Updating partition key column would cause a partition change

10-Oct-2016 10:57 by 7 Comments

Updating partition key column would cause a partition change - Webcam portal girl

Database support blog provides information on oracle database architecture and instance memory background process.

updating partition key column would cause a partition change-33

[email protected] select partition_name,high_value,interval from user_tab_partitions 2 where table_name='LINE_ITEMS' 3 ; PARTITION_ HIGH_VALUE INT ---------- ---------- --- P2014_01 NO SYS_P4268 YES SYS_P4269 YES SYS_P4270 YES SYS_P4271 YES Conclusion 1: only the necessary partitions are created (this is a feature for interval partitioned tables, but it’s valid for the child table, too) Conclusion 2: partition naming is consistent between master and child. [email protected] select table_name,partition_name,num_rows from user_tab_partitions where table_name='LINE_ITEMS'; TABLE_NAME PARTITION_ NUM_ROWS --------------- ---------- ---------- LINE_ITEMS P2014_01 9 LINE_ITEMS SYS_P4270 10 LINE_ITEMS SYS_P4271 10 LINE_ITEMS SYS_P4278 20 select partition_name,high_value from user_tab_partitions where table_name='ORDERS'; PARTITION_ HIGH_VALUE ---------- ------------------------------------------------------------------------------------- P2014_01 TO_DATE(' 2014-02-01 ', ..

[email protected] select table_name,partition_name,num_rows from user_tab_partitions where table_name='LINE_ITEMS'; TABLE_NAME PARTITION_ NUM_ROWS --------------- ---------- ---------- LINE_ITEMS P2014_01 10 LINE_ITEMS SYS_P4270 10 LINE_ITEMS SYS_P4271 10 LINE_ITEMS SYS_P4278 19 [email protected] update orders set order_date=to_date('27-FEB-2014','DD-MON-YYYY') where order_id=1; update orders set order_date=to_date('27-FEB-2014','DD-MON-YYYY') where order_id=1 * ERROR at line 1: ORA-14402: updating partition key column would cause a partition change select table_name,partition_name,num_rows from user_tab_partitions where table_name='ORDERS'; TABLE_NAME PARTITION_ NUM_ROWS --------------- ---------- ---------- ORDERS P2014_01 9 ORDERS SYS_P4270 10 ORDERS SYS_P4271 11 ORDERS SYS_P4272 10 ORDERS SYS_P4273 10 ORDERS SYS_P4274 10 ORDERS SYS_P4275 10 ORDERS SYS_P4276 9 ORDERS SYS_P4277 20 9 rows selected.

Summary: partition merge is possible with interval reference partitionierung and affects both master and child table. Let’s continue with question #2: what happens if we update a row in the master table so that it has to be moved into another partition (ROW MOVEMENT) Initial situation: select table_name,partition_name,num_rows from user_tab_partitions where table_name='ORDERS'; TABLE_NAME PARTITION_ NUM_ROWS --------------- ---------- ---------- ORDERS P2014_01 10 ORDERS SYS_P4270 10 ORDERS SYS_P4271 11 ORDERS SYS_P4272 10 ORDERS SYS_P4273 10 ORDERS SYS_P4274 10 ORDERS SYS_P4275 10 ORDERS SYS_P4276 9 ORDERS SYS_P4277 19 9 rows selected.

There are 2 solutions to this which will be explained in detail below.

example CREATE TABLE T1( TRANS_ID NUMBER, PARTITION_ID NUMBER) TABLESPACE USERS; CREATE TABLE T2( TRANS_ID NUMBER, PARTITION_ID NUMBER) PARTITION BY RANGE(PARTITION_ID) –interval (1) store IN (USERS)( PARTITION P_TRANS_BATCH_ID01 values less than (1001), PARTITION P_TRANS_BATCH_ID02 values less than (1002), PARTITION P_TRANS_BATCH_ID03 values less than (1003)) TABLESPACE USERS; INSERT INTO T2 VALUES(1,1001); — ok INSERT INTO T2 VALUES(1,1002); — ok INSERT INTO T2 VALUES(1,1003); — error You will get the error SQL Error: ORA-14400: inserted partition key does not map to any partition Solution 1CREATE TABLE T2( TRANS_ID NUMBER, PARTITION_ID NUMBER) PARTITION BY RANGE(PARTITION_ID) –interval (1) store IN (USERS)( PARTITION p_DEFAULT VALUES less than(MAXVALUE)) TABLESPACE USERS; Solution 22. In this case oracle will create the partiton for these records and insert.example CREATE TABLE T2( TRANS_ID NUMBER, PARTITION_ID NUMBER) PARTITION BY RANGE(PARTITION_ID) interval (1) store IN (USERS)( PARTITION P_TRANS_BATCH_ID01 values less than (2)) TABLESPACE USERS; INSERT INTO T2 VALUES(1,1001); –ok INSERT INTO T2 VALUES(1,1002); –ok INSERT INTO T2 VALUES(1,1003); –ok INSERT INTO T2 VALUES(1,1004); –ok INSERT INTO T2 VALUES(1,1005); –ok Both solutions will need to be analysed based on the application design and requirement before finalizing the approach.

Partitions, though get created only when data is inserted, care must be taken when selecting the partition ids (if serially increasing number partitoned table) as the partition keys get logically created. Selecting very high values for the partition key will result in the above error.

exampledrop table t CASCADE CONSTRAINTS;create table t(trans_id NUMBER) partition by range(trans_id) interval(1) ( partition p1 values less than (2)); insert into t values(1048574); — ok insert into t values(1048575); — ok insert into t values(1048576); –error SQL Error: ORA-14401: inserted partition key is outside specified partition ————————————————————————————————————————Sometimes for a partitioned table you may get the below error when trying to update the partition key.

00000 – “inserted partition key does not map to any partition”*Cause: An attempt was made to insert a record into, a Range or Composite Range object, with a concatenated partition key that is beyond the concatenated partition bound list of the last partition -OR- An attempt was made to insert a record into a List object with a partition key that did not match the literal values specified for any of the partitions.*Action: Do not insert the key.

Or, add a partition capable of accepting the key, Or add values matching the key to a partition specification The error occurs when you try to insert data into a partitioned tablewhich does not have a partition for the specific partition key.

create table orders 2 ( 3 order_id number not null constraint pk$orders primary key using index, 4 order_date date not null 5 ) 6 partition by range (order_date) interval (numtoyminterval(1,'month')) 7 ( 8 partition p2014_01 values less than (to_date('01-feb-2014','DD-MON-YYYY')) 9 ) 10 ; Table created.

[email protected] create table line_items 2 ( 3 order_id number not null, 4 pos number not null, 5 item_id number not null, 6 quantity number not null, 7 constraint pk$line_items primary key (order_id, pos) using index, 8 constraint fk$order_id foreign key (order_id) references orders (order_id) 9 ) 10 partition by reference (fk$order_id); Table created.

As often, I use this blog to answer questions of training participants.

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