SQL submission

Solution to consoleGames.txt

(1) percentage of global sales in North America:
SELECT (SUM(NA_Sales)
/(sum(na_sales)+sum(EU_Sales)+sum(JP_Sales)+sum(Other_Sales)))*100
 FROM consoleGames

(2)  SELECT Name FROM ConsoleGames
ORDER BY Platform ASC, Year DESC

(3) first four letter of console game publisher:
SELECT substr(Publisher,1,4) FROM ConsoleGames

(5) platforms ordered by longevity in ascending 
SELECT Platform FROM ConsoleDates 
ORDER BY FirstRetailAvailability DESC

Solution to consumersComplaints.txt

complaints sent and received on the same date =>28737
SELECT field1, COUNT(field1) FROM ConsumerComplaints WHERE field1=field14;


complaints received in New York==> 4413
SELECT field9, COUNT(field9) FROM ConsumerComplaints WHERE field9="NY";

complaints received in New York and califonia ==>13668
SELECT field9, COUNT(field9) FROM ConsumerComplaints WHERE field9="NY" OR field9 = "CA";

product with the word "Credit" ==>18564
SELECT field2, COUNT(field2) FROM ConsumerComplaints WHERE field2 LIKE "%Credit%";

issues with the word "Late" ==>312
SELECT field4, COUNT(field4) FROM ConsumerComplaints WHERE field4 LIKE "%Late%";

Solution to movies.txt

(1) sql:
	SELECT title FROM movies WHERE year ="2018";

(2) sql:
	SELECT birth FROM people WHERE name ="Emma Stone";

(3)	sql:
	SELECT title FROM movies WHERE year >= "2018" ORDER BY title;

(4) sql:
	SELECT title,COUNT(title) FROM movies JOIN ratings ON
    movies.id=ratings.movie_id WHERE rating = 10.0;

(5) sql:
	SELECT title, year FROM movies 
    WHERE title LIKE "%Harry Potter%" ORDER BY year ASC;   

(6) sql:
	SELECT rating, AVG(rating) FROM ratings JOIN
    movies On ratings.movie_id =movies.id WHERE year = 2012;    

(7) sql:
	SELECT title,rating FROM movies JOIN ratings ON
    movies.id=ratings.movie_id ORDER BY 2 DESC,1;

     OR
    SELECT title,rating FROM movies JOIN ratings ON
    movies.id=ratings.movie_id ORDER BY rating DESC,title;  

(8) sql:
	SELECT name FROM people JOIN stars ON people.id=stars.person_id
    JOIN movies ON stars.movie_id=movies.id WHERE title = "Toy Story"; 

(9) sql:
	SELECT DISTINCT name FROM people JOIN stars ON people.id=stars.person_id
    JOIN movies ON stars.movie_id=movies.id WHERE year = "2004" ORDER BY birth; 

(10) sql:
	 SELECT DISTINCT name FROM people JOIN directors ON people.id=directors.person_id
     JOIN movies ON directors.movie_id=movies.id 
     JOIN ratings ON movies.id=ratings.movie_id WHERE rating >= 9.0;       

(11) sql:
     SELECT DISTINCT title FROM people JOIN stars ON people.id=stars.person_id
     JOIN movies ON stars.movie_id=movies.id 
     JOIN ratings ON movies.id=ratings.movie_id WHERE name ="Chadwick Boseman" ORDER BY rating DESC LIMIT 5;

(12) sql:
     SELECT title FROM
     people JOIN stars ON people.id=stars.person_id
     JOIN movies ON stars.movie_id=movies.id 
     WHERE name = "Helena Bonham Carter" AND 
     movie_id IN (SELECT movie_id FROM people JOIN stars
     ON stars.person_id= people.id
     WHERE name = "Helena Bonham Carter");

(13) sql:
	 SELECT name FROM people 
     JOIN stars ON people.id=stars.person_id  WHERE movie_id
     IN
     (SELECT movie_id FROM people 
     JOIN stars ON people.id=stars.person_id 
     JOIN movies ON stars.movie_id=movies.id
     WHERE name="Kevin Bacon" AND birth =1958) AND name != "Kevin Bacon";                  	

Solution to pets.txt

(1) information on pet names and owners names side by side:

    SELECT "p9-pets".Name,"p9-owners".name FROM "p9-pets" 
    JOIN "p9-owners" on  "p9-pets".ownerid="p9-owners".ownerid

 (2)  Match up all procedures performed to their descriptions:
 
 		SELECT ProcedureType,Description FROM "p9-proceduresdetails"

 (3) Most popular procedures and how much they cost:

 	SELECT ProcedureType,Price FROM "p9-proceduresdetails"