Listen and notify hstore Recursive Common Table Expressions (CTE) Full text search JSON and other data types GIST and KNN Javascript - plv8js Python

More reasons…

First enable the hstore Postgres extension. In this example we define both a JSON and an HStore column:

class Schema < ActiveRecord::Migration
  def change
    enable_extension "hstore"
    create_table :links do |t|
      t.hstore :data
      t.column :settings, :json
    end
  end
end

Next, we specify accessors for the data that we will be stored in the JSON and HStore columns:

class Link < ActiveRecord::Base
  # Use hstore for text-only data:
  store :data, :name, :url, :description
  # Use JSON to support string, number, object, array, true, false, null
  store :settings, :update_interval, :created_at, :updated_at

We can now use the defined ActiveRecord attributes to store and access JSON and HStore data:

Link.create! name: 'Google', url: 'http://google.com', description: 'Ad company', update_interval: 1.day, created_at: Time.now.utc

Querying the data is where you'll see the biggest differences. Two examples:

# hstore
Link.where("data -> 'name' = ?", 'Google')
# json
Link.where("CAST(settings->>'update_interval' as integer) = ?", 1.day.to_s)

Notes for Postgres 9.3: * HStore can store only text. Nested data is not supported. * JSON supports the following types: string, number, object, array, true, false, null. For example, date and time types are not supported. Nested data is supported.

The recursive query:

WITH RECURSIVE tree AS (
  SELECT id, name, parent_id FROM nodes WHERE id = 25
  UNION ALL
  SELECT nodes.id, nodes.name, nodes.parent_id FROM nodes, tree WHERE nodes.parent_id = tree.id)

This is where we define a virtual table called tree:

WITH RECURSIVE tree

This SQL finds the root node and initializes the recursive function:

SELECT id, name, parent_id FROM nodes WHERE id = 25

The next part recursively fetches the descendants of the specified node:

SELECT nodes.id, nodes.name, nodes.parent_id FROM nodes, tree WHERE nodes.parent_id = tree.id)

This SQL query shows the descendants, and the parent which has an ID of 25:

SELECT * FROM tree;

In Ruby you could use this code to generate your SQL:

def tree_sql(opts={})
    table = opts.fetch(:table) # e.g. 'categories'
    cols = opts.fetch(:cols) # e.g. %w(id name)
    <<-SQL
      WITH RECURSIVE #{table}_tree AS (
        -- initialize
        SELECT
          #{cols.join(', ')}, 0 as n_depth
        FROM
          #{table}
        WHERE
          -- Use bind variables and ? here if you want
          id = #{parent_id}
        UNION ALL
          -- iterate
          SELECT
            #{cols.map { |col| "#{table}.#{col}" }.join(', ')}, n_depth +1
          FROM
            #{table}, #{table}_tree
          WHERE
            #{table}.parent_id = #{table}_tree.id)
    SQL
  end

In Rails you probably want the method to return an ActiveRecord::Relation object so you can chain e.g. pagination, order, and limit method calls. One way of doing that is shown here:

class User
  def descendants
      ids_query = User.select('users.id') # return only ids
         .joins(:memberships)
        .group('users.id') # avoid duplicates
        .where("memberships.organization_id in (#{tree_sql})", organization).to_sql
      User.where("id in (#{ids_query})")
  end
end

Details: * Use UNION instead of UNION ALL if you have duplicates, or if you're not worried about performance when Postgres has to scan for duplicates. * WITH Queries in Postgres * Recursive queries can loop indefinitely if e.g. self.parent = self. If this happens, use the ANY() function to check if path has been visited once already. Alternatively, use validations to make sure circular references don't happen, and hope for the best. * Use Arel's recursive feature to generate recursive CTE queries with Arel.

Create an organization hierarchy and display the organization from the top:

CREATE EXTENSION "tablefunc";
CREATE TABLE organizations(id text, parent_id text, pos int);

INSERT INTO organizations VALUES('row1',NULL, 0);
INSERT INTO organizations VALUES('row2','row1', 0);
INSERT INTO organizations VALUES('row3','row1', 0);
INSERT INTO organizations VALUES('row4','row2', 1);
INSERT INTO organizations VALUES('row5','row2', 0);
INSERT INTO organizations VALUES('row6','row4', 0);
INSERT INTO organizations VALUES('row7','row3', 0);
INSERT INTO organizations VALUES('row8','row6', 0);
INSERT INTO organizations VALUES('row9','row5', 0);

-- Fetch self and descendants for row1
SELECT * FROM connectby('organizations', 'id', 'parent_id', 'row1', 0, '~') AS t(id text, parent_id text, level int, branch text);

Output:

id  | parent_id | level |          branch
------+-----------+-------+--------------------------
 row1 |           |     0 | row1
 row2 | row1      |     1 | row1~row2
 row4 | row2      |     2 | row1~row2~row4
 row6 | row4      |     3 | row1~row2~row4~row6
 row8 | row6      |     4 | row1~row2~row4~row6~row8
 row5 | row2      |     2 | row1~row2~row5
 row9 | row5      |     3 | row1~row2~row5~row9
 row3 | row1      |     1 | row1~row3
 row7 | row3      |     2 | row1~row3~row7

Usually it's better to use recursive CTE queries:

WITH RECURSIVE organization_tree AS (
  SELECT 
    id, parent_id 
  FROM
    organizations
  WHERE
    id = 'row1'
UNION ALL
  SELECT 
    organizations.id, organizations.parent_id 
  FROM 
    organizations, organization_tree
  WHERE
    organizations.parent_id = organization_tree.id
) select * from organization_tree;

This SQL query selects the top 10 transactions grouped per user:

• partition the transactions by user_id
• order the transactions in each partition by the column created_at
• keep order of the rank, per partition (first row is 1, second is 2)
• select only first row, i.e. the latest

WITH latest_customer_transactions AS (
  SELECT
    *, rank() OVER (PARTITION BY user_id ORDER BY created_at desc) AS rank
  FROM
    transactions
)
SELECT
  id, customer_id
FROM
  latest_customer_transactions
WHERE
  rank = 1;

Keep your fingers crossed that it works. For alternatives, see Shtack Overflow.

Also note the differences between RANK, ROW_NUMBER, and DENSE_RANK: http://blog.jooq.org/2014/08/12/the-difference-between-row_number-rank-and-dense_rank/

Calculating the "percentage of the total" for each row with Postgres can be done with a window function:

SELECT
  *,  (value / SUM(value) OVER ()) AS "% of total"
FROM
  transactions
WHERE
  quarter = '2015-03-31' and company_id = 1;

NOTE: We're using "OVER ()", which means the sum over all rows returned by the where-clause.

Our requirement

Output the previous quarter's value for each company:

ID | NAME              | QUARTER     |  VALUE    | PREVIOUS_VALUE
1  | CORP              | 2015-03-31  |  44317.00 | 38755.00
1  | CORP              | 2014-12-31  |  38755.00 | 33617.00
1  | CORP              | 2014-09-30  |  33617.00 | 32406.00
1  | CORP              | 2014-06-30  |  32406.00 | 29909.00
1  | CORP              | 2014-03-31  |  29909.00 | 0

The solution

To calculate the change we use the window and lead functions:

WITH diffs AS (
  select
    *,
    coalesce(lead(value) over (partition by company_id order by quarter desc), 0) as previous_value
  from
    history
)
select * from diffs;

Note that the window is partitioned by company_id, and ordered by quarter.

What we have is: * a table "companies" and a column "name" * Postgres and the pgtrgm extension * a company named "Berkshéiße"

Let's enable the trigram extension:

-- $ psql app_schema -U superuser
CREATE EXTENSION pg_trgm;

Now we can search for trigrams, including shèiß

-- $ psql app_schema -U user
SELECT * FROM companies WHERE name ~* 'shèiß';

The query returns nothing, so let's do this instead: * install the unaccent extension * create an "immutable unaccent" function * apply "unaccent" and "lower" to the query * apply "unaccent" and "lower" to the index

-- $ psql app_schema -U superuser
CREATE EXTENSION unaccent;

-- $ psql app_schema -U user
--DROP INDEX companies_name_search_idx;
CREATE OR REPLACE FUNCTION f_unaccent(text)
  RETURNS text AS
$func$
SELECT unaccent('unaccent', $1)
$func$  LANGUAGE sql IMMUTABLE SET search_path = public, pg_temp;
CREATE INDEX companies_name_search_idx ON companies USING gin(f_unaccent(name) gin_trgm_ops);

Finally, the query returns what we're looking for:

-- Plain SQL
SELECT * FROM companies WHERE lower(f_unaccent(name)) LIKE ('%' || lower(f_unaccent('shèiß')) || '%');
-- With pg_trgrm syntax
SELECT * FROM companies WHERE lower(f_unaccent(name) ~* lower(f_unaccent('shéiße'));
-- Look, even this works
SELECT * FROM companies WHERE lower(f_unaccent(name) ~* lower(f_unaccent('shEiSe'));

If Postgres still doesn't use the index we created, it's probably because it's faster to scan the table than using the index.

Notes

* The WHERE-condition must match the index definition:

-- yes
lower(unaccent_text(name) ~* lower(unaccent_text('shéiße'))
-- no
name ~* lower(unaccent_text('shéiße'))
-- no
name ~* unaccent_text('shéiße')

References

* Erwin Brandstetter: Does PostgreSQL support “accent insensitive” collations? * Erwin Brandstetter: PostgreSQL accent + case insensitive search

This is one way of doing it:

pg_dump -s yamaha_development > structure.sql
pg_dump --column-inserts --data-only yamaha_development  > data.sql

You can then restore it to a different database if needed:

psql -c 'drop database yamaha_test;'
psql -c 'create database yamaha_test;'

psql yamaha_test < structure.sql
psql yamaha_test < data.sql

All without foreign key constraint, etc, errors.

This example creates random data from a predefined list of valid values:

insert into transactions(description, price, timestamp) values(
  ('{Bought cat,Bought dog,Bought house}'::text[])[ceil(random()*3)],
  ('{10,20}'::int[])[ceil(random()*2)],
  NOW() - '1 day'::INTERVAL * ROUND(RANDOM() * 100)
)