Partitions with Epoch

10gR2 on Solaris10
Hi,
I'd like to partition by month tables that contain time as UTC (EPOCH)
STARTTIME ---NUMBER(19,0) -like 1349952216835
PARTITION BY RANGE ("STARTTIME")
PARTITION "OCT2012" VALUES LESS THAN (TO_DATE(' 2012-11-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS'))
PARTITION "NOV2012" VALUES LESS THAN (TO_DATE(' 2012-12-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS'))
PARTITION "DEC2012" VALUES LESS THAN (TO_DATE(' 2013-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS'))
What are my options here?
Thx,

>
I'd like to partition by month tables that contain time as UTC (EPOCH)
What are my options here?
>
Your only option is to specify the epoch values manually in the partition clauses.
The bound elements must be: string, datetime or interval literal, number, or MAXVALUE. There is no Oracle function to convert a date to epoch and you can't use a function of your own.
You also can't use interval partitioning since the number of seconds per month are not the same.

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Hello again, WZZZ,
Here's an update. I was successful in creating two partitions on my internal drive, and in installing OS 10.6.6 on the second partition, as per your guidence. I now have it up to 10.6.8 with all the security updates and AppleWorks. A great thing.
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How to use Oracle partitioning with JPA @OneToOne reference?

Hi!
A little bit late in the project we have realized that we need to use Oracle partitioning both for performance and admin of the data. (Partitioning by range (month) and after a year we will move the oldest month of data to an archive db)
We have an object model with an main/root entity "Trans" with @OneToMany and @OneToOne relationships.
How do we use Oracle partitioning on the @OneToOne relationships?
(We'd rather not change the model as we already have millions of rows in the db.)
On the main entity "Trans" we use: partition by range (month) on a date column.
And on all @OneToMany we use: partition by reference (as they have a primary-foreign key relationship).
But for the @OneToOne key for the referenced object, the key is placed in the main/source object as the example below:
@Entity
public class Employee {
@Id
@Column(name="EMP_ID")
private long id;
@OneToOne(fetch=FetchType.LAZY)
@JoinColumn(name="ADDRESS_ID")
private Address address;
EMPLOYEE (table)
EMP_ID FIRSTNAME LASTNAME SALARY ADDRESS_ID
1 Bob Way 50000 6
2 Sarah Smith 60000 7
ADDRESS (table)
ADDRESS_ID STREET CITY PROVINCE COUNTRY P_CODE
6 17 Bank St Ottawa ON Canada K2H7Z5
7 22 Main St Toronto ON Canada L5H2D5
From the Oracle documentation: "Reference partitioning allows the partitioning of two tables related to one another by referential constraints. The partitioning key is resolved through an existing parent-child relationship, enforced by enabled and active primary key and foreign key constraints."
How can we use "partition by reference" on @OneToOne relationsships or are there other solutions?
Thanks for any advice.
/Mats

Crospost! How to use Oracle partitioning with JPA @OneToOne reference?

Can't boot into Windows after splitting Mac partition with Disk Utility

Hi everyone,
I installed Windows 8 with BOOTCAMP, creating a large 871 GB partition with a smaller 127 GB partition for Windows. After installing Windows, I then went back to Disk Utility to shrink the Mac partition and add two new partitions, one for storage and another blank, hoping to use it to install Linux into one day.
I did that, and now Windows doesn't boot. A Windows blue screen tells me I need to use the Windows DVD to repair it.
I tried to use Disk Utility to delete the two new partitions I made and grow the Mac partition back to the max size of 871 GB. But it doesn't let me do this. The Disk Utility log doesn't report an error and thinks it worked but the partition stays the same size. However if I make the partition a little smaller, like 870 GB then it works. I'm wondering if the Recovery partition is hiding there and preventing me from fully expanding the Mac partition.
What can I do?
Here's what I get if I type in sudo gpt -r -vv show disk0:
Code:
gpt show: disk0: mediasize=1000204886016; sectorsize=512; blocks=1953525168 gpt show: disk0: Suspicious MBR at sector 0 gpt show: disk0: Pri GPT at sector 1 gpt show: disk0: Sec GPT at sector 1953525167 start        size index contents 0           1         MBR 1           1         Pri GPT header 2          32         Pri GPT table 34           6 40      409600      1 GPT part - C12A7328-F81F-11D2-BA4B-00A0C93EC93B 409640 1701278816      2 GPT part - 48465300-0000-11AA-AA11-00306543ECAC 1701688456     1269536      3 GPT part - 426F6F74-0000-11AA-AA11-00306543ECAC 1702957992     1846360 1704804352   248002560      4 GPT part - EBD0A0A2-B9E5-4433-87C0-68B6B72699C7 1952806912      718223 1953525135          32         Sec GPT table 1953525167           1         Sec GPT header
Here's what I get if I type sudo fdisk /dev/rdisk0:
Code:
Disk: /dev/rdisk0     geometry: 121601/255/63 [1953525168 sectors] Signature: 0xAA55 Starting       Ending #: id cyl hd sec - cyl hd sec [     start -       size] ------------------------------------------------------------------------ 1: EE 1023 254 63 - 1023 254 63 [         1 -     409639] <Unknown ID> 2: AF 1023 254 63 - 1023 254 63 [    409640 - 1701278816] HFS+ 3: AB 1023 254 63 - 1023 254 63 [1701688456 -    1269536] Darwin Boot *4: 07 1023 254 63 - 1023 254 63 [1704804352 - 248002560] HPFS/QNX/AUX
Here's what I get with gdisk:
Code:
Disk /dev/rdisk0: 1953525168 sectors, 931.5 GiB Logical sector size: 512 bytes Disk identifier (GUID): 6075110F-7CEF-4604-85EE-6231B850E2AE Partition table holds up to 128 entries First usable sector is 34, last usable sector is 1953525134 Partitions will be aligned on 8-sector boundaries Total free space is 2564589 sectors (1.2 GiB) Number Start (sector)    End (sector) Size       Code Name 1              40          409639   200.0 MiB   EF00 EFI System Partition 2          409640      1701688455   811.2 GiB   AF00 1 TB APPLE HDD HTS54101 3      1701688456      1702957991   619.9 MiB   AB00 Recovery HD 4      1704804352      1952806911   118.3 GiB   0700 BOOTCAMP

Okay, the code I typed didn't display properly in the forum. Let me try again.
Here's what I get when I type
sudo gpt -r -vv show disk0:
gpt show: disk0: mediasize=1000204886016; sectorsize=512; blocks=1953525168
gpt show: disk0: Suspicious MBR at sector 0
gpt show: disk0: Pri GPT at sector 1
gpt show: disk0: Sec GPT at sector 1953525167
       start        size index contents
           0           1         MBR
           1           1         Pri GPT header
           2          32         Pri GPT table
          34           6
          40      409600      1 GPT part - C12A7328-F81F-11D2-BA4B-00A0C93EC93B
      409640 1701278816      2 GPT part - 48465300-0000-11AA-AA11-00306543ECAC
1701688456     1269536      3 GPT part - 426F6F74-0000-11AA-AA11-00306543ECAC
1702957992     1846360
1704804352   248002560      4 GPT part - EBD0A0A2-B9E5-4433-87C0-68B6B72699C7
1952806912      718223
1953525135          32         Sec GPT table
1953525167           1         Sec GPT header
And when I type fdisk /dev/rdisk0 I get:
Disk: /dev/rdisk0          geometry: 121601/255/63 [1953525168 sectors]
Signature: 0xAA55
         Starting       Ending
#: id cyl hd sec - cyl hd sec [     start -       size]
1: EE 1023 254 63 - 1023 254 63 [         1 -     409639] <Unknown ID>
2: AF 1023 254 63 - 1023 254 63 [    409640 - 1701278816] HFS+
3: AB 1023 254 63 - 1023 254 63 [1701688456 -    1269536] Darwin Boot
*4: 07 1023 254 63 - 1023 254 63 [1704804352 - 248002560] HPFS/QNX/AUX
Kappy,
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On a regular basis, I back up chunks of work on projects to external discs, and special backup areas on one of my drives. I don't like using time machine because I am running many operating environments on my mac pro and I don't ever want to be tied to one operating environment for important functionality, and I want to maximize the open space on my hard drives.
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Is there a bootable disc on the installer disk I just made? Can I use the disk utilities to do a low level erase on the lion volume that will remove the recovery partition?
Can I go back to booting from Snow Leopard and erasing the Recovery partition with the drive partition that way? Will the Snow Leopard utility take out the Lion HD recovery partition?
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I assume the unix pdisk command will show you want is going on.
You may not have a big worry. There have always been a lot of hidden partitions. Disk Utility under 10.4.11 reports this drive has three (my now report four ) partitions, when pdisk reports that there are 15.
Macintosh-HD -> Applications -> Utilities -> Terminal
Press return when done typing sudo pdisk -l
-l include a lower case L
The sudo command will ask for your administration password. No characters will appear when typing your password. Press return when done typing. sudo stands for super user do. It's just like root. Be careful.
mac $ sudo pdisk -l
Password:
Partition map (with 512 byte blocks) on '/dev/rdisk0'
#:                type name                    length   base      ( size )
1: Apple_partition_map Apple                       63 @ 1
2:      Apple_Driver43*Macintosh                   56 @ 64
3:      Apple_Driver43*Macintosh                   56 @ 120
4:    Apple_Driver_ATA*Macintosh                   56 @ 176
5:    Apple_Driver_ATA*Macintosh                   56 @ 232
6:      Apple_FWDriver Macintosh                  512 @ 288
7: Apple_Driver_IOKit Macintosh                  512 @ 800
8:       Apple_Patches Patch Partition            512 @ 1312
9:     Apple_Bootstrap untitled                  1954 @ 149319048
10:           Apple_HFS Apple_HFS_Untitled_1   2254440 @ 263968    ( 1.1G)
11:     Apple_UNIX_SVR2 untitled               6617188 @ 149321002 ( 3.2G)
12:           Apple_HFS Apple_HFS_Untitled_2 146538496 @ 2780552   ( 69.9G)
13:     Apple_UNIX_SVR2 swap                    363298 @ 155938190 (177.4M)
14:          Apple_Free Extra                   262144 @ 1824      (128.0M)
15:          Apple_Free Extra                   262144 @ 2518408   (128.0M)
Device block size=512, Number of Blocks=156301488 (74.5G)
DeviceType=0x0, DeviceId=0x0
Drivers-
1: 23 @ 64, type=0x1
2: 36 @ 120, type=0xffff
3: 21 @ 176, type=0x701
4: 34 @ 232, type=0xf8ff

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OS X partition
Windows 7 partition
blank NTFS partition (no OS)
blank NTFS partition (no OS)
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Hi,
your Topic says '...adding a third partition' whereas in your message you don't mention this third partition...
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Regards
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How to Implement 30 days Range Partitioning with Date column. Not Interval

Hi,
I am using the db:
Oracle Database 11g Enterprise Edition Release 11.2.0.2.0 - 64bit
Current table structure is:
CREATE TABLE A
a NUMBER,
CreationDate DATE
PARTITION BY RANGE (CreationDate)
INTERVAL ( NUMTODSINTERVAL (30, 'DAY') )
(PARTITION P_FIRST
VALUES LESS THAN (TIMESTAMP ' 2001-01-01 00:00:00'))
How can I define virtual column based partitioning with RANGE partitioning without using INTERVAL partitioning.
And that is with Intervals of 30 days.
For monthly I am trying as
CREATE TABLE A
a NUMBER,
CreationDate DATE,
monthly_interval date as (to_char(CreationDate,'MM-YYYY')) VIRTUAL
PARTITION BY RANGE (monthly_interval)
partition p_AUG12 values less than (to_date('08-2012','mm-yyyy')),
partition p_SEP12 values less than (to_date('09-2012','mm-yyyy')),
partition p_OCT12 values less than (to_date('10-2012','mm-yyyy'))
Enable ROw Movement
BUT CAN'T INSERT the data even for that:
Insert into a (a, CreationDate)
Values (1, '12-10-2012')
Insert into a (a, CreationDate)
Values (1, '12-10-2012')
Please suggest..

Hi rp,
Interval Partitioned to Range. Created Daily Partitions from Monthly Part. got complicated so I am posting here.
Basically,
I know Interval Partitioning is a kind of Range partitioning. But explicitly for Interval Partitioned tables XML Indexes are not allowed as discussed here:
XMLIndexes on an Interval Partitioned Table??
I can do monthly partitions as :
CREATE TABLE A
a NUMBER,
CreationDate DATE,
monthly_interval varchar2(8) as (to_char(CreationDate,'MM-YYYY')) VIRTUAL
PARTITION BY RANGE (monthly_interval)
partition p_AUG12 values less than ('09-2012'),
partition p_SEP12 values less than ('10-2012'),
partition p_OCT12 values less than ('11-2012')
) Enable ROw Movement
Insert into a (a, CreationDate)
Values (1, '12-SEP-2012')
Insert into a (a, CreationDate)
Values (1, '14-SEP-2012')
Select * from A partition (p_SEP12)
Select * from A partition (p_AUG12)
Select * from A partition (p_OCT12)
Can we do it for 30 days partitions, instead of the monthly partitions. ANY suggestions..
Thanks..

Reference partition with Trigger

This post is an extended version if my previous post -Dynamic Reference partition by Trigger on Feb 24
I need to create a Dynamic Reference partition with Trigger.
Requirement:-
1. It should be a list partition
2. It should be a referencing Partition (There are 2 tables with Parent Child relation)
3. It should be dynamic(trigger) (Whenever new value enter to the List, It should be create a New partition by trigger, As per my understanding interval partitioning is dynamic but it will work only for range partition. )
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Almost I have completed this task, but facing some issues. Please find the example i have done. (Please note that it is just an example and not my actual requirement)
==============================================================================================================
--TABLE 1
create table customers
cust_id number primary key,
cust_name varchar2(200),
rating varchar2(1) not null
partition by list (rating)
partition pA values ('A'),
partition pB values ('B')
--TABLE 2
create table sales
sales_id number primary key,
cust_id number not null,
sales_amt number,
constraint fk_sales_01
foreign key (cust_id)
references customers
partition by reference (fk_sales_01);
--TRIGGER
CREATE OR REPLACE TRIGGER CUST_INSERT
BEFORE INSERT ON CUSTOMERS FOR EACH ROW
DECLARE
V_PARTITION_COUNT NUMBER;
V_PART_NAME VARCHAR2 (100) := :NEW.RATING;
PRAGMA AUTONOMOUS_TRANSACTION;
BEGIN
SELECT COUNT (PARTITION_NAME)
INTO V_PARTITION_COUNT
FROM USER_TAB_PARTITIONS
WHERE TABLE_NAME = 'CUSTOMERS' AND PARTITION_NAME = V_PART_NAME;
IF V_PARTITION_COUNT = 0 THEN
EXECUTE IMMEDIATE 'ALTER TABLE CUSTOMERS ADD PARTITION '||V_PART_NAME||' VALUES ('''||V_PART_NAME||''')' ;
END IF;
END;
--INSERT TO CUSTOMER
insert into customers values (111, 'OOO', 'C');-- A and B are already in the Customer table List along with the create script, so I am inserting new row with 'C'
==============================================================================================================
When i am inserting the new value 'C' to the customer table it supposed to create a new partition along with new row to the table.
But I am getting an error like "ORA-14400: inserted partition key does not map to any partition", like partition is not exist.
But If I execute the same insert statement again it is inserting the row.
That means Even though it is a before insert trigger, insert is trying to happen before the partition creation.
Since it is before insert trigger My expectation is
a) Create a Partition first
b) Insert the record second
But actual behavior is
a) Try to insert first and failing
b) Creating the partition second
That is the reason when I am trying second time the record is getting inserted.
Can anyone help me on this, Thanks in advance.
Shijo

You can't do this with a trigger. And you really, really don't want to.
As you've discovered, Oracle performs a number of checks to determine whether an insert is valid before any triggers are executed. It is going to discover that there is no appropriate partition before your trigger runs. Even if you somehow worked around that problem, it's likely that you'd end up in a rather problematic position from a lock standpoint since you'd be trying to do DDL on the object in the autonomous transaction while doing an insert likely from a stored procedure that would itself be made invalid because of the DDL in another-- that's quite likely to cause an error anyway. And that's before we get into the effects of doing DDL on other sessions running at the same time.
Why do you believe that you need to add partitions dynamically? If you are doing ETL into a data warehouse, it makes sense to add any partitions you need at the beginning of the load before you start inserting the data, not in a trigger. If this is some sort of OLTP application, it doesn't make sense to do DDL while the application is running. I could see potentially having a default partition and then having a nightly job that does a split partition to add whatever list partitions you need but that would strike me as a corner case at best. If you really don't know what the possible values are and users are constantly creating new ones, list partitioning seems like a poor choice to me-- wouldn't a hash partition make more sense?
Justin

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Hi,
When I am importing csv file system is showing me the following error:
[Admin-ProcessPartition]The 'Partition' with 'ID' = 'FAC2CRDN' doesn't exist in the collection.
I checked my OLAP database. Partition with id = FAC2CRDN exists.
Please help me to come our from this issue and Thanks.
Regards,
Stanislav

I would suggest you the following steps in order to see if your problem is fixed or if you still have some errors:
1) full optimize your application
2) try to full process the OLAP database directly from the Analysis Services management console.
Regards,
Simmaco

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To Delete boot camp partition with lion
follow steps below : -
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Phewwwwwww gone

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Is SSD partitioning with boot camp stable and recommended?

Is SSD partitioning with boot camp stable and recommended?

The OS as already stated does not know the difference. The SSD will perform as it does with any operating system. SSDs can be partitioned as can HDDs.
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Range interval partitioning with null value

Hello,
I have question, how can I create table with partition and after that I will insert null value in the partition key like this kind of
range interval partitioning with null value.
I have Oracle 11g.
Thanks
Mohammed

user13777053 wrote:
Thank you, but my question is about range interval not for range?http://docs.oracle.com/cd/E11882_01/server.112/e25523/partition.htm#VLDBG00220
A MAXVALUE literal can be defined for the highest partition. MAXVALUE represents a virtual infinite value that sorts higher than any other possible value for the partitioning key, *including the NULL value.*
So since you're trying to partition by interval (which doesn't support a MAXVALUE clause) you're out of luck.
There's probably fancy hacks and what not you could try to implement, but i'm not about to recommend any of those here :)
Cheers,

Create Partition with Maxl

Hi All
I was trying to create a Partitioning with Maxl script and in the process Essbase went down. I logged back and try to create one partition then it is giving me error saying object database name does not exist.
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I tried to drop it by the drop command but it doesnt work.
Can someone shed some light how to remove the object?
Regards
Celvin

I surmise, some Temporary partitioning file are created under Database folder.
Just get into Database folder and remove the following files in background if persists.
ddb - Partitioning definition file
ddm - Temporary partitioning file
ddn - Temporary partitioning file
Regards
Cnee

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Hello All,
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I have 136GB of Hard Drive.
My RAM is 3GB.
Could anyone suggest me idol file system/partition with file type & size?
I can create partition on my SLES installation.

We are using raid5.
But I can see only one hard drive.
Do you think its ok to have only three partitions?
what about /? (root)
what file type I should used? ext 3, ext 2 or Resiser?

Partitions with Epoch

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