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PHP Integrating PHP with Databases Using Relational Tables Querying Multiple Tables with JOIN

SAHIL SANWAL
SAHIL SANWAL
6,116 Points
Posted by SAHIL SANWAL
SAHIL SANWAL
6,116 Points

getting unable to retreat

function single_item_array($id) {
  include("connection.php");

  try{
    $results = $db->query(
      "SELECT title, category,img, format, year,genre,publisher,isbn
      FROM Media
      JOIN Genre ON Media.genre_id = Genres.genre_id
      LEFT OUTER JOIN Books ON Media.media_id=Books.media_id
      WHERE Media.media_id = $id"
    );
    echo "retrieved results";
  } catch (Exception $e){
    echo "unable to retrieve results";
    exit;
  }

  $catalog = $results->fetch();
    return $catalog;
}

var_dump( single_item_array(1));```
Siddhant Mehta
Siddhant Mehta
Courses Plus Student 478 Points

will you please paste link of your workspace snapshot but in your query you are joining Genre and comparing Genres.genre_id try to correct this is error still exist paste your workspace screenshot.

jonathan wise
jonathan wise
51 Points

try{ $results = $db->query( "SELECT Media.media_id, title, category, img, format, year, genre, publisher, isbn FROM Media JOIN Genres ON Media.genre_id = Genres.genre_id LEFT OUTER JOIN Books ON Media.media_id = Books.media_id WHERE Media.media_id = $id" ); }catch (Exception $e) { echo "Unable to retrieve results"; exit; }

Try this :)

1 Answer

Dennis Amiel Domingo
Dennis Amiel Domingo
15,357 Points
Dennis Amiel Domingo
Dennis Amiel Domingo
15,357 Points

You forgot to add an "s" on the word "Genre" in your JOIN statement.

"Genre"... 

JOIN Genre ON Media.genre_id = Genres.genre_id

should be "Genres"

JOIN Genres ON Media.genre_id = Genres.genre_id