The simplest way to see if your database is part of a data guard setup

Execute this query:

select * from v$dataguard_config;

DB_UNIQUE_NAME	PARENT_DBUN	DEST_ROLE	CURRENT_SCN	CON_ID
proddb01	NONE	PRIMARY DATABASE	238029367683	0
proddb01_stby1	proddb01	PHYSICAL STANDBY	238029353772	0

In this example, I am logged into the database proddb01, the primary database.

The columnn PARENT_DBUN is interesting. It displays "DB_UNIQUE_NAME of the parent database, also known as the database that supplies live redo to the destination."

If executed on the standby database, the output will be reversed, but the principle the same.

Documentation for the view V$DATAGUARD_CONFIG can be found here

How to troubleshoot query rewrites

1. Run the script $ORACLE_HOME/rdbms/admin/utlxrw.sql logged in as the schema who owns the materialized view.

2. Analyze the query by logging on as the schema owning the materialized view and execute:

set serveroutput on

begin
      DBMS_MVIEW.EXPLAIN_REWRITE('select dato, max(sekvensnummer)'||
            'from DATE_TO_SEQNUM where dato <= :B1 group by dato','SEQ_AGGREGATED','x');
end;
/

where the first argument is the one you expect to use query rewrite, and the second argument is the name of the materialized view. 3. The potential reasons for the failed rewriting can be found in the table REWRITE_TABLE:

select message from rewrite_table;

What are Index-organized tables and how are they created?

An example of creating an IOT table from my own experience:

CREATE TABLE DATE_TO_SEQNUM (
    DATO DATE, 
    SEQNUM NUMBER(19) NOT NULL,
CONSTRAINT PK_DTS_IOT PRIMARY KEY (DATO,SEQNUM)
)
ORGANIZATION INDEX
TABLESPACE USERS;

If you expect a lot of repeating entries in the IOT, you can use index compression, like this:

CREATE TABLE DATE_TO_SEQNUM (
    DATO DATE, 
    SEQNUM NUMBER(19) NOT NULL,
CONSTRAINT PK_DTS_IOT PRIMARY KEY (DATO,SEQNUM)
)
ORGANIZATION INDEX
TABLESPACE USERS 
COMPRESS;

In my case, the IOT was really this simple. There are some other important directives when creating an IOT, too, which may be applicable in other cases. They are the use of OVERFLOW partition, the INCLUDING keyword, and the PCTTHRESHOLD-clause. I will come back to them when I have had a chance to put them into practice. In the mean time, read about them here.

Here are some statements from the Oracle documentation which defines what an IOT is, and the potential benefits the can provide:

* Index-organized tables are tables stored in an index structure.

* In an index-organized table, rows are stored in an index defined on the primary key for the table. Each index entry in the B-tree also stores the non-key column values. Thus, the index is the data, and the data is the index.

* Index-organized tables provide faster access to table rows by primary key or a valid prefix of the key. The presence of non-key columns of a row in the leaf block avoids an additional data block I/O.

* Index-organized tables are useful when related pieces of data must be stored together or data must be physically stored in a specific order.


Sources: Oracle Database 12.2 documentation

How to create a LIST-RANGE sub-partitioned table in PostgreSQL

Create the table:

CREATE TABLE orders(
  id            integer,
  country_code  VARCHAR(5),
  customer_id   integer,
  order_date    DATE,
  order_total   numeric(8,2)
)
PARTITION BY LIST (country_code);

Create partition for your table. I have defined two. Define it so that it is ready to be further diveded into sub-partitions:

CREATE TABLE orders_NO
 PARTITION OF orders 
 FOR VALUES IN ('no') 
 PARTITION BY RANGE (order_total);

CREATE TABLE orders_SE
 PARTITION OF orders 
 FOR VALUES IN ('se') 
 PARTITION BY RANGE (order_total);

Finally, create your sub-partitions and set their boundaries:

CREATE TABLE small_orders_no
 PARTITION OF orders_NO FOR VALUES FROM (0) TO (2000);

CREATE TABLE medium_orders_no
 PARTITION OF orders_NO FOR VALUES FROM (2001) TO (4000);

CREATE TABLE large_orders_no
 PARTITION OF orders_NO FOR VALUES FROM (4001) to (maxvalue);

CREATE TABLE small_orders_se
PARTITION OF orders_SE FOR VALUES FROM (0) TO (2000);

CREATE TABLE medium_orders_se
 PARTITION OF orders_SE FOR VALUES FROM (2001) TO (4000);

CREATE TABLE large_orders_se
 PARTITION OF orders_SE FOR VALUES FROM (4001) to (maxvalue);

How to avoid ERROR: no partition of relation ... found for row in PostgreSQL

Building on my previous post, I would now like to insert the following rows into my LIST partitioned table:

INSERT INTO ORDERS values (1,'NO',100,'01.02.2018',1000);
INSERT INTO ORDERS values (2,'SE',101,'01.03.2018',1200);

So one rows that I expect to go into the partition for Norway, and another row for the partition for Sweden.

I execute the following:

INSERT INTO ORDERS values (1,'NO',100,'01.02.2018',1000);

which results in an error:

Error: ERROR: no partition of relation "orders" found for row
  Detail: Partition key of the failing row contains (country_code) = (NO).

Contrary to Oracle, with PostgreSQL you need to explicitly create your partition first, like this:

CREATE TABLE orders_NO
 PARTITION OF orders FOR VALUES IN ('NO');

After this is done, PostgreSQL will accept your INSERT-statement.

You can also enforce contraints at the same time, if desirable:

CREATE TABLE orders_NO
 PARTITION OF orders (
 CONSTRAINT orders_id_nn CHECK (id is not null)
) FOR VALUES IN ('NO');

How to create a LIST-partitioned table with a primary key in PostgreSQL

In my experiements with a new RDBMS, I have tried to replicate what I am familiar with from Oracle, to Postgres.

Postgres supports partitioning: RANGE, LIST and HASH.

Here are some findings regarding LIST partitioning, which is extensively used in Oracle by my customers:

To create a list partitioned table in Postgres:

CREATE TABLE orders(
  order_id      integer,
  country_code  VARCHAR(5),
  customer_id   integer,
  order_date    DATE,
  order_total   numeric(8,2)
)
PARTITION BY LIST (country_code);

Add the desired partitions:

CREATE TABLE orders_NO
 PARTITION OF orders FOR VALUES IN ('NO');

CREATE TABLE orders_SE
 PARTITION OF orders FOR VALUES IN ('SE');

When working with Oracle, the equivalent table in Oracle could have been defined with a primary key on the column order_id:

CREATE TABLE orders(
  order_id      number primary key,
  country_code  varchar2(5),
  customer_id   number,
  order_date    date,
  order_total   number(8,2)
)
PARTITION BY LIST (country_code);

With PostgreSQL however, doing so will result in an error:

Error: ERROR: insufficient columns in PRIMARY KEY constraint definition Detail: PRIMARY KEY constraint on table "orders" lacks column "country_code" which is part of the partition key

If the primary key is changed to use the partitioning column, you'll get a valid primary key:

CREATE TABLE orders(
  id            integer,
  country_code  VARCHAR(5) primary key,
  customer_id   integer,
  order_date    DATE,
  order_total   numeric(8,2)
)
PARTITION BY LIST (country_code);

But this is not what I want, since my table is designed to store multiple values for each country.

To create a partitioned table including a valid primary key:

CREATE TABLE orders(
  id            integer,
  country_code  VARCHAR(5),
  customer_id   integer,
  order_date    DATE,
  order_total   numeric(8,2),
PRIMARY KEY (id,country_code)
)
PARTITION BY LIST (country_code);

As long as the primary key includes the partition key, the syntax is accepted by the postgres server.
This is different from Oracle, where you define the primary key independently of the partition key.

As with Oracle, a primary key constraint will automatically create a unique B-tree index.

Note that

PRIMARY KEY (id,country_code)

can alternatively be replaced by

UNIQUE (id,country_code)

The result is the same, either way.

How to connect to a postgres database using jdbc

Using SquirrelSQL I connected to a postgres database as can be seen below:




The entire connection string is

jdbc:postgresql://mypostgresserver01:5432/vegdb01