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Databases Querying Relational Databases Subqueries Review and Practice

Donnie Driskell
Donnie Driskell
2,243 Points
Posted by Donnie Driskell
Donnie Driskell
2,243 Points

Relational Data Bases PlayGround Challenge: books north and south

-- Generate a report that lists the book titles from both locations and count the total number of books with the same title.

This is as far as I can get. I have racked my brain trying to figure out how to get a total with the same title. Thanks. Donnie

SELECT * FROM books_north AS n LEFT OUTER JOIN (SELECT * FROM books_south) AS s ON n.title = s.title GROUP BY n.id, s.id;

4 Answers

Steven Parker
Steven Parker
231,184 Points
Steven Parker
Steven Parker
231,184 Points

Here's a few hints:

  • you don't need a derived table to perform the JOIN
  • you don't actually need a JOIN for this task at all
  • but you might want to use a UNION (perhaps a UNION ALL)
  • for this task you only need the "title" column from each table
  • remember to include a COUNT
Donnie Driskell
Donnie Driskell
2,243 Points
Donnie Driskell
Donnie Driskell
2,243 Points

Thank you again Steven. Sorry for the late response. I will try that again. Donnie

Amitai Blickstein
Amitai Blickstein
6,863 Points
Amitai Blickstein
Amitai Blickstein
6,863 Points

Please help me improve my kludgy code! I am particularly interested in how I could

  • Clean and simplify
  • Use aliases more effectively
  • Reach the solution in alternative ways

THANKS in advance, students and gurus!

After much effort, for:

Generate a report that lists the book titles from both locations and count the total number of books with the same title.

I decided that meant:

.---------.-------------------.
| titles  |    isDuplicate    |
:---------+-------------------:
| UNION   |                   |
:---------+-------------------:
| "Total" | COUNT(duplicates) |
'---------'-------------------'

and got:

-- Book Titles indicating/counting duplicates

SELECT *,
       CASE
         WHEN all_titles.title
         IN (
              SELECT title FROM books_north
               INTERSECT
              SELECT title FROM books_south
              GROUP BY title
            )
         THEN "yes"
         ELSE "no"
       END
       AS is_duplicate
FROM   (
  SELECT title FROM books_north
   UNION
  SELECT title FROM books_south
  ) AS all_titles
   UNION
  SELECT ".total number of duplicates",
         Count(title)
  FROM   (SELECT title FROM books_north
           INTERSECT
          SELECT title FROM books_south);

For

Generate a report that lists a patron's first name, email and loan count for loans that haven't been returned.

I got:

-- User Loans

SELECT 
  p.first_name, 
  p.email, 
  COUNT(*) AS "#books_out"
FROM (
  SELECT * FROM loans_north
  UNION 
  SELECT * FROM loans_south
) AS all_loans
INNER JOIN 
  patrons AS p 
ON
  all_loans.patron_id = p.id
WHERE 
  returned_on IS NULL
GROUP BY 
  patron_id;
Daniel Arnost
Daniel Arnost
7,190 Points
Daniel Arnost
Daniel Arnost
7,190 Points 
SELECT first_name
       , last_name
       , patron_id
       , count(tl.patron_id) as loan_count
       , email
FROM patrons p
INNER JOIN (
       SELECT  patron_id, returned_on FROM loans_north 
       UNION ALL
       SELECT  patron_id, returned_on FROM loans_south ) as tl
on p.id = tl.patron_id
WHERE tl.returned_on Is Null
Group by patron_id
Order By patron_id;