point out that i have created it again.
And I have compact my question to avoid confusion.

Data Blocks - Different data source for Query and DML

Would anyone please tell me if it is possible in a BLOCK to define a stored procedure for queriying and a table for DML operations or vice-versa at the SAME time.
If possible how do you do it using DATA BLOCK WIZARD? If you select your block to be based on stored procedure then it gives only stored procedure options for Querying and DML operations in next few screens.
Also, suppose I used a stored procedure defined in a block for insert operations, is it implicitly called by INSERT RECORD function key in default menu? How does it take values from different fields on its own?

Hi Deepon,
We get data from both BSIK(open items)and BSAK (cleared items). Obviously if a accounting doc is not cleared it would be in open items and vice versa..
Go through the help link for more information..
http://help.sap.com/saphelp_bw33/helpdata/en/90/10e73a86e99c77e10000000a114084/frameset.htm
Regards
Manga(Assign points if it helps)
Message was edited by: Manga

Relationship between a data block and latch

Hello Experts,
I have just read an article about indexes and full table scan by Richard Foote. http://richardfoote.wordpress.com/2009/04/16/indexes-on-small-tables-part-i-one-of-the-few/
I thought each block required only one latch for each operation. However, I don't undersand that why a single block required 2 latch? The qoute is the following,
Note that yes indeed, there were 4 consistent gets performed and that none of the consistent gets were the “cheaper” consistent gets examinations. Therefore, the 4 consistent gets used in performing the FTS of the one block table required 4 x 2 = 8 latches.
Can anyone explain the above quote please?
Thanks in advance.

Jonathan Lewis wrote:
Solomon Yakobson wrote:
Read Concepts manual. Database Buffer Cache in particular:Every buffer has an access mode: pinned or free (unpinned).
I hope the manual doesn't use the word "free" as a synonym for "unpinned", since a buffer (header) with state = "free" is one that doesn't hold a copy of a datablock and is therefore immediately available for re-use.
Regards
Jonathan Lewis
Hello Sir,
I have just read Memory Architecture document, buffer states part, I know the meaning of dirty buffers. However, when we just SELECT any table, the corresponding data blocks that read into memory from disk, are they considered as clean buffers? Because as far as I understand from dirty buffers, these are modified buffers. And maybe the manual should have used "clean" instead of "free"?
The database uses a sophisticated algorithm to make buffer access efficient. Pointers to dirty and nondirty buffers exist on the same least recently used (LRU) list, which has a hot end and cold end. A coldbuffer is one that has not been recently used. A hot buffer is frequently accessed and has been recently used
Also the above statement is excerpt from Oracle Memory Architecture document, does the nondirty buffers refer to data blocks that move to the memory in SELECT query?
Thanks a lot for your remarkable thoughts.

Segments, extents and data blocks

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