Difference between revisions of "JOIN Quiz 2"

JOIN Quiz - part 2

movie
Field name Type Notes
id INTEGER An arbitrary unique identifier
title CHAR(70) The name of the film - usually in the language of the first release.
yr DECIMAL(4) Year of first release.
director INT A reference to the actor table.
budget INTEGER How much the movie cost to make (in a variety of currencies unfortunately).
gross INTEGER How much the movie made at the box office.
actor
Field name Type Notes
id INTEGER An arbitrary unique identifier
name CHAR(36) The name of the actor (the term actor is used to refer to both male and female thesps.)
casting
Field name Type Notes
movieid INTEGER A reference to the movie table.
actorid INTEGER A reference to the actor table.
ord INTEGER The ordinal position of the actor in the cast list. The

star of the movie will have ord value 1 the co-star will have

value 2, ...

<quiz shuffle=none display=simple> {Which of the following statements lists the unfortunate directors of the movies which have caused financial loses? (gross < budget) |type="()"} - SELECT JOIN(name FROM actor, movie ON actor.id:director WHERE gross < budget) GROUP BY name - SELECT name FROM actor INNER JOIN movie BY actor.id = director HAVING gross < budget + SELECT name FROM actor INNER JOIN movie ON actor.id = director WHERE gross < budget - SELECT name FROM actor INNER JOIN movie ON actor.id:director WHERE gross < budget - SELECT name FROM director INNER JOIN movie ON movie.id = director.id WHERE gross < budget

{Select the correct example of JOINing three tables |type="()"} - SELECT * FROM actor JOIN casting BY actor.id = actorid JOIN movie BY movie.id = movieid - SELECT * FROM actor JOIN casting ON actor.id = actorid AND JOIN movie ON movie.id = movieid - SELECT * FROM actor JOIN casting JOIN movie ON actor.id = actorid AND movie.id = movieid - SELECT * FROM actor JOIN casting ON actor.id = actorid AND movie ON movie.id = movieid + SELECT * FROM actor JOIN casting ON actor.id = actorid JOIN movie ON movie.id = movieid

</quiz>