8 done today

Author: mrinal1704

Medium/Active Businesses.sql

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+-- Question 65
+-- Table: Events
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | business_id   | int     |
+-- | event_type    | varchar |
+-- | occurences    | int     | 
+-- +---------------+---------+
+-- (business_id, event_type) is the primary key of this table.
+-- Each row in the table logs the info that an event of some type occured at some business for a number of times.
+ 
+
+-- Write an SQL query to find all active businesses.
+
+-- An active business is a business that has more than one event type with occurences greater than the average occurences of that event type among all businesses.
+
+-- The query result format is in the following example:
+
+-- Events table:
+-- +-------------+------------+------------+
+-- | business_id | event_type | occurences |
+-- +-------------+------------+------------+
+-- | 1           | reviews    | 7          |
+-- | 3           | reviews    | 3          |
+-- | 1           | ads        | 11         |
+-- | 2           | ads        | 7          |
+-- | 3           | ads        | 6          |
+-- | 1           | page views | 3          |
+-- | 2           | page views | 12         |
+-- +-------------+------------+------------+
+
+-- Result table:
+-- +-------------+
+-- | business_id |
+-- +-------------+
+-- | 1           |
+-- +-------------+ 
+-- Average for 'reviews', 'ads' and 'page views' are (7+3)/2=5, (11+7+6)/3=8, (3+12)/2=7.5 respectively.
+-- Business with id 1 has 7 'reviews' events (more than 5) and 11 'ads' events (more than 8) so it is an active business.
+
+-- Solution
+select c.business_id
+from(
+select *
+from events e
+join
+(select event_type as event, round(avg(occurences),2) as average from events group by event_type) b
+on e.event_type = b.event) c
+where c.occurences>c.average
+group by c.business_id
+having count(*) > 1

Medium/Apples & Oranges.sql

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+-- Question 66
+-- Table: Sales
+
+-- +---------------+---------+
+-- | Column Name   | Type    |
+-- +---------------+---------+
+-- | sale_date     | date    |
+-- | fruit         | enum    | 
+-- | sold_num      | int     | 
+-- +---------------+---------+
+-- (sale_date,fruit) is the primary key for this table.
+-- This table contains the sales of "apples" and "oranges" sold each day.
+ 
+
+-- Write an SQL query to report the difference between number of apples and oranges sold each day.
+
+-- Return the result table ordered by sale_date in format ('YYYY-MM-DD').
+
+-- The query result format is in the following example:
+
+ 
+
+-- Sales table:
+-- +------------+------------+-------------+
+-- | sale_date  | fruit      | sold_num    |
+-- +------------+------------+-------------+
+-- | 2020-05-01 | apples     | 10          |
+-- | 2020-05-01 | oranges    | 8           |
+-- | 2020-05-02 | apples     | 15          |
+-- | 2020-05-02 | oranges    | 15          |
+-- | 2020-05-03 | apples     | 20          |
+-- | 2020-05-03 | oranges    | 0           |
+-- | 2020-05-04 | apples     | 15          |
+-- | 2020-05-04 | oranges    | 16          |
+-- +------------+------------+-------------+
+
+-- Result table:
+-- +------------+--------------+
+-- | sale_date  | diff         |
+-- +------------+--------------+
+-- | 2020-05-01 | 2            |
+-- | 2020-05-02 | 0            |
+-- | 2020-05-03 | 20           |
+-- | 2020-05-04 | -1           |
+-- +------------+--------------+
+
+-- Day 2020-05-01, 10 apples and 8 oranges were sold (Difference  10 - 8 = 2).
+-- Day 2020-05-02, 15 apples and 15 oranges were sold (Difference 15 - 15 = 0).
+-- Day 2020-05-03, 20 apples and 0 oranges were sold (Difference 20 - 0 = 20).
+-- Day 2020-05-04, 15 apples and 16 oranges were sold (Difference 15 - 16 = -1).
+
+-- Solution
+Select sale_date, sold_num-sold as diff
+from 
+((select *
+from sales
+where fruit = 'apples') a
+join 
+(select sale_date as sale, fruit, sold_num as sold
+from sales
+where fruit = 'oranges') b
+on a.sale_date = b.sale) 

Medium/Unpopular Books.sql

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+-- Question 64
+-- Table: Books
+
+-- +----------------+---------+
+-- | Column Name    | Type    |
+-- +----------------+---------+
+-- | book_id        | int     |
+-- | name           | varchar |
+-- | available_from | date    |
+-- +----------------+---------+
+-- book_id is the primary key of this table.
+-- Table: Orders
+
+-- +----------------+---------+
+-- | Column Name    | Type    |
+-- +----------------+---------+
+-- | order_id       | int     |
+-- | book_id        | int     |
+-- | quantity       | int     |
+-- | dispatch_date  | date    |
+-- +----------------+---------+
+-- order_id is the primary key of this table.
+-- book_id is a foreign key to the Books table.
+ 
+
+-- Write an SQL query that reports the books that have sold less than 10 copies in the last year, excluding books that have been available for less than 1 month from today. Assume today is 2019-06-23.
+
+-- The query result format is in the following example:
+
+-- Books table:
+-- +---------+--------------------+----------------+
+-- | book_id | name               | available_from |
+-- +---------+--------------------+----------------+
+-- | 1       | "Kalila And Demna" | 2010-01-01     |
+-- | 2       | "28 Letters"       | 2012-05-12     |
+-- | 3       | "The Hobbit"       | 2019-06-10     |
+-- | 4       | "13 Reasons Why"   | 2019-06-01     |
+-- | 5       | "The Hunger Games" | 2008-09-21     |
+-- +---------+--------------------+----------------+
+
+-- Orders table:
+-- +----------+---------+----------+---------------+
+-- | order_id | book_id | quantity | dispatch_date |
+-- +----------+---------+----------+---------------+
+-- | 1        | 1       | 2        | 2018-07-26    |
+-- | 2        | 1       | 1        | 2018-11-05    |
+-- | 3        | 3       | 8        | 2019-06-11    |
+-- | 4        | 4       | 6        | 2019-06-05    |
+-- | 5        | 4       | 5        | 2019-06-20    |
+-- | 6        | 5       | 9        | 2009-02-02    |
+-- | 7        | 5       | 8        | 2010-04-13    |
+-- +----------+---------+----------+---------------+
+
+-- Result table:
+-- +-----------+--------------------+
+-- | book_id   | name               |
+-- +-----------+--------------------+
+-- | 1         | "Kalila And Demna" |
+-- | 2         | "28 Letters"       |
+-- | 5         | "The Hunger Games" |
+-- +-----------+--------------------+
+
+-- Solution
+select b.book_id, name
+from
+(select *
+from books
+where available_from < '2019-05-23') b
+left join
+(select *
+from orders 
+where dispatch_date > '2018-06-23') a
+on a.book_id = b.book_id
+group by b.book_id, name
+having coalesce(sum(quantity),0)<10