Mark-McAdam / sql-training

Just a play to store my sql playgrounds

#sql-training

sample chinook queries below

-- 2020-03-16 11:44:52.1270 SELECT "rowid" as rowid,* FROM "albums" ORDER BY 1 LIMIT 300 OFFSET 0;

-- 2020-03-16 11:44:52.1300 SELECT COUNT(*) as count FROM "albums";

-- 2020-03-16 12:08:45.7820 SELECT *

-- 2020-03-16 12:12:12.4560 SELECT COUNT(*) as count FROM "albums";

-- 2020-03-16 12:12:23.8590 SELECT COUNT(*) as count FROM "albums";

-- 2020-03-16 12:12:51.6040 SELECT COUNT(*) as count FROM "albums";

these were in my sql table reader when I opened it so I figured I would include it here.

place holder ryan meckings method

David dodds technique ...

SELECT count (DISTINCT charactercreator_character.character_id) AS total_characters ,COUNT (DISTINCT charactercreator_mage.character_ptr_id) - COUNT (DISTINCT charactercreator_necromancer.mage_ptr_id) AS just_mages ,COUNT (DISTINCT charactercreator_necromancer.mage_ptr_id) AS necromancers ,COUNT (DISTINCT charactercreator_thief.character_ptr_id) AS theives ,COUNT (DISTINCT charactercreator_cleric.character_ptr_id) AS clerics ,COUNT (DISTINCT charactercreator_fighter.character_ptr_id) AS fighters FROM charactercreator_character ,charactercreator_mage ,charactercreator_necromancer ,charactercreator_thief ,charactercreator_cleric ,charactercreator_fighter

Chris Jacuk

This is chris jacuks method for trying this out

Question 2: How many of each specific subclass?

charachter_types_items = ['charactercreator_mage', 'charactercreator_thief', 'charactercreator_cleric', 'charactercreator_fighter'] subclasses = ['mage','thief','cleric','fighter'] print("How many of each specific subclass?") i = 0 for items in charachter_types_items: subclass_query = f""" SELECT count(distinct character_ptr_id) as subclass_count FROM {items} """ subclass_result = cursor.execute(subclass_query).fetchone() print(f"There are {subclass_result[0]} charcters of subclass {subclasses[i]}") i += 1