Hw1, hw3

Week1/solutions/myscript.sh

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+#!/bin/bash
+echo "Welcome to the DaveCoffeeshop!"
+
+if [ -f "$HOME"/davecoffeeshop.db ] 
+then
+	echo "$HOME/davecoffeeshop.db, using existing database"
+else 
+	echo "$HOME/davecoffeeshop.db missing => creating new file"
+	touch "$HOME/davecoffeeshop.db"
+fi
+echo "To log order, enter LOG"
+echo "To end session, enter EOS"
+echo "To end business day, enter EOD"
+
+flag=$(( 0 ))
+LOG="LOG"
+EOD="EOD"
+EOS="EOS"
+
+while [ $flag -lt 1 ]
+do
+	read -r str
+	if [ "$str" = "$LOG" ]
+	then
+		echo -n "Barista name: "
+		read -r name
+		echo -n "Drinking: "
+		read -r drink
+		if [ "$drink" = "RAF" ]
+		then
+			echo "$name;$drink;200" >> "$HOME"/davecoffeeshop.db
+			echo "Order was saved!"
+		elif [ "$drink" = "CAP" ]
+		then
+			echo "$name;$drink;150" >> "$HOME"/davecoffeeshop.db
+			echo "Order was saved!"
+		elif [ "$drink" = "AME" ]
+		then
+			echo "$name;$drink;100" >> "$HOME"/davecoffeeshop.db
+			echo "Order was saved!"
+		elif [ "$drink" = "ESP" ]
+		then
+			echo "$name;$drink;120" >> "$HOME"/davecoffeeshop.db
+			echo "Order was saved!"
+		elif [ "$drink" = "LAT" ]
+		then
+			echo "$name;$drink;170" >> "$HOME"/davecoffeeshop.db
+			echo "Order was saved!"
+		fi
+	elif [ "$str" = "$EOS" ]
+	then
+		flag+=1
+		echo "End of session"
+	elif [ "$str" = "$EOD" ]
+	then
+		echo "End of day. Result:"
+		awk -F ";" '{
+			array[$1]+=$3;
+		} END {
+			for (i in array)
+				print i";"array[i]}' "$HOME"/davecoffeeshop.db | sort -t ";" -k2 -n -r
+	else 
+		echo "Bad command. Try again"
+	fi
+done

Week3/solutions/hw2_task1.sql

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+-- Задание 1. Вывести все альбомы со средней длинной трека, большей 250 секунд.
+-- Вывести название альбома и количество треков.
+-- (album_name, cnt).
+
+select distinct title as album_name, count(t.album_id) as cnt
+from album
+         join track t on album.album_id = t.album_id
+group by title
+having AVG(t.milliseconds) > 250000;

Week3/solutions/hw2_task10.sql

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+--Задание 10. Для каждого альбома тип медиа определяется как наибольшая суммарный размер треков по жанру.
+-- Для каждого исполнителя вывести тип медиа, соответствующий наибольшему количеству типов медиа по альбомам,
+-- где они участвуют. Вывести имя исполнителя, тип медиа, количество альбомов с данными типом,
+-- общее количество альбомов.
+-- (artist_name, media_type_name, media_cnt, whole_cnt)
+
+with Media_type_Media_cnt as (select media_type.name as med_type, count(distinct t.album_id) as media_cnt
+                              from media_type
+                                       join track t on media_type.media_type_id = t.media_type_id
+                                       join album a on a.album_id = t.album_id
+                              where a.title is not null
+                              group by media_type.name)
+select Artist_name_Cnt_a_t_Media_type.artist_name as artist_name,
+       Artist_name_Cnt_a_t_Media_type.media_type  as media_type,
+       m_t_m_c.media_cnt                          as media_cnt,
+       (select count(*) from album)               as whole_cnt
+from (select distinct on (a2.name) a2.name                             as artist_name,
+                                   count(distinct a.title)             as a_t,
+                                   Album_name_Sum_bytes_Media_type.m_t as media_type
+      from (select distinct on (album.title) album.title as a_t, sum(t.bytes) as sum_b, mt.name as m_t
+            from album
+                     join track t on album.album_id = t.album_id
+                     join media_type mt on t.media_type_id = mt.media_type_id
+                     join artist a on a.artist_id = album.artist_id
+            group by album.title, mt.name
+            order by album.title, sum_b DESC) as Album_name_Sum_bytes_Media_type
+               join album a on a.title = a_t
+               join track t on a.album_id = t.album_id
+               join artist a2 on a2.artist_id = a.artist_id
+      group by a2.name, Album_name_Sum_bytes_Media_type.m_t
+      order by a2.name, a_t DESC) as Artist_name_Cnt_a_t_Media_type
+         join Media_type_Media_cnt m_t_m_c on m_t_m_c.med_type = Artist_name_Cnt_a_t_Media_type.media_type

Week3/solutions/hw2_task2.sql

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+-- Задание 2. Для каждого набора жанр, тип медиа вывести количество треков.
+-- Если для набора жанр, тип медиа треков нет, то такой набор выводить не требуется.
+-- (genre_name, media_type_name, cnt)
+
+select genre.name as genre_name, mt.name as media_type_name, count(t.name) as cnt
+from genre
+         join track t on genre.genre_id = t.genre_id
+         join media_type mt on t.media_type_id = mt.media_type_id
+group by genre.name, mt.name;

Week3/solutions/hw2_task3.sql

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+-- Задание 3. Вывести сотрудников, которые старше своего менеджера, но были наняты позже своего менеджера.
+-- Вывести полное имя сотрудника(Фамилия<Пробел>Имя), день рождения, дата найма, полное имя менеджера, день рождения,
+-- дата найма.
+-- (employee_full_name, employee_birth_date, employee_hire_date, manager_full_name, manager_birth_date, manager_hire_date)
+
+select employee.last_name || ' ' || employee.first_name as employee_full_name,
+       employee.birth_date                              as employee_birth_date,
+       employee.hire_date                               as employee_hire_date,
+       manager.last_name || ' ' || manager.first_name   as manager_full_name,
+       manager.birth_date                               as manager_birth_date,
+       manager.hire_date                                as manager_hire_date
+
+from employee
+         join employee as manager on manager.employee_id = employee.reports_to
+where (manager.birth_date < employee.birth_date)
+  AND (manager.hire_date < employee.hire_date);
+

Week3/solutions/hw2_task4.sql

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+--Задание 4. Для всех треков жанра Rock и Metal, для каждого артиста подсчитать количество треков.
+-- Вывести имя артиста, количество треков.
+-- (artist_name, qty)
+
+
+select art.name as artist_name, count(track.name) as qty
+from track
+         join genre g on g.genre_id = track.genre_id
+         join album alb on alb.album_id = track.album_id
+         join artist art on art.artist_id = alb.artist_id
+where (g.name = 'Rock')
+   Or (g.name = 'Metal')
+group by art.name;

Week3/solutions/hw2_task5.sql

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+--Задание 5. Для каждого сотрудника подсчитать количество клиентов, количестов выполненных заказов,
+--           количество уникальных адресов доставки и общее число приобретенных треков и вывести полное имя сотрудника.
+-- (employee_full_name, invoices_count, customers_count, addresses_count, tracks_count)
+
+select (employee.first_name || ' ' || employee.last_name) as employee_full_name,
+       count(distinct i.invoice_id)                       as invoice_count,
+       count(distinct cust.last_name)                     as customers_count,
+       count(distinct i.billing_address)                  as addresses_count,
+       sum(il.quantity)                                   as tracks_count
+from employee
+         join customer cust on employee.employee_id = cust.support_rep_id
+         join invoice i on cust.customer_id = i.customer_id
+         join invoice_line il on i.invoice_id = il.invoice_id
+group by employee_full_name;

Week3/solutions/hw2_task6.sql

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+--Задание 6. Выполнить сегментацию клиентов по количеству букв а в его фамилии - Группа 'A' - 0 букв, 'B' - 1 буква,
+-- 'C' - 2 буквы, 'D' - 3 и более букв.
+-- Для каждой группы вывести количество уникальных типов медиа,
+-- купленных клиентами по трекам из различных жанров метала.
+-- Вывести имя группы и количество уникальных типов медиа (segment_group, cnt)
+
+select distinct CASE
+                    when (SELECT length(customer.last_name) - length(REPLACE(lower(customer.last_name), 'a', '')) = 0)
+                        then 'A'
+                    when (SELECT length(customer.last_name) - length(REPLACE(lower(customer.last_name), 'a', '')) = 1)
+                        then 'B'
+                    when (SELECT length(customer.last_name) - length(REPLACE(lower(customer.last_name), 'a', '')) = 2)
+                        then 'C'
+                    when (SELECT length(customer.last_name) - length(REPLACE(lower(customer.last_name), 'a', '')) >= 3)
+                        then 'D'
+                    end                 as segment_group,
+                count(distinct mt.name) as cnt
+from customer
+         join invoice i on customer.customer_id = i.customer_id
+         join invoice_line il on i.invoice_id = il.invoice_id
+         join track t on il.track_id = t.track_id
+         join genre g on t.genre_id = g.genre_id
+         join media_type mt on mt.media_type_id = t.media_type_id
+where g.name like '%Metal'
+group by customer.last_name;

Week3/solutions/hw2_task7.sql

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+--Задание 7. Вывести компании, которые приобрели больше всех композиций по каждому из жанров.
+-- При этом указать самую популярную композицию по жанру у данной компании.
+-- (популярность определяется по количеству покупок)
+-- (company_name, track_name)
+
+
+with table_1 as (select distinct on (genre.name) genre.name as name, c.company, sum(il.quantity) as sum
+                 from genre
+                          join track t on genre.genre_id = t.genre_id
+                          join invoice_line il on t.track_id = il.track_id
+                          join invoice i on i.invoice_id = il.invoice_id
+                          join customer c on c.customer_id = i.customer_id
+                 where c.company is not null
+                 group by genre.name, c.company
+                 order by genre.name, sum desc, c.company),
+     table_2 as (
+         select c.company, g.name as name, track.name as track, sum(il.quantity) as sum
+         from track
+                  join genre g on g.genre_id = track.genre_id
+                  join invoice_line il on track.track_id = il.track_id
+                  join invoice i on i.invoice_id = il.invoice_id
+                  join customer c on c.customer_id = i.customer_id
+         where c.company is not null
+         group by c.company, g.name, track.name
+         order by c.company, sum desc, c.company)
+select table_3.company, table_3.track from (select distinct on (table_1.name) table_1.company, table_1.name, t_2.track
+      from table_1
+               join table_2 t_2 on t_2.name = table_1.name
+      where table_1.company = t_2.company
+        and t_2.name = table_1.name
+      order by  table_1.name,table_1.company, t_2.track) as table_3

Week3/solutions/hw2_task8.sql

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+--Задание 8. Вывести треки по каждому жанру, которые входять в самые популярные 25% с точки зрения количества покупок.
+-- Если у нескольких треков одиннаковоко количество приобретенных композиций,
+-- популярность определяется в лексикографическом порядке
+-- (Например, трек, начинающийся с 'A' будет более популярен, чем трек, начинающийся с 'B').
+-- Для каждого трека указать популярность в процентах от общего количества покупок по жанру.
+-- (track_name, popularity)
+with table_1 as (select genre.name      as genre_name,
+                        sum(i.quantity) as sum_buy_genre
+                 from genre
+                          join track t on genre.genre_id = t.genre_id
+                          join invoice_line i on t.track_id = i.track_id
+                 group by genre.name),
+     table_2 as (
+         select g.name                                                          as genre_name,
+                t.name                                                          as track_name,
+                sum(il.quantity)                                                as sum_buy,
+                100.0 * sum(il.quantity) / table_1.sum_buy_genre                as popularity,
+                row_number() over (partition by g.name order by g.name, t.name) as track_pos
+         from table_1
+                  join genre g on g.name = table_1.genre_name
+                  join track t on g.genre_id = t.genre_id
+                  join invoice_line il on t.track_id = il.track_id
+         group by g.name, t.name, table_1.sum_buy_genre
+         order by g.name, t.name),
+ table_3 as (select table_2.track_name    as track_name,
+       table_2.genre_name as genre_name,
+       sum(t_2_2.popularity) as buf
+from table_2
+         join table_2 t_2_2 on t_2_2.genre_name = table_2.genre_name and table_2.track_pos >= t_2_2.track_pos
+group by table_2.track_name, table_2.genre_name)
+select table_3.track_name, t_2.popularity from table_3
+join table_2 t_2 on t_2.track_name = table_3.track_name
+where buf <=25

Week3/solutions/hw2_task9.sql

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+--Задание 9. Для каждого альбома жанр определяется как наибольшая суммарная продолжительность треков по жанру.
+-- Вывести альбом, жанр, и общее количество купленных песен по нему.
+-- (album_name, genre_name, cnt)
+
+
+with first_table as (select distinct on (album.title) album.title         as a_n,
+                                                      sum(t.milliseconds) as s_t,
+                                                      g.name              as g_n
+                     from album
+                              join track t on album.album_id = t.album_id
+                              join genre g on t.genre_id = g.genre_id
+                     group by album.title, g.name
+                     order by album.title, s_t desc),
+     second_table as (select album.title, count(il.quantity) as num
+                      from album
+                               join track t on album.album_id = t.album_id
+                               join invoice_line il on t.track_id = il.track_id
+                      group by album.title)
+select first_table.a_n as album_name, first_table.g_n as genre_name, s_t.num
+from first_table
+         join second_table s_t on s_t.title = first_table.a_n;
+