IMDB+ansswers.sql

USE imdb;

/* Now that you have imported the data sets, let’s explore some of the tables. 
 To begin with, it is beneficial to know the shape of the tables and whether any column has null values.
 Further in this segment, you will take a look at 'movies' and 'genre' tables.*/



-- Segment 1:




-- Q1. Find the total number of rows in each table of the schema?
-- Type your code below:

SELECT 
    COUNT(movie_id) AS Rows_in_dirctMapping
FROM
    director_mapping;
SELECT 
    COUNT(movie_id) AS Rows_in_Genre
FROM
    genre;
SELECT 
    COUNT(id) AS Rows_in_movie
FROM
    movie;
SELECT 
    COUNT(id) AS Rows_in_names
FROM
    names;
SELECT 
    COUNT(movie_id) AS Rows_in_ratings
FROM
    ratings;
SELECT 
    COUNT(name_id) AS Rows_in_role_mapping
FROM
    role_mapping;



-- Q2. Which columns in the movie table have null values?
-- Type your code below:


SELECT 
    SUM(CASE
        WHEN id IS NULL THEN 1
        ELSE 0
    END) AS Nulls_in_id,
    SUM(CASE
        WHEN title IS NULL THEN 1
        ELSE 0
    END) AS Nulls_in_Title,
    SUM(CASE
        WHEN year IS NULL THEN 1
        ELSE 0
    END) AS Nulls_in_Year,
    SUM(CASE
        WHEN date_published IS NULL THEN 1
        ELSE 0
    END) AS Nulls_in_Date_Published,
    SUM(CASE
        WHEN duration IS NULL THEN 1
        ELSE 0
    END) AS Nulls_in_Duration,
    SUM(CASE
        WHEN country IS NULL THEN 1
        ELSE 0
    END) AS Nulls_in_Country,
    SUM(CASE
        WHEN worlwide_gross_income IS NULL THEN 1
        ELSE 0
    END) AS Nulls_in_WorldWide,
    SUM(CASE
        WHEN languages IS NULL THEN 1
        ELSE 0
    END) AS Nulls_in_Language,
    SUM(CASE
        WHEN production_company IS NULL THEN 1
        ELSE 0
    END) AS Nulls_in_Prod_Company
FROM
    movie;




-- Now as you can see four columns of the movie table has null values. Let's look at the at the movies released each year. 
-- Q3. Find the total number of movies released each year? How does the trend look month wise? (Output expected)

/* Output format for the first part:

+---------------+-------------------+
| Year			|	number_of_movies|
+-------------------+----------------
|	2017		|	2134			|
|	2018		|		.			|
|	2019		|		.			|
+---------------+-------------------+


Output format for the second part of the question:
+---------------+-------------------+
|	month_num	|	number_of_movies|
+---------------+----------------
|	1			|	 134			|
|	2			|	 231			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:

SELECT year,
       Count(id) AS number_of_movies
FROM   movie
GROUP  BY year;

SELECT Month(date_published) AS month_num,
       Count(id)             AS number_of_movies
FROM   movie
GROUP  BY Month(date_published)
ORDER  BY month_num; 


/*The highest number of movies is produced in the month of March.
So, now that you have understood the month-wise trend of movies, let’s take a look at the other details in the movies table. 
We know USA and India produces huge number of movies each year. Lets find the number of movies produced by USA or India for the last year.*/
  
-- Q4. How many movies were produced in the USA or India in the year 2019??
-- Type your code below:


SELECT Count(id) AS No_of_movies
FROM   movie
WHERE  ( country LIKE '%USA%'
          OR country LIKE '%India%' )
       AND year = '2019'; 



/* USA and India produced more than a thousand movies(you know the exact number!) in the year 2019.
Exploring table Genre would be fun!! 
Let’s find out the different genres in the dataset.*/

-- Q5. Find the unique list of the genres present in the data set?
-- Type your code below:

SELECT DISTINCT genre
FROM   genre
ORDER  BY genre; 


/* So, RSVP Movies plans to make a movie of one of these genres.
Now, wouldn’t you want to know which genre had the highest number of movies produced in the last year?
Combining both the movie and genres table can give more interesting insights. */

-- Q6.Which genre had the highest number of movies produced overall?
-- Type your code below:

SELECT g.genre,
       Count(m.id) AS No_of_movies
FROM   genre AS g
       INNER JOIN movie AS m
               ON g.movie_id = m.id
GROUP  BY genre
ORDER  BY no_of_movies DESC
LIMIT  1; 

/* So, based on the insight that you just drew, RSVP Movies should focus on the ‘Drama’ genre. 
But wait, it is too early to decide. A movie can belong to two or more genres. 
So, let’s find out the count of movies that belong to only one genre.*/

-- Q7. How many movies belong to only one genre?
-- Type your code below:

WITH movies_of_genre
     AS (SELECT movie_id,
                Count(genre)
         FROM   genre
         GROUP  BY movie_id
         HAVING Count(genre) = 1)
SELECT Count(movie_id) AS Movies_with_One_Genre
FROM   movies_of_genre; 



/* There are more than three thousand movies which has only one genre associated with them.
So, this figure appears significant. 
Now, let's find out the possible duration of RSVP Movies’ next project.*/

-- Q8.What is the average duration of movies in each genre? 
-- (Note: The same movie can belong to multiple genres.)


/* Output format:

+---------------+-------------------+
| genre			|	avg_duration	|
+-------------------+----------------
|	thriller	|		105			|
|	.			|		.			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:


SELECT 
    g.Genre, ROUND(AVG(m.duration), 2) AS Avg_Duration
FROM
    genre AS g
        INNER JOIN
    movie AS m ON g.movie_id = m.id
GROUP BY genre
ORDER BY avg_duration DESC;





/* Now you know, movies of genre 'Drama' (produced highest in number in 2019) has the average duration of 106.77 mins.
Lets find where the movies of genre 'thriller' on the basis of number of movies.*/

-- Q9.What is the rank of the ‘thriller’ genre of movies among all the genres in terms of number of movies produced? 
-- (Hint: Use the Rank function)


/* Output format:
+---------------+-------------------+---------------------+
| genre			|		movie_count	|		genre_rank    |	
+---------------+-------------------+---------------------+
|drama			|	2312			|			2		  |
+---------------+-------------------+---------------------+*/
-- Type your code below:

WITH genre_rank
     AS (SELECT genre,
                Count(DISTINCT movie_id)                    AS movie_count,
                Rank()
                  OVER (
                    ORDER BY Count(DISTINCT movie_id) DESC) AS genre_rank
         FROM   genre
         GROUP  BY genre)
SELECT *
FROM   genre_rank
WHERE  genre = 'Thriller';



/*Thriller movies is in top 3 among all genres in terms of number of movies
 In the previous segment, you analysed the movies and genres tables. 
 In this segment, you will analyse the ratings table as well.
To start with lets get the min and max values of different columns in the table*/




-- Segment 2:




-- Q10.  Find the minimum and maximum values in  each column of the ratings table except the movie_id column?
/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+-----------------+
| min_avg_rating|	max_avg_rating	|	min_total_votes   |	max_total_votes 	 |min_median_rating|min_median_rating|
+---------------+-------------------+---------------------+----------------------+-----------------+-----------------+
|		0		|			5		|	       177		  |	   2000	    		 |		0	       |	8			 |
+---------------+-------------------+---------------------+----------------------+-----------------+-----------------+*/
-- Type your code below:


SELECT 
    MIN(avg_rating) AS min_avg_rating,
    MAX(avg_rating) AS max_avg_rating,
    MIN(total_votes) AS min_total_votes,
    MAX(total_votes) AS max_total_votes,
    MIN(median_rating) AS min_median_rating,
    MAX(median_rating) AS max_median_rating
FROM
    ratings;

    

/* So, the minimum and maximum values in each column of the ratings table are in the expected range. 
This implies there are no outliers in the table. 
Now, let’s find out the top 10 movies based on average rating.*/

-- Q11. Which are the top 10 movies based on average rating?
/* Output format:
+---------------+-------------------+---------------------+
| title			|		avg_rating	|		movie_rank    |
+---------------+-------------------+---------------------+
| Fan			|		9.6			|			5	  	  |
|	.			|		.			|			.		  |
|	.			|		.			|			.		  |
|	.			|		.			|			.		  |
+---------------+-------------------+---------------------+*/
-- Type your code below:
-- It's ok if RANK() or DENSE_RANK() is used too

WITH movie_rank AS
(
           SELECT     m.title,
                      r.avg_rating,
                      Dense_rank () OVER (ORDER BY r.avg_rating DESC) AS movie_rank
           FROM       movie                                           AS m
           INNER JOIN ratings                                         AS r
           ON         m.id = r.movie_id
           ORDER BY   avg_rating DESC )
SELECT *
FROM   movie_rank limit 10;





/* Do you find you favourite movie FAN in the top 10 movies with an average rating of 9.6? If not, please check your code again!!
So, now that you know the top 10 movies, do you think character actors and filler actors can be from these movies?
Summarising the ratings table based on the movie counts by median rating can give an excellent insight.*/

-- Q12. Summarise the ratings table based on the movie counts by median ratings.
/* Output format:

+---------------+-------------------+
| median_rating	|	movie_count		|
+-------------------+----------------
|	1			|		105			|
|	.			|		.			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:
-- Order by is good to have

SELECT 
    median_rating, COUNT(movie_id) AS movie_count
FROM
    ratings
GROUP BY (median_rating)
ORDER BY movie_count DESC;



/* Movies with a median rating of 7 is highest in number. 
Now, let's find out the production house with which RSVP Movies can partner for its next project.*/

-- Q13. Which production house has produced the most number of hit movies (average rating > 8)??
/* Output format:
+------------------+-------------------+---------------------+
|production_company|movie_count	       |	prod_company_rank|
+------------------+-------------------+---------------------+
| The Archers	   |		1		   |			1	  	 |
+------------------+-------------------+---------------------+*/
-- Type your code below:


WITH prod_company
     AS (SELECT m.production_company,
                Count(r.movie_id)                   AS movie_count,
                Rank()
                  OVER (
                    ORDER BY Count(r.movie_id)DESC) AS prod_company_rank
         FROM   movie AS m
                INNER JOIN ratings AS r
                        ON m.id = r.movie_id
         WHERE  avg_rating > 8
                AND production_company IS NOT NULL
         GROUP  BY m.production_company)
SELECT *
FROM   prod_company
WHERE  prod_company_rank = 1; 



-- It's ok if RANK() or DENSE_RANK() is used too
-- Answer can be Dream Warrior Pictures or National Theatre Live or both

-- Q14. How many movies released in each genre during March 2017 in the USA had more than 1,000 votes?
/* Output format:

+---------------+-------------------+
| genre			|	movie_count		|
+-------------------+----------------
|	thriller	|		105			|
|	.			|		.			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:

SELECT 
    g.genre, COUNT(m.id) AS movie_count
FROM
    genre AS g
        INNER JOIN
    movie AS m ON g.movie_id = m.id
        INNER JOIN
    ratings AS r ON m.id = r.movie_id
WHERE
    MONTH(date_published) = 3
        AND m.year = '2017'
        AND m.country LIKE '%USA%'
        AND r.total_votes > 1000
GROUP BY genre
ORDER BY movie_count DESC;



-- Lets try to analyse with a unique problem statement.
-- Q15. Find movies of each genre that start with the word ‘The’ and which have an average rating > 8?
/* Output format:
+---------------+-------------------+---------------------+
| title			|		avg_rating	|		genre	      |
+---------------+-------------------+---------------------+
| Theeran		|		8.3			|		Thriller	  |
|	.			|		.			|			.		  |
|	.			|		.			|			.		  |
|	.			|		.			|			.		  |
+---------------+-------------------+---------------------+*/
-- Type your code below:

SELECT m.title,
       r.avg_rating,
       g.genre
FROM   genre AS g
       INNER JOIN movie AS m
               ON m.id = g.movie_id
       INNER JOIN ratings AS r
               ON m.id = r.movie_id
WHERE  m.title LIKE "the%"
       AND r.avg_rating > 8
ORDER  BY r.avg_rating DESC; 




-- You should also try your hand at median rating and check whether the ‘median rating’ column gives any significant insights.
-- Q16. Of the movies released between 1 April 2018 and 1 April 2019, how many were given a median rating of 8?
-- Type your code below:


SELECT 
    median_rating, count(title) as No_of_Movies
FROM
    movie AS m
        INNER JOIN
    ratings AS r ON m.id = r.movie_id
WHERE
    (m.date_published BETWEEN '2018-04-01' AND '2019-04-01')
        AND median_rating = '8'
GROUP BY median_rating;




-- Once again, try to solve the problem given below.
-- Q17. Do German movies get more votes than Italian movies? 
-- Hint: Here you have to find the total number of votes for both German and Italian movies.
-- Type your code below:

SELECT 
    Country, SUM(total_votes) AS Total_Votes
FROM
    movie AS m
        INNER JOIN
    ratings AS r ON m.id = r.movie_id
WHERE
    country = 'Germany' OR country = 'Italy'
GROUP BY country;


-- Answer is Yes

/* Now that you have analysed the movies, genres and ratings tables, let us now analyse another table, the names table. 
Let’s begin by searching for null values in the tables.*/




-- Segment 3:



-- Q18. Which columns in the names table have null values??
/*Hint: You can find null values for individual columns or follow below output format
+---------------+-------------------+---------------------+----------------------+
| name_nulls	|	height_nulls	|date_of_birth_nulls  |known_for_movies_nulls|
+---------------+-------------------+---------------------+----------------------+
|		0		|			123		|	       1234		  |	   12345	    	 |
+---------------+-------------------+---------------------+----------------------+*/
-- Type your code below:
SELECT 
    SUM(CASE
        WHEN n.name IS NULL THEN '1'
        ELSE '0'
    END) AS name_nulls,
    SUM(CASE
        WHEN n.height IS NULL THEN '1'
        ELSE '0'
    END) AS height_nulls,
    SUM(CASE
        WHEN n.date_of_birth IS NULL THEN '1'
        ELSE '0'
    END) AS date_of_birth_nulls,
    SUM(CASE
        WHEN n.known_for_movies IS NULL THEN '1'
        ELSE '0'
    END) AS known_for_movies_nulls
FROM
    names AS n;



/* There are no Null value in the column 'name'.
The director is the most important person in a movie crew. 
Let’s find out the top three directors in the top three genres who can be hired by RSVP Movies.*/

-- Q19. Who are the top three directors in the top three genres whose movies have an average rating > 8?
-- (Hint: The top three genres would have the most number of movies with an average rating > 8.)
/* Output format:

+---------------+-------------------+
| director_name	|	movie_count		|
+---------------+-------------------|
|James Mangold	|		4			|
|	.			|		.			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:
WITH top_genres AS
(
           SELECT     Count(title),
                      genre
           FROM       ratings AS r
           INNER JOIN movie   AS m
           ON         m.id = r.movie_id
           INNER JOIN genre AS g
           ON         g.movie_id = m.id
           WHERE      avg_rating >8
           GROUP BY   genre
           ORDER BY   Count(title) DESC limit 3 )
SELECT     n.NAME            AS director_name ,
           Count(d.movie_id) AS movie_count
FROM       director_mapping  AS d
INNER JOIN genre G
using      (movie_id)
INNER JOIN names AS n
ON         n.id = d.name_id
INNER JOIN top_genres
using      (genre)
INNER JOIN ratings
using      (movie_id)
WHERE      avg_rating > 8
GROUP BY   NAME
ORDER BY   movie_count DESC limit 3;

/* James Mangold can be hired as the director for RSVP's next project. Do you remeber his movies, 'Logan' and 'The Wolverine'. 
Now, let’s find out the top two actors.*/

-- Q20. Who are the top two actors whose movies have a median rating >= 8?
/* Output format:

+---------------+-------------------+
| actor_name	|	movie_count		|
+-------------------+----------------
|Christain Bale	|		10			|
|	.			|		.			|
+---------------+-------------------+ */
-- Type your code below:

SELECT 
    n.name AS actor_name, COUNT(movie_id) as No_of_Movies
FROM
    names AS n
        INNER JOIN
    role_mapping AS rm ON n.id = rm.name_id
        INNER JOIN
    ratings AS r USING (movie_id)
WHERE
    median_rating >= 8
        AND rm.category = 'actor'
GROUP BY name
ORDER BY COUNT(movie_id) DESC
LIMIT 2;


/* Have you find your favourite actor 'Mohanlal' in the list. If no, please check your code again. 
RSVP Movies plans to partner with other global production houses. 
Let’s find out the top three production houses in the world.*/

-- Q21. Which are the top three production houses based on the number of votes received by their movies?
/* Output format:
+------------------+--------------------+---------------------+
|production_company|vote_count			|		prod_comp_rank|
+------------------+--------------------+---------------------+
| The Archers		|		830			|		1	  		  |
|	.				|		.			|			.		  |
|	.				|		.			|			.		  |
+-------------------+-------------------+---------------------+*/
-- Type your code below:

SELECT     production_company,
           Sum(total_votes)                           AS vote_count,
           Rank() OVER(ORDER BY Sum(total_votes)DESC) AS prod_comp_rank
FROM       movie                                      AS m
INNER JOIN ratings                                    AS r
ON         m.id = r.movie_id
GROUP BY   production_company limit 3;


/*Yes Marvel Studios rules the movie world.
So, these are the top three production houses based on the number of votes received by the movies they have produced.

Since RSVP Movies is based out of Mumbai, India also wants to woo its local audience. 
RSVP Movies also wants to hire a few Indian actors for its upcoming project to give a regional feel. 
Let’s find who these actors could be.*/

-- Q22. Rank actors with movies released in India based on their average ratings. Which actor is at the top of the list?
-- Note: The actor should have acted in at least five Indian movies. 
-- (Hint: You should use the weighted average based on votes. If the ratings clash, then the total number of votes should act as the tie breaker.)

/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+
| actor_name	|	total_votes		|	movie_count		  |	actor_avg_rating 	 |actor_rank	   |
+---------------+-------------------+---------------------+----------------------+-----------------+
|	Yogi Babu	|			3455	|	       11		  |	   8.42	    		 |		1	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
+---------------+-------------------+---------------------+----------------------+-----------------+*/
-- Type your code below:

SELECT n.NAME,
       Sum(total_votes),
       Count(m.id)                                                AS movie_count
       ,
       Round(Sum(avg_rating * total_votes) / Sum(total_votes), 2) AS
       actor_avg_rating,
       Rank()
         OVER (
           ORDER BY Sum(avg_rating * total_votes) / Sum(total_votes) DESC, Sum(
         total_votes) DESC)                                       AS actor_rank
FROM   names AS n
       INNER JOIN role_mapping AS rm
               ON rm.name_id = n.id
       INNER JOIN movie AS m
               ON m.id = rm.movie_id
       INNER JOIN ratings AS r
               ON m.id = r.movie_id
WHERE  rm.category = 'actor'
       AND m.country = 'India'
GROUP  BY n.NAME
HAVING Count(m.id) >= 5; 



-- Top actor is Vijay Sethupathi

-- Q23.Find out the top five actresses in Hindi movies released in India based on their average ratings? 
-- Note: The actresses should have acted in at least three Indian movies. 
-- (Hint: You should use the weighted average based on votes. If the ratings clash, then the total number of votes should act as the tie breaker.)
/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+
| actress_name	|	total_votes		|	movie_count		  |	actress_avg_rating 	 |actress_rank	   |
+---------------+-------------------+---------------------+----------------------+-----------------+
|	Tabu		|			3455	|	       11		  |	   8.42	    		 |		1	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
+---------------+-------------------+---------------------+----------------------+-----------------+*/
-- Type your code below:

SELECT     n.NAME                                                                       AS actress_name,
           Sum(total_votes)                                                             AS total_votes,
           Count(m.id)                                                                  AS movie_count,
           Round(Sum(avg_rating                 * total_votes) / Sum(total_votes),2)    AS actress_avg_rating,
           Rank() OVER (ORDER BY Sum(avg_rating * total_votes) / Sum(total_votes) DESC) AS actress_rank
FROM       names                                                                        AS n
INNER JOIN role_mapping                                                                 AS rm
ON         rm.name_id = n.id
INNER JOIN movie AS m
ON         m.id = rm.movie_id
INNER JOIN ratings AS r
ON         m.id = r.movie_id
WHERE      rm.category='actress'
AND        m.country='India'
AND        languages LIKE "%hindi%"
GROUP BY   n.NAME
HAVING     Count(m.id) >=3 limit 5;



/* Taapsee Pannu tops with average rating 7.74. 
Now let us divide all the thriller movies in the following categories and find out their numbers.*/


/* Q24. Select thriller movies as per avg rating and classify them in the following category: 

			Rating > 8: Superhit movies
			Rating between 7 and 8: Hit movies
			Rating between 5 and 7: One-time-watch movies
			Rating < 5: Flop movies
--------------------------------------------------------------------------------------------*/
-- Type your code below:


SELECT title,
       avg_rating AS Rating,
       CASE
         WHEN avg_rating > 8 THEN 'Superhit movies'
         WHEN avg_rating BETWEEN 7 AND 8 THEN "hit movies"
         WHEN avg_rating BETWEEN 5 AND 7 THEN "one-time_watch movies"
         ELSE "flop movies"
       END        AS Category
FROM   ratings AS r
       INNER JOIN movie AS m
               ON r.movie_id = m.id
       INNER JOIN genre AS g
               ON g.movie_id = m.id
WHERE  g.genre = 'Thriller'
ORDER  BY rating DESC; 



/* Until now, you have analysed various tables of the data set. 
Now, you will perform some tasks that will give you a broader understanding of the data in this segment.*/

-- Segment 4:

-- Q25. What is the genre-wise running total and moving average of the average movie duration? 
-- (Note: You need to show the output table in the question.) 
/* Output format:
+---------------+-------------------+---------------------+----------------------+
| genre			|	avg_duration	|running_total_duration|moving_avg_duration  |
+---------------+-------------------+---------------------+----------------------+
|	comdy		|			145		|	       106.2	  |	   128.42	    	 |
|		.		|			.		|	       .		  |	   .	    		 |
|		.		|			.		|	       .		  |	   .	    		 |
|		.		|			.		|	       .		  |	   .	    		 |
+---------------+-------------------+---------------------+----------------------+*/
-- Type your code below:

SELECT genre,
       round(Avg(duration),2)                                AS avg_duration,
		SUM(Round(Avg(duration), 2))
         over(
           ORDER BY genre ) AS running_total_duration,
       Avg(Round(Avg(duration), 2))
         over(
           ORDER BY genre )        AS moving_avg_duration
FROM   genre AS g
       inner join movie AS m
               ON g.movie_id = m.id
GROUP  BY genre; 

-- Round is good to have and not a must have; Same thing applies to sorting


-- Let us find top 5 movies of each year with top 3 genres.

-- Q26. Which are the five highest-grossing movies of each year that belong to the top three genres? 
-- (Note: The top 3 genres would have the most number of movies.)

/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+
| genre			|	year			|	movie_name		  |worldwide_gross_income|movie_rank	   |
+---------------+-------------------+---------------------+----------------------+-----------------+
|	comedy		|			2017	|	       indian	  |	   $103244842	     |		1	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
+---------------+-------------------+---------------------+----------------------+-----------------+*/
-- Type your code below:

-- Top 3 Genres based on most number of movies

WITH top3_genre AS
(
         SELECT   genre,
                  Count(movie_id) AS movie_count
         FROM     genre
         GROUP BY genre
         ORDER BY movie_count DESC limit 3 ), movie_sum AS
(
           SELECT     g.genre,
                      m.year,
                      m.title                                                                    AS movie_name,
                                 Concat("$ ",m.worlwide_gross_income)                            AS world_gross_income,
                      Dense_rank () OVER (partition BY year ORDER BY worlwide_gross_income DESC) AS movie_rank
           FROM       (
                               SELECT   title ,
                                        id,
                                        year,
                                        Sum(worlwide_gross_income) AS worlwide_gross_income
                               FROM     (
                                               SELECT title,
                                                      id,
                                                      year,
                                                      CASE
                                                             WHEN currency="US" THEN modified_income
                                                             WHEN currency="IND" THEN modified_income/80
                                                             ELSE 0
                                                      END AS worlwide_gross_income
                                               FROM   (
                                                             SELECT title,
                                                                    id,
                                                                    year,
                                                                    worlwide_gross_income,
                                                                    CASE
                                                                           WHEN worlwide_gross_income LIKE "$%" THEN "US"
                                                                           WHEN worlwide_gross_income LIKE "INR%" THEN "IND"
                                                                           ELSE "nor"
                                                                    END                                                                                AS currency,
                                                                    Cast(Replace(Replace(Ifnull(worlwide_gross_income,0),'INR',''),'$','') AS DECIMAL) AS modified_income
                                                             FROM   movie) as p ) as q
                               GROUP BY title,
                                        year,
                                        id ) AS m
           INNER JOIN genre                  AS g
           ON         m.id=g.movie_id
           WHERE      g.genre IN
                      (
                             SELECT genre
                             FROM   top3_genre)
                      )
SELECT   *
FROM     movie_sum
WHERE    movie_rank<=5
ORDER BY year;


-- Finally, let’s find out the names of the top two production houses that have produced the highest number of hits among multilingual movies.
-- Q27.  Which are the top two production houses that have produced the highest number of hits (median rating >= 8) among multilingual movies?
/* Output format:
+-------------------+-------------------+---------------------+
|production_company |movie_count		|		prod_comp_rank|
+-------------------+-------------------+---------------------+
| The Archers		|		830			|		1	  		  |
|	.				|		.			|			.		  |
|	.				|		.			|			.		  |
+-------------------+-------------------+---------------------+*/
-- Type your code below:


SELECT     production_company,
           Count(id)                             AS movie_count,
           Rank() OVER (ORDER BY Count(id) DESC) AS prod_comp_rank
FROM       movie                                 AS m
INNER JOIN ratings                               AS r
ON         m.id = r. movie_id
WHERE      median_rating >=8
AND        position("," IN languages) >0
AND        production_company IS NOT NULL
GROUP BY   production_company limit 2;



-- Multilingual is the important piece in the above question. It was created using POSITION(',' IN languages)>0 logic
-- If there is a comma, that means the movie is of more than one language


-- Q28. Who are the top 3 actresses based on number of Super Hit movies (average rating >8) in drama genre?
/* Output format:
+---------------+-------------------+---------------------+----------------------+-----------------+
| actress_name	|	total_votes		|	movie_count		  |actress_avg_rating	 |actress_rank	   |
+---------------+-------------------+---------------------+----------------------+-----------------+
|	Laura Dern	|			1016	|	       1		  |	   9.60			     |		1	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
|		.		|			.		|	       .		  |	   .	    		 |		.	       |
+---------------+-------------------+---------------------+----------------------+-----------------+*/
-- Type your code below:

SELECT     n.NAME                                                    AS actress_name,
           Sum(total_votes)                                          AS total_votes,
           Count(r.movie_id)                                         AS movie_count,
           Round(Sum(avg_rating * total_votes) / Sum(total_votes),2) AS actress_avg_rating ,
           Rank() OVER (ORDER BY Count(r.movie_id) DESC)             AS actress_rank
FROM       movie                                                     AS m
INNER JOIN role_mapping                                              AS rm
ON         rm.movie_id = m.id
INNER JOIN names AS n
ON         rm.name_id = n.id
INNER JOIN genre AS g
ON         g.movie_id = m.id
INNER JOIN ratings AS r
ON         m.id = r.movie_id
WHERE      r.avg_rating>8
AND        rm.category = "ACTRESS"
AND        g.genre ="DRAMA"
GROUP BY   n.NAME limit 3 ;


/* Q29. Get the following details for top 9 directors (based on number of movies)
Director id
Name
Number of movies
Average inter movie duration in days
Average movie ratings
Total votes
Min rating
Max rating
total movie durations

Format:
+---------------+-------------------+---------------------+----------------------+--------------+--------------+------------+------------+----------------+
| director_id	|	director_name	|	number_of_movies  |	avg_inter_movie_days |	avg_rating	| total_votes  | min_rating	| max_rating | total_duration |
+---------------+-------------------+---------------------+----------------------+--------------+--------------+------------+------------+----------------+
|nm1777967		|	A.L. Vijay		|			5		  |	       177			 |	   5.65	    |	1754	   |	3.7		|	6.9		 |		613		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
|	.			|		.			|			.		  |	       .			 |	   .	    |	.		   |	.		|	.		 |		.		  |
+---------------+-------------------+---------------------+----------------------+--------------+--------------+------------+------------+----------------+

--------------------------------------------------------------------------------------------*/
-- Type you code below:

 
WITH movie_info AS
(
         SELECT   dm.name_id,
                  NAME,
                  dm.movie_id,
                  m.date_published,
                  Lead(date_published, 1) OVER(partition BY dm.name_id ORDER BY date_published, dm.movie_id) AS next_movie_date
         FROM     director_mapping dm
         JOIN     names AS n
         ON       dm.name_id=n.id
         JOIN     movie AS m
         ON       dm.movie_id=m.id ), date_difference AS
(
       SELECT *,
              Datediff(next_movie_date, date_published) AS date_diff
       FROM   movie_info ), avg_inter_days AS
(
         SELECT   name_id,
                  Avg(date_diff) AS avg_inter_movie_days
         FROM     date_difference
         GROUP BY name_id ), final AS
(
         SELECT   dm.name_id                                          AS director_id,
                  NAME                                                AS director_name,
                  Count(dm.movie_id)                                  AS number_of_movies,
                  Round(avg_inter_movie_days)                         AS inter_movie_days,
                  Round(Avg(avg_rating),2)                            AS avg_rating,
                  Sum(total_votes)                                    AS total_votes,
                  Min(avg_rating)                                     AS min_rating,
                  Max(avg_rating)                                     AS max_rating,
                  Sum(duration)                                       AS total_duration,
                  Row_number() OVER(ORDER BY Count(dm.movie_id) DESC) AS director_row_rank
         FROM     names                                               AS n
         JOIN     director_mapping                                    AS dm
         ON       n.id=dm.name_id
         JOIN     ratings AS r
         ON       dm.movie_id=r.movie_id
         JOIN     movie AS m
         ON       m.id=r.movie_id
         JOIN     avg_inter_days AS a
         ON       a.name_id=dm.name_id
         GROUP BY director_id )
SELECT *
FROM   final LIMIT 9;