Find range partition key information
Hello, I try to insert a row in a table and I get this msg: "inserted partition key is beyond highest legal partition key". This table has a range partitioning key but I don't know on which colmn(s) this partioning is working.
Is there a way to find this information?
- which column
- which are the values
Thx in advance,
Pascal
Look at the following views, you should be able to find the information
USER_IND_PARTITIONS
USER_IND_SUBPARTITIONS
USER_LOB_PARTITIONS
USER_LOB_SUBPARTITIONS
USER_TAB_PARTITIONS
USER_TAB_SUBPARTITIONS
Similar Messages
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Modifing range partition key values
I have table in oracle 10g with range partiotion , now I want to change or
modify the partition key values . How i can do thatDepending on the level of change???
If your just rejiging boundaries in the range then you can merge, re split partitions? Without specific information on the change its hard to guess what is in your mind.
I have used this method in the past when some of the partitions remain empty and I wish to rebalance the data skew to even up the distribution of a key? Is this what you want to do?
Or do you intend adding columns to a key without recreating the object? I don't know any way to change this without creating a new object; sorry…. If you have the space create table as select with new range specification would do. Otherwise create csv and sqlload into new defined object after drop and recreate. Hope this helps.
Kind regards -
Column Containing Partition Key Condition Information
I am using 11.2.0.2 on AIX 6.1.
I am creating a range partition table as below.
CREATE TABLE SALES_RANGE
( prod_id NUMBER(6)
, cust_id NUMBER
, time_id DATE
, channel_id NUMBER(3)
PARTITION BY RANGE (time_id)
( PARTITION sales_q1_2012 VALUES LESS THAN (TO_DATE('01-APR-2012','dd-MON-yyyy'))
, PARTITION sales_q2_2012 VALUES LESS THAN (TO_DATE('01-JUL-2012','dd-MON-yyyy'))
, PARTITION sales_q3_2012 VALUES LESS THAN (TO_DATE('01-OCT-2012','dd-MON-yyyy'))
, PARTITION sales_q4_2012 VALUES LESS THAN (TO_DATE('01-JAN-2013','dd-MON-yyyy'))
);I can find all the information related to this table in dictionary tables like below.
SQL> SELECT OWNER,
TABLE_NAME,
PARTITIONING_TYPE,
SUBPARTITIONING_TYPE,
PARTITION_COUNT,
PARTITIONING_KEY_COUNT,
STATUS,
DEF_TABLESPACE_NAME
FROM DBA_PART_TABLES
WHERE OWNER='SCOTT';
OWNER TABLE_NAME PARTITION SUBPARTIT PARTITION_COUNT PARTITIONING_KEY_COUNT STATUS DEF_TABLESPACE_NAME
SCOTT SALES_RANGE RANGE NONE 4 1 VALID USERS
SQL> SELECT TABLE_OWNER,
TABLE_NAME,
COMPOSITE,
PARTITION_NAME,
PARTITION_POSITION,
TABLESPACE_NAME,
LAST_ANALYZED
FROM DBA_TAB_PARTITIONS
WHERE TABLE_OWNER='SCOTT'
ORDER BY PARTITION_POSITION;
TABLE_OWNER TABLE_NAME COM PARTITION_NAME PARTITION_POSITION TABLESPACE_NAME LAST_ANAL
SCOTT SALES_RANGE NO SALES_Q1_2012 1 USERS 17-FEB-12
SCOTT SALES_RANGE NO SALES_Q2_2012 2 USERS 17-FEB-12
SCOTT SALES_RANGE NO SALES_Q3_2012 3 USERS 17-FEB-12
SCOTT SALES_RANGE NO SALES_Q4_2012 4 USERS 17-FEB-12
SQL> SELECT * FROM DBA_PART_KEY_COLUMNS WHERE OWNER='SCOTT';
OWNER NAME OBJEC COLUMN_NAME COLUMN_POSITION
SCOTT SALES_RANGE TABLE TIME_ID 1But where can i find the partition key condition i.e. *"VALUES LESS THAN (TO_DATE('01-APR-2012','dd-MON-yyyy'))"* in any dictionary table ?Thanks for your quick reply....
I have already checked and included in my example given.
SQL> SELECT * FROM DBA_PART_KEY_COLUMNS WHERE OWNER='SCOTT';
OWNER NAME OBJEC COLUMN_NAME COLUMN_POSITION
SCOTT SALES_RANGE TABLE TIME_ID 1It's only giving the column name on which partitioning is done, not the partitioning condition.
My question was, in which table i can check the criteria on which i have range partitioned a table ? Is there an dictionary table which holds information like:
PARTITION_NAME PARTITION_CONDITION
SALES_Q1_2012 VALUES LESS THAN (TO_DATE('01-APR-2012','dd-MON-yyyy')I hope i am clear in my question...... -
I partitioned my MacBook Pro and installed Windows 8 in it.....now I can't find my partition when I pressed option key when booting up.....so how can I do to find the partition to boot...I can see it in the disk manager in Mac....any ideas? thanks..!..
Hi LamboMong,
just try this:
http://refit.sourceforge.net
install, reboot twice and choose the Windows-Partition on next startup.
(The apple-bootloader doesn't display all bootable harddisks.
With rEFIt you can boot from all partitions/harddisks on your computer.)
I hope that will solve your problem.
Daniel Fernau -
How to find out the min & max partition_id in a range partition?
Hi we have a table set up by range partition
In the table creation script. It goes something like:
PARTITION PARTD_CUST_3_1_NEW VALUES LESS THAN ('39500')
so how we can find out this number '39500' from some of the data_dictionary view?
What we want to do is to set up parallel processing (do-it-yourself) based on partition.
So the number of parallel process will be based on this ID-range.
I can find out how many partitions are there by using something like this:
SELECT COUNT(*)
FROM all_tab_partitions WHERE table_owner='OWNER_NAME' AND table_name='TABLE_NAME'
We have 5 partitions now but the table structure can change in the near future even this table has moer than 130Million rows
I do not want to hardcode this "39500" just in case the table structure changes (partition structure changes)
Any idea ??vxwo0owxv wrote:
Hi we have a table set up by range partition
In the table creation script. It goes something like:
PARTITION PARTD_CUST_3_1_NEW VALUES LESS THAN ('39500')
so how we can find out this number '39500' from some of the data_dictionary view?
What we want to do is to set up parallel processing (do-it-yourself) based on partition.
So the number of parallel process will be based on this ID-range.
I can find out how many partitions are there by using something like this:
SELECT COUNT(*)
FROM all_tab_partitions WHERE table_owner='OWNER_NAME' AND table_name='TABLE_NAME'
We have 5 partitions now but the table structure can change in the near future even this table has moer than 130Million rows
I do not want to hardcode this "39500" just in case the table structure changes (partition structure changes)
Any idea ??query DBA_TAB_PARTITIONS.HIGH_VALUE -
NULL partition key in RANGE partition
All,
This is regarding partitioning a table using RANGE partition method. But the partition key contains null. How do I handle this situation? This is because there is no DEFAULT partition in RANGE partition though its present in LIST partition. Will rows with NULL partition key fall in MAXVALUE partition? Seeking your guidence.
Thanks,
...NULLS would fit into the MAXVAL partition yes.
http://download.oracle.com/docs/cd/B19306_01/server.102/b14220/partconc.htm#sthref2590
Thanks
Paul -
Unique key on range-partitioned table
Hi,
We are using a composite range-hash interval partitioned table
Uses index - trying to make this have same tablespace as the partitions i.e. local but not liking it
alter table RETAILER_TRANSACTION_COMP_POR
add constraint RETAILER_TRANSACTION_COMP_PK primary key (DWH_NUM)
using index
LOCAL
ora-14039: partitioning columns must form a subset of key columns of a unique index
Without local then fine but doesn't have same tablespace as the partitions and don't want to make this part of partition key.
Tbale range partitioned - this is just a UK to prevent duplicates[oracle@localhost ~]$ oerr ora 14039
14039, 00000, "partitioning columns must form a subset of key columns of a UNIQUE index"
// *Cause: User attempted to create a UNIQUE partitioned index whose
// partitioning columns do not form a subset of its key columns
// which is illegal
// *Action: If the user, indeed, desired to create an index whose
// partitioning columns do not form a subset of its key columns,
// it must be created as non-UNIQUE; otherwise, correct the
// list of key and/or partitioning columns to ensure that the index'
// partitioning columns form a subset of its key columns -
TIMESTAMP(6) Partitioned Key - Range partitioned table ddl needed
What is DDL syntax for TIMESTAMP(6) Partitioned Key, Range partitioned table
Edited by: oracletune on Jan 11, 2013 10:26 AM>
What is DDL syntax for TIMESTAMP(6) Partitioned Key, Range partitioned table
>
Not sure what you are asking. Are you asking how to create a partitioned table using a TIMESTAMP(6) column for the key?
CREATE TABLE TEST1
USERID NUMBER,
ENTRYCREATEDDATE TIMESTAMP(6)
PARTITION BY RANGE (ENTRYCREATEDDATE) INTERVAL(NUMTOYMINTERVAL(1, 'MONTH'))
PARTITION P0 VALUES LESS THAN (TO_DATE('1-1-2013', 'DD-MM-YYYY'))
)See my reply Posted: Jan 10, 2013 9:56 PM if you need to do it on a TIMESTAMP with TIME ZONE column. You need to add a virtual column.
Creating range paritions automatically -
We have tables that are interval range partitioned on a DATE column, with a partition for each day - all very standard and straight out of Oracle doc.
A 3rd party application queries the tables to find number of rows based on date range that is on the column used for the partition key.
This application uses date range specified relative to current date - i.e. for last two days would be "..startdate > SYSDATE -2 " - but partition pruning does not take place and the explain plan shows that every partition is included.
By presenting the query using the date in a variable partition pruning does table place, and query obviously performs much better.
DB is 11.2.0.3 on RHEL6, and default parameters set - i.e. nothing changed that would influence optimizer behavior to something unusual.
I can't work out why this would be so. It very easy to reproduce with simple test case below.
I'd be very interested to hear any thoughts on why it is this way and whether anything can be done to permit the partition pruning to work with a query including SYSDATE as it would be difficult to get the application code changed.
Furthermore to make a case to change the code I would need an explanation of why querying using SYSDATE is not good practice, and I don't know of any such information.
1) Create simple partitioned table
CREATETABLE part_test
(id NUMBER NOT NULL,
starttime DATE NOT NULL,
CONSTRAINT pk_part_test PRIMARY KEY (id))
PARTITION BY RANGE (starttime) INTERVAL (NUMTODSINTERVAL(1,'day')) (PARTITION p0 VALUES LESS THAN (TO_DATE('01-01-2013','DD-MM-YYYY')));
2) Populate table 1million rows spread between 10 partitions
BEGIN
FOR i IN 1..1000000
LOOP
INSERT INTO part_test (id, starttime) VALUES (i, SYSDATE - DBMS_RANDOM.value(low => 1, high => 10));
END LOOP;
END;
EXEC dbms_stats.gather_table_stats('SUPER_CONF','PART_TEST');
3) Query the Table for data from last 2 days using SYSDATE in clause
EXPLAIN PLAN FOR
SELECT count(*)
FROM part_test
WHERE starttime >= SYSDATE - 2;
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time | Pstart| Pstop |
| 0 | SELECT STATEMENT | | 1 | 8 | 7895 (1)| 00:00:01 | | |
| 1 | SORT AGGREGATE | | 1 | 8 | | | | |
| 2 | PARTITION RANGE ITERATOR| | 111K| 867K| 7895 (1)| 00:00:01 | KEY |1048575|
|* 3 | TABLE ACCESS FULL | PART_TEST | 111K| 867K| 7895 (1)| 00:00:01 | KEY |1048575|
Predicate Information (identified by operation id):
3 - filter("STARTTIME">=SYSDATE@!-2)
4) Now do the same query but with SYSDATE - 2 presented as a literal value.
This query returns the same answer but very different cost.
EXPLAIN PLAN FOR
SELECT count(*)
FROM part_test
WHERE starttime >= (to_date('23122013:0950','DDMMYYYY:HH24MI'))-2;
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time | Pstart| Pstop |
| 0 | SELECT STATEMENT | | 1 | 8 | 131 (0)| 00:00:01 | | |
| 1 | SORT AGGREGATE | | 1 | 8 | | | | |
| 2 | PARTITION RANGE ITERATOR| | 111K| 867K| 131 (0)| 00:00:01 | 356 |1048575|
|* 3 | TABLE ACCESS FULL | PART_TEST | 111K| 867K| 131 (0)| 00:00:01 | 356 |1048575|
Predicate Information (identified by operation id):
3 - filter("STARTTIME">=TO_DATE(' 2013-12-21 09:50:00', 'syyyy-mm-dd hh24:mi:ss'))
thanks in anticipation
JimAs Jonathan has already pointed out there are situations where the CBO knows that partition pruning will occur but is unable to identify those partitions at parse time. The CBO will then use a dynamic pruning which means determine the partitions to eliminate dynamically at run time. This is why you see the KEY information instead of a known partition number. This is to occur mainly when you compare a function to your partition key i.e. where partition_key = function. And SYSDATE is a function. For the other bizarre PSTOP number (1048575) see this blog
http://hourim.wordpress.com/2013/11/08/interval-partitioning-and-pstop-in-execution-plan/
Best regards
Mohamed Houri -
"ORA-14400: inserted partition key does not map to any partition"
Hi Experts,
While loading from DSO to Infocube, we are facing the below issue.,
Database error text........: "ORA-14400: inserted partition key does no
any partition"
Internal call code.........: "[RSQL/INSR//BIC/FZORIMB08C ]"
Please check the entries in the system log (Transaction SM21).
"DBIF_RSQL_SQL_ERROR" "CX_SY_OPEN_SQL_DB"
"GP4JNZKXF93GTRTVQO7A5J3Z2VD" or "GP4JNZKXF93GTRTVQO7A5J3Z2VD"
"WRITE_ICFACT"
Information on where terminated:
Termination occurred in the ABAP program "GP4JNZKXF93GTRTVQO7A5J3Z2VD" - in
"WRITE_ICFACT".
The main program was "RSBATCH_EXECUTE_PROZESS ".
In the source code you have the termination point in line 5185
of the (Include) program "GP4JNZKXF93GTRTVQO7A5J3Z2VD".
The program "GP4JNZKXF93GTRTVQO7A5J3Z2VD" was started as a background job.
Job Name....... "BIDTPR_1999057_1"
Job Initiator.. "ALE_POS"
Job Number..... 09074400
The termination is caused because exception "CX_SY_OPEN_SQL_DB" occurred in
procedure "WRITE_ICFACT" "(FORM)", but it was neither handled locally nor
declared
in the RAISING clause of its signature.
The procedure is in program "GP4JNZKXF93GTRTVQO7A5J3Z2VD "; its source code
begins in line
5147 of the (Include program "GP4JNZKXF93GTRTVQO7A5J3Z2VD ".
Can anyone help me.
Regards,
AaryanHi Aaryan,
From an Oracle point of view the error ORA-14400 is due to the partition range where an insert rows with a value out-of-bound of partition range
The system displays the error message if you want to load data into a partitioned table where the partitions have not been defined correctly.
Please run the RSRV test "Entries Not Used in the Dimension of an InfoCube " for the relevant Infocube and afterwards repair any errors with "Correct error" button.
Please have a look at the following notes:
#509660 - ORACLE ERROR 14400 during update to the InfoCube
#590370 - too many uncompressed request (f table partitions)
Finally please do not leave the infocube uncompressed.
Rgds,
Colum -
Hello everyone.
Oracle9i Enterprise Edition Release 9.2.0.5.0 - 64bit Production
18:44:45 161:tcs237427@EBPP> desc rs_ebpp.rs_mail_job
Name Null? Type
JOB_KEY NOT NULL VARCHAR2(50)
MOF_KEY NOT NULL VARCHAR2(50)
MOF_PATH VARCHAR2(255)
REPORT_ID NOT NULL NUMBER(38)
PROFILE_NAME VARCHAR2(25)
FILE_PATH VARCHAR2(1024)
JOB_DATE DATE
JOB_FINISH_DATE DATE
SCHEDULED NUMBER(38)
USER_ID VARCHAR2(25)
STATUS VARCHAR2(25)
COMMENTS VARCHAR2(255)
MAIL_FROM VARCHAR2(255)
MAIL_TO VARCHAR2(255)
MAIL_CC VARCHAR2(255)
MAIL_BCC VARCHAR2(255)
MAIL_SUBJECT VARCHAR2(255)
MAIL_MESSAGE VARCHAR2(512)
DATA7 VARCHAR2(50)
DATA8 VARCHAR2(50)
DATA9 VARCHAR2(50)
DATA20 DATE
DATA21 NUMBER
DATA22 NUMBER
DATA23 NUMBER
DATA24 NUMBER
DATA25 NUMBER
DATA26 FLOAT(126)
DATA27 FLOAT(126)
DATA28 FLOAT(126)
DATA29 FLOAT(126)
DATA30 FLOAT(126)
DATA10 VARCHAR2(50)
DATA11 VARCHAR2(50)
DATA12 VARCHAR2(50)
DATA13 VARCHAR2(50)
DATA14 VARCHAR2(50)
DATA1 VARCHAR2(50)
DATA15 VARCHAR2(50)
DATA16 DATE
DATA2 VARCHAR2(50)
DATA17 DATE
DATA3 VARCHAR2(50)
DATA4 VARCHAR2(50)
DATA18 DATE
DATA5 VARCHAR2(50)
DATA19 DATE
DATA6 VARCHAR2(50)
I tried to create range partition on table whose partitioning key (JOB_KEY) was a VARCHAR2. I was able to create the partitions till 30th Jan.
16:56:32 SQL> SELECT PARTITION_NAME,TABLE_NAME,HIGH_VALUE FROM DBA_TAB_PARTITIONS WHERE TABLE_NAME='RS_MAIL_JOB_NEW' ORDER BY 1;
Partition
Name TABLE_NAME HIGH_VALUE
P110101 RS_MAIL_JOB_NEW TO_DATE('2011-01-02 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110102 RS_MAIL_JOB_NEW TO_DATE('2011-01-03 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110103 RS_MAIL_JOB_NEW TO_DATE('2011-01-04 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110104 RS_MAIL_JOB_NEW TO_DATE('2011-01-05 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110105 RS_MAIL_JOB_NEW TO_DATE('2011-01-06 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110106 RS_MAIL_JOB_NEW TO_DATE('2011-01-07 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110107 RS_MAIL_JOB_NEW TO_DATE('2011-01-08 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110108 RS_MAIL_JOB_NEW TO_DATE('2011-01-09 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110109 RS_MAIL_JOB_NEW TO_DATE('2011-01-10 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110110 RS_MAIL_JOB_NEW TO_DATE('2011-01-11 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110111 RS_MAIL_JOB_NEW TO_DATE('2011-01-12 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110112 RS_MAIL_JOB_NEW TO_DATE('2011-01-13 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110113 RS_MAIL_JOB_NEW TO_DATE('2011-01-14 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110114 RS_MAIL_JOB_NEW TO_DATE('2011-01-15 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110115 RS_MAIL_JOB_NEW TO_DATE('2011-01-16 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110116 RS_MAIL_JOB_NEW TO_DATE('2011-01-17 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110117 RS_MAIL_JOB_NEW TO_DATE('2011-01-18 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110118 RS_MAIL_JOB_NEW TO_DATE('2011-01-19 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110119 RS_MAIL_JOB_NEW TO_DATE('2011-01-20 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110120 RS_MAIL_JOB_NEW TO_DATE('2011-01-21 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110121 RS_MAIL_JOB_NEW TO_DATE('2011-01-22 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110122 RS_MAIL_JOB_NEW TO_DATE('2011-01-23 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110123 RS_MAIL_JOB_NEW TO_DATE('2011-01-24 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110124 RS_MAIL_JOB_NEW TO_DATE('2011-01-25 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110125 RS_MAIL_JOB_NEW TO_DATE('2011-01-26 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110126 RS_MAIL_JOB_NEW TO_DATE('2011-01-27 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110127 RS_MAIL_JOB_NEW TO_DATE('2011-01-28 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110128 RS_MAIL_JOB_NEW TO_DATE('2011-01-29 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110129 RS_MAIL_JOB_NEW TO_DATE('2011-01-30 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
P110130 RS_MAIL_JOB_NEW TO_DATE('2011-01-31 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN
But after that I started getting below error:
16:56:40 SQL> alter table rs_ebpp.rs_mail_job_new add partition p110131 values less than (to_date('2011-02-01 00:00:00', 'SYYYY-mm-dd h
h24:mi:ss', 'nls_calendar=GREGORIAN'));
alter table rs_ebpp.rs_mail_job_new add partition p110131 values less than (to_date('2011-02-01 00:00:00', 'SYYYY-mm-dd hh24:mi:ss', 'n
ls_calendar=GREGORIAN'))
ERROR at line 1:
ORA-14074: partition bound must collate higher than that of the last partition
I checked the manual, but didn't find it written anywhere about restriction on column data type for Range partition or others.
I doubt if creating range partitions on columns with varchar2 is allowed and if yes, not sure why I get that errorTry to split first partition:
alter table rs_ebpp.rs_mail_job_new
split partition p110201 at (to_date('2011-31-01 00:00:00', 'SYYYY-mm-dd hh24:mi:ss', 'nls_calendar=GREGORIAN'))
into (partition p110131, partition p110201) ; -
How can I find Bitlocker External Key File location?
My Windows 8.1 PC includes a system drive and data drives. All the drivers were encrypted using Bitlocker with the data drives set for autounlock.
I recently decrypted the system drive (without decrypting the data drives) and reinstalled the OS, after which my data drives required the Bitlocker recovery key to unlock.
However, I had "backed up" the recovery keys to my Microsoft account but now I can find only the recovery keys for the system drive. The recovery keys for the data drives cannot be found on my Microsoft account.
I have tried to use the "manage-bde" command at the console to obtain the recovery password but I am only getting the Numerical Password ID and the External Key File Name. Can anyone provide advice on how I can retrieve the passwords or the
External Key File location?
Thanks.Hi Ridgewood,
As my point of viewer, the BitLocker Automatic unlock volume is also protected by BitLocker Disk Encryption. The user encrypted information is stored in the registry and volume metadata. After a user unlocks the operating system volume, BitLocker uses the
encrypted information to unlock the data volume automatically.
After the reinstallation of the system, the encrypted information is lost and BitLocker can’t unlock the data volume automatically.
Every volume has own recovery key.
As mentioned in your post, the data drive require the recovery key to unlock.
I suggest you to double-check the OneDrive and try to find out where did you store the recovery key.
If you can’t find the recovery key, we can’t help you to decrypt the data volume.
Best regards,
Fangzhou CHEN
Fangzhou CHEN
TechNet Community Support -
Find the partition for the fact table
Oracle version : Oracle 10.2
I have one fact table with daily partitions.
I am inserting some test data in this table for old date 20100101.
I am able to insert this record in this table as below
insert into fact_table values (20100101,123,456);
However I observed that the partition for this date does not exist in the table (all_tab_partitions) morever I am not able to select the data using
select * from facT_table partition(d_20100101)
but I am able to extract the data using
select * from facT_table where date_id=20100101
could some one please let me know the way to find the partition in which this data might be inserted
and if the partition for date 20100101 is not present then why insert for that date is working ?user507531 wrote:
However I observed that the partition for this date does not exist in the table (all_tab_partitions) morever I am not able to select the data using
select * from facT_table partition(d_20100101)Wrong approach.
but I am able to extract the data using
select * from facT_table where date_id=20100101Correct approach.
could some one please let me know the way to find the partition in which this data might be inserted
and if the partition for date 20100101 is not present then why insert for that date is working ?Who says that the date is invalid..? This is a range partition - which means that each partition covers a range. And if you bothered to read up in the SQL Reference Guide on how a range partition is defined, you will notice that each partition is defined with the end value of the range it covers. There is no start value - as the previous partition's end value is the "+border+" between this and the prior partition.
I suggest that before you use a database feature you first familiarise yourself with it. Else incorrectly using it, and making the wrong assumptions about it, more than likely results. -
What Oracle Table contains Partition Key Field Name?
What Oracle table/view maintains the partition key field name?
All_Tab_Partitions does not appear to maintain such information.
When I use Toad -> Schema -> Tables -> Partitions, it lists the partition key field name that the partition is based.
Thank Youall_part_key_columns
or
USER_part_key_columns
Edited by: OrionNet on Dec 5, 2008 3:56 PM -
Is there any plan of allowing non-range partitioning of spatial indexes
our application is really lacking the ability of spatial index being partitioned by list. We are binning information in our data warehouse by spatial locations, and when we query it (actually, MapViewer is generating a tile), entire geometry set is searched because table is list-partitioned and we cannot have local spatial index.
Hi,
I'm not in a position to answer the question about list partitioning, but would a pseudo-list partition implemented as a range partition work? For example, something like this:
CREATE TABLE customers (
first_name VARCHAR2(20),
last_name VARCHAR2(20),
address_1 VARCHAR2(20),
address_2 VARCHAR2(20),
city VARCHAR2(20),
state_abrv VARCHAR2(3),
postal_code VARCHAR2(10),
customer_id NUMBER NOT NULL UNIQUE)
PARTITION BY RANGE (state_abrv)
PARTITION ALASKA VALUES LESS THAN ('AKz'),
PARTITION ALABAMA VALUES LESS THAN ('ALz'),
PARTITION ARKANSAS VALUES LESS THAN ('ARz'),
PARTITION AMERICANSAMOA VALUES LESS THAN ('ASz'),
PARTITION ARIZONA VALUES LESS THAN ('AZz'),
PARTITION CALIFORNIA VALUES LESS THAN ('CAz'),
PARTITION WYOMING VALUES LESS THAN ('WYz')
);I've used this method in the past, although I agree a list partitioning solution is much preferred.
Dan
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