Difference between revisions of "More JOIN operations"

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   .
 
   .
 
</p>
 
</p>
  <div class='schema'>movie(<b>id</b>, title, yr, <i>director</i>)<br/> actor(<b>id</b>, name)<br/> casting(<b><i>movieid</i>, <i>actorid</i></b>, ord)<br/></div>
+
  <div class='sc'></div>
 +
<div class = 'ref_section'>
 +
<table class = 'db_ref'>
 +
<tr><th>'''movie'''</th><th>'''actor'''</th><th>'''casting'''</th></tr>
 +
<tr><td>id</td><td>id</td><td>movieid</td></tr>
 +
<tr><td>title</td><td>name</td><td>actorid</td></tr>
 +
<tr><td>yr</td><td></td><td>ord</td></tr>
 +
<tr><td>director</td><td></td><td></td></tr>
 +
<tr><td>budget</td><td></td><td></td></tr>
 +
<tr><td>gross</td><td></td><td></td></tr>
 +
<tr><td></td><td></td><td></td></tr>
 +
</table>
 +
</div>
 
   <p>
 
   <p>
 
     [[More details about the database.]]
 
     [[More details about the database.]]
 
   </p>
 
   </p>
  
{{JOIN tables 2}}
 
  
<div class="summary">Summary</div>
+
<div class="progress_panel"><div>
 
+
  <div class="summary">Summary</div>
<div class="progressbarbg">
+
  <div class="progressbarbg">
  <div class="progressbar"></div>
+
    <div class="progressbar"></div>
</div>
+
  </div>
 +
</div></div>
  
 
<h2>Let's go to work.</h2>
 
<h2>Let's go to work.</h2>
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<div class='qu'>
 
<div class='qu'>
List all of the Star Trek movies, include the '''id title''' and '''yr'''. (All of these movies include the words Star Trek in the title.)  
+
List all of the Star Trek movies, include the '''id''', '''title''' and '''yr''' (all of these movies include the words Star Trek in the title). Order results by year.
 
<source lang='sql' class='def'>
 
<source lang='sql' class='def'>
 
</source>
 
</source>
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<h2>Harder Questions</h2>
 
<h2>Harder Questions</h2>
 
<div class='qu'>
 
<div class='qu'>
Which were the busiest years for 'John Travolta'. Show the number of movies he made for each year.       
+
Which were the busiest years for 'John Travolta', show the year and the number of movies he made each year for any year in which he made at least 2 movies.       
 
<source lang='sql' class='def'>
 
<source lang='sql' class='def'>
 
SELECT yr,COUNT(title) FROM
 
SELECT yr,COUNT(title) FROM
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<div class='qu'>
 
<div class='qu'>
List the film title and the leading actor for all of 'Julie Andrews' films.   
+
List the film title and the leading actor for all of the films 'Julie Andrews' played in.   
 
<source lang='sql' class='def'>
 
<source lang='sql' class='def'>
 
</source>
 
</source>
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<div class='qu'>
 
<div class='qu'>
Obtain a list of actors in who have had at least 30 starring roles.         
+
Obtain a list in alphabetical order of actors who've had at least 30 starring roles.         
 
<source lang='sql' class='def'>
 
<source lang='sql' class='def'>
 
</source>
 
</source>
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<source lang='sql' class='ans'>
 
<source lang='sql' class='ans'>
SELECT title, COUNT(actorid)
+
  SELECT title, COUNT(actorid)
   FROM casting, movie
+
   FROM casting,movie              
 
   WHERE yr=1978
 
   WHERE yr=1978
    AND movieid=movie.id
+
        AND movieid=movie.id
 
   GROUP BY title
 
   GROUP BY title
 
   ORDER BY 2 DESC
 
   ORDER BY 2 DESC
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</source>
 
</source>
 
</div>
 
</div>
 
+
<div>
[[JOIN Quiz 2]]
+
<div class="lsclear">Clear your results</div>
 
+
<p><div class="quizlink">[[JOIN Quiz 2]]</div></p>
 +
</div>
 
[http://sqlzoo.net/w/index.php/Using_Null That is definitely enough. Students should, under no circumstances look at the next tutorial, concerning outer joins.]
 
[http://sqlzoo.net/w/index.php/Using_Null That is definitely enough. Students should, under no circumstances look at the next tutorial, concerning outer joins.]

Revision as of 12:18, 27 March 2013

Contents

Movie Database

This tutorial introduces the notion of a join. The database consists of three tables movie , actor and casting .

movieactorcasting
ididmovieid
titlenameactorid
yrord
director
budget
gross

More details about the database.


Summary

Let's go to work.

Limbering up

List the films where the yr is 1962 [Show id, title]

SELECT id, title
 FROM movie
 WHERE yr=1962
SELECT id, title
 FROM movie
 WHERE yr=1962

Give year of 'Citizen Kane'.

 
SELECT yr 
FROM movie 
WHERE title='Citizen Kane'

List all of the Star Trek movies, include the id, title and yr (all of these movies include the words Star Trek in the title). Order results by year.

 
SELECT id,title, yr FROM movie
 WHERE title LIKE 'Star Trek%'
 ORDER BY yr

Looking at the id field.

What are the titles of the films with id 11768, 11955, 21191

 
SELECT title 
FROM movie 
WHERE id IN (11768, 11955, 21191)

What id number does the actor 'Glenn Close' have?

 
SELECT id FROM actor
  WHERE name= 'Glenn Close'


What is the id of the film 'Casablanca'

 
SELECT id 
FROM movie 
WHERE title='Casablanca'

Get to the point

Obtain the cast list for 'Casablanca'. Use the id value that you obtained in the previous question.

 
SELECT name
  FROM casting, actor
  WHERE movieid=(SELECT id 
             FROM movie 
             WHERE title='Casablanca')
    AND actorid=actor.id

Obtain the cast list for the film 'Alien'

 
SELECT name
  FROM movie, casting, actor
  WHERE title='Alien'
    AND movieid=movie.id
    AND actorid=actor.id

List the films in which 'Harrison Ford' has appeared

 
SELECT title
  FROM movie, casting, actor
 WHERE name='Harrison Ford'
    AND movieid=movie.id
    AND actorid=actor.id

List the films where 'Harrison Ford' has appeared - but not in the star role. [Note: the ord field of casting gives the position of the actor. If ord=1 then this actor is in the starring role]

 
SELECT title
  FROM movie, casting, actor
 WHERE name='Harrison Ford'
    AND movieid=movie.id
    AND actorid=actor.id
  AND ord<>1

List the films together with the leading star for all 1962 films.

 
SELECT title, name
  FROM movie, casting, actor
 WHERE yr=1962
    AND movieid=movie.id
    AND actorid=actor.id
    AND ord=1

Harder Questions

Which were the busiest years for 'John Travolta', show the year and the number of movies he made each year for any year in which he made at least 2 movies.

SELECT yr,COUNT(title) FROM
  movie JOIN casting ON movie.id=movieid
         JOIN actor   ON actorid=actor.id
WHERE name='John Travolta'
GROUP BY yr
HAVING COUNT(title)=(SELECT MAX(c) FROM
(SELECT yr,COUNT(title) AS c FROM
   movie JOIN casting ON movie.id=movieid
         JOIN actor   ON actorid=actor.id
 WHERE name='John Travolta'
 GROUP BY yr) AS t
)
SELECT yr,COUNT(title) FROM
  movie JOIN casting ON movie.id=movieid
         JOIN actor   ON actorid=actor.id
WHERE name='John Travolta'
GROUP BY yr
HAVING COUNT(title)=(SELECT MAX(c) FROM
(SELECT yr,COUNT(title) AS c FROM
movie JOIN casting ON movie.id=movieid 
JOIN actor   ON actorid=actor.id
 WHERE name='John Travolta'
 GROUP BY yr) AS t
)

List the film title and the leading actor for all of the films 'Julie Andrews' played in.

 
SELECT title, name
  FROM movie, casting, actor
  WHERE movieid=movie.id
    AND actorid=actor.id
    AND ord=1
    AND movieid IN
    (SELECT movieid FROM casting, actor
     WHERE actorid=actor.id
     AND name='Julie Andrews')

Obtain a list in alphabetical order of actors who've had at least 30 starring roles.

 
SELECT name
    FROM casting JOIN actor
      ON  actorid = actor.id
    WHERE ord=1
    GROUP BY name
    HAVING COUNT(movieid)>=30

List the 1978 films by order of cast list size.

 
  SELECT title, COUNT(actorid)
  FROM casting,movie                
  WHERE yr=1978
        AND movieid=movie.id
  GROUP BY title
  ORDER BY 2 DESC

List all the people who have worked with 'Art Garfunkel'.

 
SELECT DISTINCT d.name
FROM actor d JOIN casting a ON (a.actorid=d.id)
   JOIN casting b ON (a.movieid=b.movieid)
   JOIN actor c ON (b.actorid=c.id 
                AND c.name='Art Garfunkel')
  WHERE d.id!=c.id
Clear your results

That is definitely enough. Students should, under no circumstances look at the next tutorial, concerning outer joins.

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