completes lab

publications.sqbpro

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+<?xml version="1.0" encoding="UTF-8"?><sqlb_project><db path="/Users/test/Desktop/ironhack_labs/lab-sql-basics/files_for_lab/lab1_bank.sqlite" readonly="0" foreign_keys="1" case_sensitive_like="0" temp_store="0" wal_autocheckpoint="1000" synchronous="2"/><attached/><window><main_tabs open="structure browser pragmas query" current="3"/></window><tab_structure><column_width id="0" width="300"/><column_width id="1" width="0"/><column_width id="2" width="100"/><column_width id="3" width="2280"/><column_width id="4" width="0"/><expanded_item id="0" parent="1"/><expanded_item id="1" parent="1"/><expanded_item id="2" parent="1"/><expanded_item id="3" parent="1"/></tab_structure><tab_browse><table title="account" custom_title="0" dock_id="4" table="4,7:mainaccount"/><dock_state state="000000ff00000000fd0000000100000002000004490000033ffc0100000004fb000000160064006f0063006b00420072006f00770073006500310100000000ffffffff0000000000000000fb000000160064006f0063006b00420072006f00770073006500320100000000000004490000000000000000fb000000160064006f0063006b00420072006f00770073006500330100000000ffffffff0000000000000000fb000000160064006f0063006b00420072006f00770073006500340100000000ffffffff0000010100ffffff0000027e0000000000000004000000040000000800000008fc00000000"/><default_encoding codec=""/><browse_table_settings/></tab_browse><tab_sql><sql name="SQL 1">-- Get the `id` values of the first 5 clients from `district_id` with a value equals to 1.
+
+SELECT client_id
+FROM client
+WHERE district_id = 1
+LIMIT 5;
+
+-- In the `client` table, get an `id` value of the last client where the `district_id` equals to 72.
+
+SELECT client_id
+FROM client
+WHERE district_id = 72
+ORDER BY district_id DESC
+LIMIT 1;
+
+-- Get the 3 lowest amounts in the `loan` table.
+
+SELECT amount
+FROM loan
+ORDER BY amount ASC
+LIMIT 3;
+
+--What are the possible values for `status`, ordered alphabetically in ascending order in the `loan` table?
+
+SELECT status
+FROM loan
+ORDER BY status ASC
+
+-- What is the `loan_id` of the highest payment received in the `loan` table?
+
+SELECT loan_id
+FROM loan
+WHERE payment = (SELECT MAX(payment) FROM loan);
+
+
+-- What is the loan `amount` of the lowest 5 `account_id`s in the `loan` table? Show the `account_id` and the corresponding `amount`
+
+SELECT amount, account_id
+FROM loan
+ORDER BY account_id ASC
+LIMIT 5;
+
+-- What are the `account_id`s with the lowest loan `amount` that have a loan `duration` of 60 in the `loan` table?
+
+SELECT  account_id, duration
+FROM loan
+WHERE duration == 60
+ORDER BY amount ASC
+LIMIT 5;
+
+-- What are the unique values of `k_symbol` in the `order` table?
+-- Note: There shouldn't be a table name `order`, since `order` is reserved from the `ORDER BY` clause. You have to use backticks to escape the `order` table name.
+
+SELECT DISTINCT k_symbol FROM `order`;
+
+-- In the `order` table, what are the `order_id`s of the client with the `account_id` 34?
+
+SELECT  order_id
+FROM `order`
+WHERE account_id == 34
+
+-- In the `order` table, which `account_id`s were responsible for orders between `order_id` 29540 and `order_id` 29560 (inclusive)?
+
+SELECT account_id
+FROM `order`
+WHERE order_id BETWEEN 29540 AND 29560;
+
+-- In the `order` table, what are the individual amounts that were sent to (`account_to`) id 30067122?
+
+SELECT amount
+FROM `order`
+WHERE account_to = 30067122;
+
+-- In the `trans` table, show the `trans_id`, `date`, `type` and `amount` of the 10 first transactions from `account_id` 793 in chronological order, from newest to oldest.
+
+SELECT trans_id, `date`, type, amount
+FROM trans
+WHERE account_id = 793
+ORDER BY date DESC
+LIMIT 10;
+
+-- In the `client` table, of all districts with a `district_id` lower than 10, how many clients are from each `district_id`? Show the results sorted by the `district_id` in ascending order.
+
+SELECT district_id, COUNT(*) AS client_count
+FROM client
+WHERE district_id &lt; 10
+GROUP BY district_id
+ORDER BY district_id ASC;
+
+-- In the `card` table, how many cards exist for each `type`? Rank the result starting with the most frequent `type`.
+
+SELECT 
+  type, 
+  COUNT(*) AS type_count,
+  RANK() OVER (ORDER BY COUNT(*) DESC) AS Rank_no
+FROM card
+GROUP BY type
+ORDER BY Rank_no;
+
+-- Using the `loan` table, print the top 10 `account_id`s based on the sum of all of their loan amounts.
+
+SELECT account_id, SUM(loan_amount) AS total_loan_amount
+FROM loan
+GROUP BY account_id
+ORDER BY total_loan_amount DESC
+LIMIT 10;
+
+-- In the `loan` table, retrieve the number of loans issued for each day, before (excl) 930907, ordered by date in descending order.
+--  what is the number of loans for each day before 930907, ordered by date in descending order
+
+SELECT 
+  loan_id, 
+  COUNT(*) AS loan_count,
+FROM loan
+ORDER BY date DESC
+
+SELECT date, COUNT(*) AS loan_count
+FROM loan
+WHERE date &lt; 930907
+GROUP BY date
+ORDER BY date DESC;
+
+-- In the `loan` table, for each day in December 1997, count the number of loans issued for each unique loan duration, ordered by date and duration, both in ascending order. You can ignore days without any loans in your output.
+
+SELECT date, duration, COUNT(*) AS loan_count
+FROM loan
+WHERE date BETWEEN 971201 AND 971231
+GROUP BY date, duration
+ORDER BY date ASC, duration ASC;
+
+-- In the `trans` table, for `account_id` 396, sum the amount of transactions for each type (`VYDAJ` = Outgoing, `PRIJEM` = Incoming). 
+-- Your output should have the `account_id`, the `type` and the sum of amount, named as `total_amount`. Sort alphabetically by type.
+
+SELECT account_id, type, SUM(amount) AS total_amount
+FROM trans
+WHERE account_id = 396
+GROUP BY account_id, type
+ORDER BY type ASC;</sql><current_tab id="0"/></tab_sql></sqlb_project>

solutions.sql

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+-- 1
+SELECT
+	authors.au_id AS "Author ID", authors.au_lname AS "Last name", authors.au_fname AS "First name",
+	titles.title AS "Title",
+	publishers.pub_name AS "Publisher"
+
+FROM
+	authors, titles, publishers, titleauthor
+
+WHERE
+	authors.au_id == titleauthor.au_id AND
+	titleauthor.title_id == titles.title_id AND
+	titles.pub_id == publishers.pub_id
+
+-- 2	
+SELECT 
+    a.au_id AS AUTHOR_ID,
+    a.au_lname AS LAST_NAME,
+    a.au_fname AS FIRST_NAME,
+    p.pub_name AS PUBLISHER,
+    COUNT(ta.title_id) AS TITLE_COUNT
+FROM titleauthor ta
+JOIN authors a ON ta.au_id = a.au_id
+JOIN titles t ON ta.title_id = t.title_id
+JOIN publishers p ON t.pub_id = p.pub_id
+GROUP BY a.au_id, a.au_lname, a.au_fname, p.pub_name
+ORDER BY a.au_lname, a.au_fname, p.pub_name;
+
+-- Who are the top 3 authors who have sold the highest number of titles? Write a query to find out.
+
+SELECT 
+    a.au_id AS AUTHOR_ID,
+    a.au_lname AS LAST_NAME,
+    a.au_fname AS FIRST_NAME,
+    COUNT(ta.title_id) AS TITLES_SOLD
+FROM titleauthor ta
+JOIN authors a ON ta.au_id = a.au_id
+GROUP BY a.au_id, a.au_lname, a.au_fname
+ORDER BY TITLES_SOLD DESC
+LIMIT 3;
+
+-- Now modify your solution in Challenge 3 so that the output will display all 23 authors instead of the top 3. Note that the authors who have sold 0 titles should also appear in your output (ideally display `0` instead of `NULL` as the `TOTAL`). Also order your results based on `TOTAL` from high to low.
+
+SELECT 
+    a.au_id AS AUTHOR_ID,
+    a.au_lname AS LAST_NAME,
+    a.au_fname AS FIRST_NAME,
+    COUNT(ta.title_id) AS TOTAL
+FROM authors a
+LEFT JOIN titleauthor ta ON a.au_id = ta.au_id
+GROUP BY a.au_id, a.au_lname, a.au_fname
+ORDER BY TOTAL DESC;