Difference between revisions of "JOIN Quiz 2"

From SQLZOO
Jump to: navigation, search
 
Line 56: Line 56:
  
 
{Select the result that would be obtained from the following code: SELECT title, yr FROM movie, casting, actor WHERE name='Robert De Niro' AND movieid=movie.id AND actorid=actor.id AND ord = 3
 
{Select the result that would be obtained from the following code: SELECT title, yr FROM movie, casting, actor WHERE name='Robert De Niro' AND movieid=movie.id AND actorid=actor.id AND ord = 3
−
<table style='float:left'><caption>Table-C</caption><tr><td>A Bronx Tale</td><td>1993</td><td>3</td></tr><tr><td>Bang the Drum Slowly</td><td>1973</td><td>3</td></tr><tr><td>Limitless</td><td>2011</td><td>3</td></tr></table>
+
<table style='float:left'><caption>Table-A</caption><tr><td>A Bronx Tale</td><td>1993</td><td>3</td></tr><tr><td>Bang the Drum Slowly</td><td>1973</td><td>3</td></tr><tr><td>Limitless</td><td>2011</td><td>3</td></tr></table>
−
<table style='float:left'><caption>Table-C</caption><tr><td>A Bronx Tale</td><td>1993</td></tr><tr><td>Bang the Drum Slowly</td><td>1973</td></tr><tr><td>Limitless</td><td>2011</td></tr></table>
+
<table style='float:left'><caption>Table-B</caption><tr><td>A Bronx Tale</td><td>1993</td></tr><tr><td>Bang the Drum Slowly</td><td>1973</td></tr><tr><td>Limitless</td><td>2011</td></tr></table>
 
<table style='float:left'><caption>Table-C</caption><tr><td>A Bronx Tale</td><td>3</td></tr><tr><td>Bang the Drum Slowly</td><td>3</td></tr><tr><td>Limitless</td><td>3</td></tr></table>
 
<table style='float:left'><caption>Table-C</caption><tr><td>A Bronx Tale</td><td>3</td></tr><tr><td>Bang the Drum Slowly</td><td>3</td></tr><tr><td>Limitless</td><td>3</td></tr></table>
−
<table style='float:left'><caption>Table-C</caption><tr><td>A Bronx Tale</td></tr><tr><td>Bang the Drum Slowly</td></tr><tr><td>Limitless</td></tr></table>
+
<table style='float:left'><caption>Table-D</caption><tr><td>A Bronx Tale</td></tr><tr><td>Bang the Drum Slowly</td></tr><tr><td>Limitless</td></tr></table>
−
<table style='float:left'><caption>Table-C</caption><tr><td>A Bronx Tale</td><td>Robert De Niro</td><td>1993</td></tr><tr><td>Bang the Drum Slowly</td><td>Robert De Niro</td><td>1973</td></tr><tr><td>Limitless</td><td>Robert De Niro</td><td>2011</td></tr></table>
+
<table style='float:left'><caption>Table-E</caption><tr><td>A Bronx Tale</td><td>Robert De Niro</td><td>1993</td></tr><tr><td>Bang the Drum Slowly</td><td>Robert De Niro</td><td>1973</td></tr><tr><td>Limitless</td><td>Robert De Niro</td><td>2011</td></tr></table>
 
|type="()"}
 
|type="()"}
 
- Table-A
 
- Table-A

Latest revision as of 20:59, 7 September 2012

JOIN Quiz - part 2

1. Select the statement which lists the unfortunate directors of the movies which have caused financial loses (gross < budget)

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

2. Select the correct example of JOINing three tables

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

3. Select the statement that shows the list of actors called 'John' by order of number of movies in which they acted

SELECT name, COUNT(movieid) FROM actor JOIN casting ON actorid=actor.id WHERE name IN 'John %' GROUP BY name ORDER BY 2
SELECT name, COUNT(movieid) FROM actor JOIN casting ON actorid=actor.id WHERE name LIKE 'J%' GROUP BY name ORDER BY 2 DESC
SELECT name, COUNT(movieid) FROM casting JOIN actor ON actorid=actor.id WHERE name LIKE 'John %' GROUP BY name ORDER BY 2 DESC
SELECT name, COUNT(movieid) FROM casting JOIN actor WHERE (actorid ON actor.id) AND name LIKE 'John %' GROUP BY name ORDER BY 2 DESC
SELECT name, COUNT(movieid) FROM casting JOIN actor WHERE name LIKE 'John %' GROUP BY name ORDER BY COUNT(movieid) DESC

4. Select the result that would be obtained from the following code: SELECT title FROM movie, casting, actor WHERE name='Paul Hogan' AND movieid=movie.id AND actorid=actor.id AND ord = 1

Table-A
"Crocodile" Dundee1
Crocodile Dundee in Los Angeles1
Flipper1
Lightning Jack1
Table-B
"Crocodile" Dundee
Crocodile Dundee in Los Angeles
Flipper
Lightning Jack
Table-C
"Crocodile" Dundee
Paul Hogan
1
Table-D
"Crocodile" DundeePaul Hogan1
Crocodile Dundee in Los AngelesPaul Hogan1
FlipperPaul Hogan1
Lightning JackPaul Hogan1
Table-E
"Crocodile" DundeePaul Hogan
Crocodile Dundee in Los AngelesPaul Hogan
FlipperPaul Hogan
Lightning JackPaul Hogan
Table-A
Table-B
Table-C
Table-D
Table-E

5. Select the statement that lists all the actors that starred in movies directed by Ridley Scott

SELECT name FROM movie JOIN casting AND actor ON movie.id = movieid AND actor.id = actorid WHERE ord = NULL AND director = (SELECT id FROM actor WHERE name LIKE 'Ridley Scott')
SELECT name FROM movie JOIN casting JOIN actor ON movie.id = movieid OR actor.id = actorid WHERE ord = 1 AND director = (SELECT id FROM actor WHERE name LIKE 'Ridley Scott')
SELECT name FROM movie JOIN casting ON movie.id = movieid JOIN actor ON actor.id = actorid WHERE ord = 1 AND actorid = (SELECT id FROM actor WHERE name LIKE 'Ridley Scott')
SELECT name FROM movie JOIN casting ON movie.id = movieid JOIN actor ON actor.id = actorid WHERE ord = 1 AND director = (SELECT id FROM actor WHERE name LIKE 'Ridley Scott')
SELECT name FROM movie JOIN casting ON movie.id = actorid JOIN actor ON actor.id = movieid WHERE director = (SELECT id FROM actor WHERE name LIKE 'Ridley Scott')

6. Select the statement showing all movies with budget bigger than ALL movie with Harrison Ford is incorrect

SELECT title FROM movie WHERE budget > ALL (SELECT budget FROM movie JOIN casting JOIN actor ON movie.id = movieid AND actor.id = actorid WHERE name = 'Harrison Ford')
SELECT title FROM movie WHERE budget > ALL (SELECT budget FROM movie JOIN casting JOIN actor ON movie.id = movieid AND actor.id = actorid OR name = 'Harrison Ford')
SELECT title FROM movie WHERE budget > (SELECT MAX(budget) FROM movie JOIN casting JOIN actor ON movie.id = movieid AND actor.id = actorid WHERE name = 'Harrison Ford')
SELECT title FROM movie WHERE budget > ALL (SELECT budget FROM movie JOIN casting ON movie.id = movieid JOIN actor ON actor.id = actorid WHERE name = 'Harrison Ford')
SELECT title FROM movie WHERE budget > (SELECT MAX(budget) FROM movie JOIN casting JOIN actor ON movie.id = movieid AND actor.id = actorid AND name = 'Harrison Ford')

7. Select the result that would be obtained from the following code: SELECT title, yr FROM movie, casting, actor WHERE name='Robert De Niro' AND movieid=movie.id AND actorid=actor.id AND ord = 3

Table-A
A Bronx Tale19933
Bang the Drum Slowly19733
Limitless20113
Table-B
A Bronx Tale1993
Bang the Drum Slowly1973
Limitless2011
Table-C
A Bronx Tale3
Bang the Drum Slowly3
Limitless3
Table-D
A Bronx Tale
Bang the Drum Slowly
Limitless
Table-E
A Bronx TaleRobert De Niro1993
Bang the Drum SlowlyRobert De Niro1973
LimitlessRobert De Niro2011
Table-A
Table-B
Table-C
Table-D
Table-E

Your score is 0 / 0
Retrieved from "http://sqlzoo.net/w/index.php?title=JOIN_Quiz_2&oldid=4013"
Personal tools
Namespaces

Variants
Actions
Reference
Toolbox
Google AdSense