Hive table dynamic partition load using Sqoop - hive

I have a MySQL table called dept which I load to Hive EXTERNAL table dept table as incremental append check-column last-value based on a time stamp column ts.
This is working fine and now I want to add dynamic partition to my hive table dept based on dept_name and ts columns.
I know that I can create table dynamic partition outside sqoop but need help in loading data to dynamic partitions. I saw one example using --hive-partition-key & --hive-partition-value but there the value is hard coded. Without hard coding the hive-partition-values, can I load data automatically to corresponding partitions using SQOOP say (dept1,timestamp1/dept1,timestamp2..etc) and (dept2,timestamp1/dept2,timestamp2...etc)?

Yes, you can load the data from the external table you have created into the dynamic partitioned table without hardcoding the partitioned values.
Assuming that you already have your external table externalDept with the values and also that you have created your partitioned table dept, you could use something like this:
INSERT OVERWRITE TABLE dept
PARTITION (dept_name, ts)
SELECT ..., ed.dept_name, ed.ts
FROM externalDept ed
It's not part of your question, but you could also mix both static and dynamic partitions. For instance, if you wanted to specify a dynamic value for the timestamp but a static value for the dept, you could do something like this:
INSERT OVERWRITE TABLE dept
PARTITION (dept_name = 'Math', ts)
SELECT ..., ed.dept_name, ed.ts
FROM externalDept ed
WHERE ed.dept_name = 'Math'

Related

How to modify CTAS query to append query results to table based on if new partition doesn't exist? - Athena

I have a query that I want to execute daily that's to be partitioned by the date it's executed. The results of this query should be appended to a the same table.
My idea was ideally having something similar to the CREATE TABLE IF NOT EXISTS command for adding data by a new partition every day to the existing table if the partition doesn't already exist, but I can't figure out how I'd be able to integrate this in my query.
My query:
CREATE TABLE IF NOT EXISTS db_name.table_name
WITH (
external_location = 's3://my-query-results-location/',
format = 'PARQUET',
parquet_compression = 'SNAPPY',
partitioned_by = ARRAY['date_executed'])
AS
SELECT
{columns_that_I_am_selecting_here_including_'date_executed'}
What this does is create a new table for the first day it's executed but nothing happens for subsequent days, I'm assuming because of the CREATE TABLE IF NOT EXISTS validating that the table already exists and not proceeding with the logic.
Is there a way to modify my query to create a table for the first day executed and append the results by a new partition for each subsequent day?
I'm quite sure ALTER TABLE table_name ADD [IF NOT EXISTS] PARTITION would not apply to my use case here as I'm running a CTAS query.
You can simply use INSERT INTO existing_table SELECT....
Presumably your table is already partitioned, so include that partition column in the SELECT and Amazon Athena will automatically put the data in the correct directory.
For example, you might include hte column like this: SELECT ... CURRENT_DATE as date_executed
See: INSERT INTO - Amazon Athena

How do I partition a table by all values?

I have an external table, now I want to add partitions to it. I have 224 unique city id's and I want to just write alter table my_table add partition (cityid) location /path; but hive complains, saying that I don't provide anything for the city id value, it should be e.g. alter table my_table add partition (cityid=VALUE) location /path;, but I don't want to run alter table commands for every value of city id, how can I do it for all id's in one go?
This is what hive command line looks like:
hive> alter table pavel.browserdata add partition (cityid) location '/user/maria_dev/data/cityidPartition';
FAILED: ValidationFailureSemanticException table is not partitioned but partition spec exists: {cityid=null}
Partition on physical level is a location (separate location for each value, usually looks like key=value) with data files. If you already have partitions directory structure with files, all you need is to create partitions in Hive metastore, then you can point your table to the root directory using ALTER TABLE SET LOCATION, then use MSCK REPAIR TABLE command. The equivalent command on Amazon Elastic MapReduce (EMR)'s version of Hive is: ALTER TABLE table_name RECOVER PARTITIONS. This will add Hive partitions metadata. See manual here: RECOVER PARTITIONS
If you have only not-partitioned table with data in it's location, then adding partitions will not work because the data needs to be reloaded, you need to:
Create another partitioned table and use insert overwrite to load partition data using dynamic partition load:
set hive.exec.dynamic.partition=true;
set hive.exec.dynamic.partition.mode=nonstrict;
insert overwrite table2 partition(cityid)
select col1, ... colN,
cityid
from table1; --partitions columns should be last in the select
This is quite efficient way to reorganize your data.
After this you can delete source table and rename your target table.

Hive table creation with a default value

I have a table in RDBMS like so:
create table test (sno number, entry_date date default sysdate).
Now I want to create a table in hive with a structure as adding a default value to a column.
Hive currently doesn't support the feature of adding default value to any column while creating a table.
As a workaround load data into a temporary table and use the insert overwrite table statement to add the current date and time into the main table.
Create a temporary table:
create table test (sno number);
Load data into the table:
Create final table:
create table final_table (sno number, createDate string);
Finally load the data from temp test table to the final table:
insert overwrite table final_table select sno, FROM_UNIXTIME( UNIX_TIMESTAMP(), 'dd/MM/YYYY' ) from test;
Hive doesn't support DEFAULT fields
Doesn't mean you can't do it, though. Just a two step process of creating one "staging" table, then inserting into a second table and selecting that "default" value.
Adding a default value to a column while creating table in hive
Since you mention,
I've table in RDBMS
You could also use your existing table, and use Sqoop to import the data into Hive.

How can I create a partitioned table 'like' an unpartitioned table with Hive HQL?

I've got a table with two weeks worth of entries, and I would like to copy those entries into a table partitioned by date (creating it if it does not exist).
I'm writing a luigi task to do this, and I would love for it to be independent of the table schema--i.e. I wouldn't have to specify column names and types, and it would CREATE TABLE IF NOT EXISTS when necessary.
I was hoping I could use:
CREATE TABLE IF NOT EXISTS test_part
COMMENT 'This is a test table to see if partitioning works in this case'
PARTITIONED BY (event_date string)
AS select *, '2014-12-15' from source_db.source_table
where event_at <'2014-12-16' and event_at >='2014-12-15';
But this of course fails with: FAILED: SemanticException [Error 10068]: CREATE-TABLE-AS-SELECT does not support partitioning in the target table
I tried again with "like" with basically the same results. Is there a way to do this that I am missing? It doesn't have to be atomic. Multiple sequential commands are fine.
You do not do a create table as.
You create a table first using describe source_table and then you make an insert into table partition (event_date string)
2 steps it works better.

How can I copy a Redshift table but add a sortkey to a column?

I'm currently working on a project that uses a Redshift table with 51 columns. However, the person who made the table forgot to add a sortkey to our time column which will hurt performance for our use case if we don't add it.
How can I make a version of the table with our time column as the sortkey? I'm aware that you can't make a column a sortkey if its a member of an existing table, but I was hoping there's a way to do it that doesn't involve writing out the CREATE TABLE syntax by hand; for example, something like this would be nice:
timecube=# CREATE TABLE foo (like bar) sortkey(time);
ERROR: CREATE TABLE LIKE is not supported with DISTSTYLE, DISTKEY(), or SORTKEY() clauses
but as you can see its not supported. Is there another way? As we're still developing we don't need any of existing data.
Using traditional tools like pgdump didn't work well because they don't include any of the Redshift extras like encoding.
Redshift supports specifying the DIST and SORT keys as part of CREATE TABLE AS statements, as per the docs.
CREATE TABLE table_name
DISTSTYLE KEY
DISTKEY ( column )
SORTKEY ( column )
AS
(SELECT *
FROM source_table)
;
First step you need to do use get create table statement for existing table. Then create new table this time add sort key to new table.
Check encoding for old table ( when you load data using copy command it automatically adds compression encodings)
select "column", type, encoding
from pg_table_def where tablename = 'old_table'
When creating new table add encoding type for each column. Create table with Sort key .
Once new table is created use below command
insert into new table ( select * from old table order by time asc)