diff --git a/.idea/.gitignore b/.idea/.gitignore
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+++ b/.idea/.gitignore
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+# Default ignored files
+/shelf/
+/workspace.xml
diff --git a/.idea/databases-are-friendly-Anujangalapalli.iml b/.idea/databases-are-friendly-Anujangalapalli.iml
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diff --git a/.idea/inspectionProfiles/Project_Default.xml b/.idea/inspectionProfiles/Project_Default.xml
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diff --git a/.idea/misc.xml b/.idea/misc.xml
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diff --git a/.idea/vcs.xml b/.idea/vcs.xml
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diff --git a/ph1-solutions.sql b/ph1-solutions.sql
index e77f788..a62f804 100644
--- a/ph1-solutions.sql
+++ b/ph1-solutions.sql
@@ -4,3 +4,50 @@
USE sakila;
-- Your solutions...
+--Phase 1:
+--1. Which actors have the first name ‘Scarlett’
+SELECT * FROM actor
+WHERE first_name = 'Scarlett';
+
+
+--2. Which actors have the last name ‘Johansson’
+SELECT * FROM actor
+WHERE last_name = 'Johansson';
+
+
+--3. How many distinct actors last names are there?
+SELECT COUNT(DISTINCT last_name) FROM actor;
+
+
+--4.Which last names are not repeated?
+SELECT last_name FROM actor
+GROUP BY last_name
+HAVING count(last_name)=1;
+
+
+--5. Which last names appear more than once?
+SELECT last_name FROM actor
+GROUP BY last_name
+HAVING count(last_name)>1;
+
+
+--6. Which actor has appeared in the most films?
+SELECT count(fa.actor_id) as NoOfMovies, a.first_name, a.last_name
+FROM actor a
+INNER JOIN film_actor fa
+ON a.actor_id = fa.actor_id
+GROUP BY fa.actor_id
+ORDER BY count(fa.actor_id) DESC LIMIT 1;
+
+
+--7.Is ‘Academy Dinosaur’ available for rent from Store 1?
+
+--8.Insert a record to represent Mary Smith renting ‘Academy Dinosaur’ from Mike Hillyer at Store 1 today .
+
+--9.When is ‘Academy Dinosaur’ due?
+
+--10.What is that average running time of all the films in the sakila DB?
+
+--11.What is the average running time of films by category?
+
+--12. Why does this query return the empty set?
diff --git a/ph2-solutions.sql b/ph2-solutions.sql
index e77f788..8bd7cd7 100644
--- a/ph2-solutions.sql
+++ b/ph2-solutions.sql
@@ -4,3 +4,200 @@
USE sakila;
-- Your solutions...
+--Phase2
+--1a. Display the first and last names of all actors from the table actor.
+SELECT first_name, last_name FROM actor;
+
+
+--1b. Display the first and last name of each actor in a single column in upper case letters. Name the column Actor Name.
+SELECT UPPER(CONCAT(first_name, " ",last_name)) as ActorName FROM actor;
+
+--2a. You need to find the ID number, first name, and last name of an actor, of whom you know only the first name, "Joe." What is one query would you use to obtain this information?
+SELECT actor_id, first_name, last_name FROM actor WHERE first_name = 'Joe';
+
+
+--2b. Find all actors whose last name contain the letters GEN:
+SELECT actor_id, first_name, last_name
+FROM actor
+WHERE last_name like '%GEN%';
+
+--2c. Find all actors whose last names contain the letters LI. This time, order the rows by last name and first name, in that order:
+SELECT last_name, first_name
+FROM actor
+WHERE last_name like '%LI%'
+ORDER BY last_name, first_name;
+
+
+--2d. Using IN, display the country_id and country columns of the following countries: Afghanistan, Bangladesh, and China:
+SELECT country_id, country
+FROM country
+WHERE country IN('Afghanistan', 'Bangladesh', 'China');
+
+
+--3a. Add a middle_name column to the table actor. Position it between first_name and last_name. Hint: you will need to specify the data type.
+ALTER TABLE actor
+ADD COLUMN middle_name VARCHAR(45) AFTER first_name;
+
+
+-- 3b. You realize that some of these actors have tremendously long last names. Change the data type of the middle_name column to blobs.
+ ALTER TABLE actor
+ MODIFY middle_name BLOB;
+
+-- 3c. Now delete the middle_name column.
+ ALTER TABLE actor
+ DROP middle_name;
+
+-- 4a. List the last names of actors, as well as how many actors have that last name.
+SELECT last_name, COUNT(last_name) FROM actor GROUP BY last_name;
+
+
+--4b. List last names of actors and the number of actors who have that last name, but only for names that are shared by at least two actors
+SELECT last_name, COUNT(last_name)
+FROM actor
+GROUP BY last_name
+HAVING COUNT(last_name)>=2;
+
+
+--4c. Oh, no! The actor HARPO WILLIAMS was accidentally entered in the actor table as GROUCHO WILLIAMS, the name of Harpo's second cousin's husband's yoga teacher. Write a query to fix the record.
+UPDATE actor
+SET first_name = 'HAPRO'
+WHERE first_name = 'GROUCHO' and last_name = 'WILLIAMS';
+
+
+--4d. Perhaps we were too hasty in changing GROUCHO to HARPO. It turns out that GROUCHO was the correct name after all! In a single query, if the first name of the actor is currently HARPO, change it to GROUCHO. Otherwise, change the first name to MUCHO GROUCHO, as that is exactly what the actor will be with the grievous error. BE CAREFUL NOT TO CHANGE THE FIRST NAME OF EVERY ACTOR TO MUCHO GROUCHO, HOWEVER! (Hint: update the record using a unique identifier.)
+UPDATE actor
+SET first_name = 'GROUCHO'
+WHERE actor_id = 172;
+
+
+--5a. You cannot locate the schema of the address table. Which query would you use to re-create it?
+SHOW CREATE TABLE address;
+
+
+--6a. Use JOIN to display the first and last names, as well as the address, of each staff member. Use the tables staff and address:
+SELECT s.first_name, s.last_name, a.address
+FROM staff s
+INNER JOIN address a
+ON a.address_id = s.address_id;
+
+--6b. Use JOIN to display the total amount rung up by each staff member in August of 2005. Use tables staff and payment.
+SELECT s.first_name, s.last_name, SUM(p.amount)
+FROM staff s
+INNER JOIN payment p
+ON s.staff_id = p.staff_id
+AND p.payment_date LIKE '2005-08-%'
+GROUP BY s.first_name, s.last_name;
+
+
+--6c. List each film and the number of actors who are listed for that film. Use tables film_actor and film. Use inner join.
+SELECT f.title, COUNT(fa.actor_id) as NumberOfActors
+FROM film f
+INNER JOIN film_actor fa
+ON f.film_id = fa.film_id
+GROUP BY f.title;
+
+
+--6d. How many copies of the film Hunchback Impossible exist in the inventory system?
+SELECT f.title, COUNT(i.film_id) as NoOfCopies
+FROM film f
+INNER JOIN inventory i
+ON f.film_id = i.film_id
+WHERE title = 'Hunchback Impossible';
+
+
+--6e. Using the tables payment and customer and the JOIN command, list the total paid by each customer. List the customers alphabetically by last name
+SELECT c.first_name, c.last_name, SUM(p.amount)
+FROM customer c
+INNER JOIN payment p
+ON c.customer_id = p.customer_id
+GROUP BY p.customer_id
+ORDER BY last_name ASC;
+
+--7a. The music of Queen and Kris Kristofferson have seen an unlikely resurgence. As an unintended consequence, films starting with the letters K and Q have also soared in popularity. Use subqueries to display the titles of movies starting with the letters K and Q whose language is English.
+SELECT title FROM film
+WHERE language_id
+IN (SELECT language_id FROM language WHERE name = 'ENGLISH')
+AND title like 'K%' OR title LIKE 'Q%';
+
+
+--7b. Use subqueries to display all actors who appear in the film Alone Trip.
+SELECT title, actors FROM film_list WHERE title="ALONE TRIP";
+
+
+--7c. You want to run an email marketing campaign in Canada, for which you will need the names and email addresses of all Canadian customers. Use joins to retrieve this information.
+ SELECT cl.name, cl.country, c.email
+ FROM customer_list cl
+ INNER JOIN customer c
+ ON cl.ID = c.customer_id
+ WHERE cl.country = 'Canada';
+
+
+--7d. Sales have been lagging among young families, and you wish to target all family movies for a promotion. Identify all movies categorized as famiy films
+ SELECT title, category
+ FROM nicer_but_slower_film_list
+ WHERE category = 'FAMILY';
+
+
+-- 7e. Display the most frequently rented movies in descending order.
+SELECT f.title, COUNT(r.inventory_id)
+FROM inventory i
+INNER JOIN rental r
+ON i.inventory_id = r.inventory_id
+INNER JOIN film_text f
+ON i.film_id = f.film_id
+GROUP BY r.inventory_id
+ORDER BY COUNT(r.inventory_id) DESC;
+
+
+--7f. Write a query to display how much business, in dollars, each store brought in.
+ SELECT store, total_sales
+ FROM sales_by_store;
+
+-- 7g. Write a query to display for each store its store ID, city, and country.
+SELECT s.store_id, city, country
+FROM store s
+INNER JOIN customer c
+ON s.store_id = c.store_id
+INNER JOIN staff st
+ON s.store_id = st.store_id
+INNER JOIN address a
+ON c.address_id = a.address_id
+INNER JOIN city ci
+ON a.city_id = ci.city_id
+INNER JOIN country coun
+ON ci.country_id = coun.country_id;
+
+
+--7h. List the top five genres in gross revenue in descending order. (Hint: you may need to use the following tables: category, film_category, inventory, payment, and rental.)
+SELECT c.name AS 'Genre', SUM(p.amount) AS 'Gross'
+FROM category c
+JOIN film_category fc
+ON (c.category_id=fc.category_id)
+JOIN inventory i
+ON (fc.film_id=i.film_id)
+JOIN rental r
+ON (i.inventory_id=r.inventory_id)
+JOIN payment p
+ON (r.rental_id=p.rental_id)
+GROUP BY c.name ORDER BY Gross LIMIT 5;
+
+
+--8a. In your new role as an executive, you would like to have an easy way of viewing the Top five genres by gross revenue. Use the solution from the problem above to create a view. If you haven't solved 7h, you can substitute another query to create a view.
+CREATE VIEW top_five_genres
+AS SELECT c.name AS 'Genre', SUM(p.amount) AS 'Gross'
+FROM category c JOIN film_category fc
+ON (c.category_id=fc.category_id)
+JOIN inventory i
+ON (fc.film_id=i.film_id)
+JOIN rental r
+ON (i.inventory_id=r.inventory_id)
+JOIN payment p
+ON (r.rental_id=p.rental_id)
+GROUP BY c.name
+ORDER BY Gross LIMIT 5;
+
+--8b. How would you display the view that you created in 8a?
+SELECT * FROM top_five_genres;
+
+--8c. You find that you no longer need the view top_five_genres. Write a query to delete it.
+DROP VIEW top_five_genres;
\ No newline at end of file