Am creating an Oracle HASH Table Partitions by using the below query
CREATE TABLE Table1 (
ID NUMBER, NAME VARCHAR2(50))
PARTITION BY HASH (ID)
PARTITIONS 25
STORE IN (Tablespace1);
Which Creates 25 HASH table partitions and also, the Database generates the 25 Unique partition names like SYS_P122, SYS_P123, SYS_P124... and so on for the partitions. Is there a way to find out this Partition lets say SYS_P123 is a system generated Partition name with the help of Oracle Catalog tables.
With the below link
http://docs.oracle.com/cd/B28359_01/server.111/b28320/statviews_2096.htm#REFRN20281
I could find the Oracle Table Partition information, but this catalog table does not have any value to say that the give Table Partition is a system generate or not. Is there any way to find out the given table partition name is system generated ?
Am using Oracle version 10 and 11.
Thanks,
Ravi,
Yes. The generated column in dba_objects gives the information.
Run the following query -
select owner, object_name, subobject_name, generated from all_objects where object_name = 'TABLE1' and object_type = 'TABLE PARTITION';
View the description for the 'generated' column in the following link - http://docs.oracle.com/cd/B28359_01/server.111/b28320/statviews_1145.htm#REFRN20146
Related
I am attempting to alter the data type of a column from a NUMBER to a VARCHAR2 in an existing database table. When running the following ALTER TABLE statement I receive the "ORA-54033: column to be modified is used in a virtual column expression" error:
ALTER TABLE table MODIFY (col1 varchar2(8));
I have already worked through the directions listed here. When looking at the SYS generated export statistics using the following query
select column_name, data_default, hidden_column
from user_tab_cols
where table_name = 'Table';
there is nothing referencing col1 in export statistics. There are about 15 hidden, SYS generated rows associated with the table and they all have a data default value of <Long>. There are no virtual columns in the DDL, nor is this column being used for any indexes or as a FK. I have also had the DBA run the following:
SELECT EXTENSION_NAME, EXTENSION, CREATOR, DROPPABLE
FROM DBA_STAT_EXTENSIONS
WHERE TABLE_NAME='Table'
and the output lines up with what I find in user_tab_cols. Where else can I look for this seemingly buried virtual column?
I am new to Hive and have some problems. I try to find a answer here and other sites but with no luck... I also tried many different querys that come to my mind, also without success.
I have my source table and i want to create new table like this.
Were:
id would be number of distinct counties as auto increment numbers and primary key
counties as distinct names of counties (from source table)
You could follow this approach.
A CTAS(Create Table As Select)
with your example this CTAS could work
CREATE TABLE t_county
ROW FORMAT DELIMITED FIELDS TERMINATED BY ','
STORED AS TEXTFILE AS
WITH t AS(
SELECT DISTINCT county, ROW_NUMBER() OVER() AS id
FROM counties)
SELECT id, county
FROM t;
You cannot have primary key or foreign keys on Hive as you have primary key on RBDMSs like Oracle or MySql because Hive is schema on read instead of schema on write like Oracle so you cannot implement constraints of any kind on Hive.
I can not give you the exact answer because of it suppose to you must try to do it by yourself and then if you have a problem or a doubt come here and tell us. But, what i can tell you is that you can use the insertstatement to create a new table using data from another table, I.E:
create table CARS (name string);
insert table CARS select x, y from TABLE_2;
You can also use the overwrite statement if you desire to delete all the existing data that you have inside that table (CARS).
So, the operation will be
CREATE TABLE ==> INSERT OPERATION (OVERWRITE?) + QUERY OPERATION
Hive is not an RDBMS database, so there is no concept of primary key or foreign key.
But you can add auto increment column in Hive. Please try as:
Create table new_table as
select reflect("java.util.UUID", "randomUUID") id, countries from my_source_table;
I need to do exchange partition. I have table_A which is partitioned by File_ID and sub-partitioned by file_type. I need to create temp table within my PL/SQL block which has the table structure as Table_A (including partitions). I need something like below:
create table temp_tab as
select * from table_A partition(SYS_P8924) where file_id=1000
partition by file_type
I know this a bad code example but somehow Can I achieve this? Appreciate help on this.
No worries guys! I got the solution:
create table temp_tab
partition by list (file_type)
(
partition values ('P1'),
partition values ('P2')
)
as
select * from table_A partition(SYS_P8924) where file_id=1000
Since my sub-partitions were 2 in number, hence it was partitioned into 2 sets.
I am a moderate user of Oracle and I had to create some of the tables in specified table space as shown below.
create table t_abc tablespace abc_tbspc as select * from abc;
create table t_xyz tablespace abc_tbspc as select * from xyz;
After running these through jobs (file containing around 5 tables to be created in a single tablespace), I could see that the table t_abc is created in abc_tbspc ; but the table t_xyz is assigned to null when I query the all_tables. Not sure why the 2nd table is not created in the specified tablespace even though there is abundant space in the table space.
TABLESPACE_NAME will be null for one of these reasons:
Temporary Temporary tables use a temporary tablespace.
Index Organized Index-organized tables store data in an index, not in a heap.
Partitioned Each partition could have a different tablespace, there is not necessarily one tablespace for the whole table.
External External tables do not store their data in a tablespace.
Your code should not meet one of the conditions above; did you leave out some details? I ran the query below to look for other cases where TABLESPACE_NAME is null but could not find any.
select *
from dba_tables
where tablespace_name is null
and (temporary is null or temporary <> 'Y') -- #1
and (iot_type is null or iot_type <> 'IOT') -- #2
and (partitioned is null or partitioned <> 'YES') -- #3
and (owner, table_name) not in -- #4
(select owner, table_name from dba_external_tables)
I'm using Amazon's Elastic MapReduce and I have a hive table created based on a series of log files stored in Amazon S3 and split in folders by day like so:
data/day=2011-09-01/log_file.tsv
data/day=2011-09-02/log_file.tsv
I am currently trying to create an additional table which filters out some unwanted activity in these log files but I can't figure out how to do this and keep getting errors such as:
FAILED: Error in semantic analysis: need to specify partition columns because the destination table is partitioned.
If my initial table create statement looks something like this:
CREATE EXTERNAL TABLE IF NOT EXISTS table1 (
... fields ...
)
PARTITIONED BY ( DAY STRING )
ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t'
LOCATION 's3://bucketname/data/';
That initial table works fine and I've been able to query it with no problems.
How then should I create a new table that shares the structure of the previous one but simply filters out data? This doesn't seem to work.
CREATE EXTERNAL TABLE IF NOT EXISTS table2 LIKE table1;
FROM table1
INSERT OVERWRITE TABLE table2
SELECT * WHERE
col1 = '%somecriteria%' AND
more criteria...
;
As I've stated above, this returns:
FAILED: Error in semantic analysis: need to specify partition columns because the destination table is partitioned.
Thanks!
This always works for me:
CREATE EXTERNAL TABLE IF NOT EXISTS table2 LIKE table1;
INSERT OVERWRITE TABLE table2 PARTITION (day) SELECT col1, col2, ..., day FROM table1;
ALTER TABLE table2 RECOVER PARTITIONS;
Notice that I've added 'day' as a column in the SELECT statement. Also notice that there is an ALTER TABLE line which is necessary for Hive to become aware of the partitions that were newly created in table2.
I have never used the like option.. so thanks for showing me that. Will that actually create all of the partitions that the first table has as well? If not, that could be the issue. You could try using dynamic partitions:
create external table if not exists table2 like table1;
insert overwrite table table2 partition(part) select col1, col2 from table1;
Might not be the best solution, as I think you have to specify your columns in the select clause (as well as the partition column in the partition clause).
And, you must turn on dynamic partitioning.
I hope this helps.