I know how partitioning in DB2 works but I am unaware about where this partition values exactly get stored. After writing a create partition query, for example:
CREATE TABLE orders(id INT, shipdate DATE, …)
PARTITION BY RANGE(shipdate)
(
STARTING '1/1/2006' ENDING '12/31/2006'
EVERY 3 MONTHS
)
after running the above query we know that partitions are created on order for every 3 month but when we run a select query the query engine refers this partitions. I am curious to know where this actually get stored, whether in the same table or DB2 has a different table where partition value for every table get stored.
Thanks,
table partitions in DB2 are stored in tablespaces.
For regular tables (if table partitioning is not used) table data is stored in a single tablespace (not considering LOBs).
For partitioned tables multiple tablespaces can used for its partitions.
This is achieved by the "" clause of the CREATE TABLE statement.
CREATE TABLE parttab
...
in TBSP1, TBSP2, TBSP3
In this example the first partition will be stored in TBSP1, the second in TBSP2, The third in TBSP3, the fourth in TBSP1 and so on.
Table partitions are named in DB2 - by default PART1 ..PARTn - and all these details can be looked up in the system catalog view SYSCAT.DATAPARTITIONS including the specified partition ranges.
See also http://www-01.ibm.com/support/knowledgecenter/SSEPGG_10.5.0/com.ibm.db2.luw.sql.ref.doc/doc/r0021353.html?cp=SSEPGG_10.5.0%2F2-12-8-27&lang=en
The column used as partitioning key can be looked up in syscat.datapartitionexpression.
There is also a long syntax for creating partitioned tables where partition names can be explizitly specified as well as the tablespace where the partitions will get stored.
For applications partitioned tables look like a single normal table.
Partitions can be detached from a partitioned table. In this case a partition is "disconnected" from the partitioned table and converted to a table without moving the data (or vice versa).
best regards
Michael
After a bit of research I finally figure it out and want to share this information with others, I hope it may come useful to others.
How to see this key values ? => For LUW (Linux/Unix/Windows) you can see the keys in the Table Object Editor or the Object Viewer Script tab. For z/OS there is an Object Viewer tab called "Limit Keys". I've opened issue TDB-885 to create an Object Viewer tab for LUW tables.
A simple query to check this values:
SELECT * FROM SYSCAT.DATAPARTITIONS
WHERE TABSCHEMA = ? AND TABNAME = ?
ORDER BY SEQNO
reference: http://www-01.ibm.com/support/knowledgecenter/SSEPGG_9.5.0/com.ibm.db2.luw.sql.ref.doc/doc/r0021353.html?lang=en
DB2 will create separate Physical Locations for each partition. So each partition will have its own Table-space. When you SELECT on this partitioned Table your SQL may directly go to a single partition or it may span across many depending on how your SQL is. Also, this may allow your SQL to run in parallel i.e. many TS can be accessed concurrently to speed up the SELECT.
Related
I have a postgres table (in postgres12) which is supposed to have thousands of partitions (200k at least) in near future.
Here is how I am creating the parent table:
create table if not exists content (
key varchar(20) not NULL,
value json not null default '[]'::json
) PARTITION BY LIST(key)
And then adding any given child tables like:
create table if not exists content_123 PARTITION OF content for VALUES in ('123');
Also I am adding an index on top of the child table for quick access (since I will be accessing the child table directly):
create index if not exists content_123_idx on content_123 using btree(key)
Here is my question: I have never in the past managed this many partitions in a postgres table so I am just wondering is there any downside of doing what I am doing? Also, (as mentioned above) I will not be querying from the parent table directly, but will read directly from individual child tables.
With these table definitions, the index is completely useless.
With 200000 partitions, query planning will become intolerably slow, and each SQL statement will need very many locks and open files. This won't work well.
Lump several keys together into a single partition (then the index might make sense).
I'm new to table partitions in Postgres and have one doubt.
Let us assume I have a table:
product_visitors
I can create multiple partitions like:
product_visitors_year_2017
product_visitors_year_2018
etc.
I can create a trigger which can redirect insertion on product_visitors to appropriate table.
My question is, what if I want to aggregate on full data of product_visitors? For example, products and their visit count
As I understand, at the moment, data resides in year wise tables instead of main table
In Postgres 10 inserts will automatically be routed to the correct partition.
If you select from the "base table" product_visitors without any condition limiting the rows to one (or more) specific partitions, Postgres will automatically read the data from all partitions.
So
select count(*)
from product_visitors;
will count the rows in all partitions.
Redshift documentation identifies time-series tables as a best practice:
http://docs.aws.amazon.com/redshift/latest/dg/c_best-practices-time-series-tables.html
However, it doesn't address any of the following issues:
how many tables within a union-all view is reasonable - hundreds? (unanswered)
any method of writing to the union-all view and having redshift direct those inserts to the correct underlying tables? (Answer: no)
most effective method of loading underlying tables? Perhaps using firehose to insert into a staging table then periodically inserting those rows into appropriate table within union-all view? (unanswered)
any way to enable redshift to eliminate some underlying partitions (tables) when querying the union-all view if their date range is outside of a query's criteria? (Answer: No)
can redshift support dropping old tables, adding new tables and rebuilding union-all view within a transaction? (unanswered)
My situation:
100 million rows added daily, which will grow to 500 million in 3 years
12 month retention desired
Estimated 99% of all queries will hit the most recent 1-7 days
Data is written to existing table via kinesis firehose to s3 which then triggers a copy to redshift table.
My proposed solution:
Create a year of daily tables with a union all view, along with a dist_key of sensor_id (100,000+ uniq values) and a sort_key of (timestamp, sensor_id).
Have firehose load into staging table
Create separate process that once an hour queries staging table to discover dates of data within table, then performs an insert into 'appropriate table' select * from where timestamp = table's timestamp.
This hourly writer can probably wrap a table rename, multiple insert-selects, and table recreate in a transaction to be invisible to firehose.
Once a month drop old tables, create next month of tables, and rebuild view.
This union-all view maintenance can probably be wrapped in a transaction to avoid impacts to users.
Once a night run the vacuum analyzer.
EDITS: added notes identifying which issues have been answered, and added some detail to the proposed solution.
Your proposed process sounds quite good! While I can't answer all your questions, here is some information:
Any method of writing to the union-all view and having redshift direct those inserts to the correct underlying tables?
Views are read-only. It is not possible to write to a view, nor is it possible to insert data while expecting Redshift to send it to an appropriate table (eg a specific table for the given day).
Any way to enable redshift to eliminate some underlying partitions (tables) when querying the union-all view if their date range is outside of a query's criteria?
Redshift will not exclude specific tables from the query, but it will avoid reading particular disk blocks through the use of Zone Maps. Each block of data written to disk is associated with a specific table and column. The block has a Zone Map, which indicates the minimum and maximum values of that field stored within the block.
If a query includes a WHERE clause, Redshift can skip blocks that do not contain relevant data. This is particularly powerful when used on the SORTKEY column, since similar ranges of data are grouped together.
Given that you are using a date as the SORTKEY, Redshift will read very few disk blocks if the query includes a WHERE clause based on that column. This is very similar to the idea of skipping tables, but it actually skips reading disk blocks.
I have a schema with one table with the majority of data, customer, and three other tables with foreign key references to customer.entry_id which is a BIGSERIAL field. The three other tables are called location, devices and urls where we store various data related to a specific entry in the customer table.
I want to partition the customer table into monthly child tables, and have that part worked out; customer will stay as-is, each month will have a table customer_YYYY_MM that inherits from the master table with the right CHECK constraint and indexes will be created on each individual child table. Data will be moved to the correct child tables while the master table stays empty.
My question is about the other three tables, as I want to partition them as well. However, they have no date information (at all), only the reference to the primary key from the master table. How can I setup the constraints on these tables? Is it even meaningful or possible without date information?
My application logic knows where to insert all the data (it's fairly trivial), but I expect to be able to do simple SELECT queries without specifying which child tables to get it from. So this should work as you would expect from non-partitioned tables:
SELECT l.*
FROM customer c
JOIN location l USING entry_id
WHERE c.date_field > '2015-01-01'
I would partition them by the reference key. The foreign key is used in join conditions and is not usually subject to change so it fulfills the following important points:
Partition by the information that is used mostly in the WHERE clauses of the queries or other parts where partitioning can be used to filter out tables that don't need to be scanned. As one guide puts it:
The objective when defining partitions should be to allow as many queries as possible to fetch data from as few partitions as possible - ideally one.
Partition by information that is not going to be changed so that rows don't constantly need to be thrown from one subtable to another
This all depends of the size of the tables too of course. If the sizes stay small then there is no need to partition.
Read more about partitioning here.
Use views:
create view customer as
select * from customer_jan_15 union all
select * from customer_feb_15 union all
select * from customer_mar_15;
create view location as
select * from location_jan_15 union all
select * from location_feb_15 union all
select * from location_mar_15;
I want to partition an external table in hive based on range of numbers. Say numbers with 1 to 100 go to one partition. Is it possible to do this in hive?
I am assuming here that you have a table with some records from which you want to load data to an external table which is partitioned by some field say RANGEOFNUMS.
Now, suppose we have a table called testtable with columns name and value. The contents are like
India,1
India,2
India,3
India,3
India,4
India,10
India,11
India,12
India,13
India,14
Now, suppose we have a external table called testext with some columns along with a partition column say, RANGEOFNUMS.
Now you can do one thing,
insert into table testext partition(rangeofnums="your value")
select * from testtable where value>=1 and value<=5;
This way all records from the testtable having value 1 to 5 will come into one partition of the external table.
The scenario is my assumption only. Please comment if this is not the scenario you have.
Achyut