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PHP Integrating PHP with Databases Using Relational Tables Understanding SQL Injections

Prince Henry
Prince Henry
4,312 Points
Posted by Prince Henry
Prince Henry
4,312 Points

I have this error "Unable to retrieve results" on my browser.

I have followed the steps to this point but since from the last video, i keep having this error message on by browser "Unable to retrieve results". Can someone help me out?

Here is my code....

<?php

function full_catalog_array() { include("connection.php");

try { $results = $db->query("SELECT media_id, title, category, img FROM Media"); } catch (Exception $e) { echo "Unable to retrieve results"; exit; }

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

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

try { $results = $db->query( "SELECT 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; }

$catalog = $results->fetch(); return $catalog; } var_dump(single_item_array(1));

function get_item_html($id,$item) { $output = "<li><a href='details.php?id=" . $id . "'><img src='" . $item["img"] . "' alt='" . $item["title"] . "' />" . "<p>View Details</p>" . "</a></li>"; return $output; }

function array_category($catalog,$category) { $output = array();

foreach ($catalog as $id => $item) {
    if ($category == null OR strtolower($category) == strtolower($item["category"])) {
        $sort = $item["title"];
        $sort = ltrim($sort,"The ");
        $sort = ltrim($sort,"A ");
        $sort = ltrim($sort,"An ");
        $output[$id] = $sort;            
    }
}

asort($output);
return array_keys($output);

}

Siddhant Mehta
Siddhant Mehta
Courses Plus Student 478 Points

paste your code here

Siddhant Mehta
Siddhant Mehta
Courses Plus Student 478 Points

Hey Prince Henry, In your function single_item_array 

try { 
$results = $db->query( "SELECT 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" 
                                       ); 
}

Here in your query you are joining two tables with media table i.e. Genres or Books & media_id is common in Media & Books .So while selecting the feild media_id you need to denote the table name from where you are capturing media_id.

try { 
$results = $db->query( "SELECT Media.media_id/Genres.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" 
                                       ); 
}

1 Answer

Leonieke Paalvast
Leonieke Paalvast
11,905 Points
Leonieke Paalvast
Leonieke Paalvast
11,905 Points

Try replacing this code: try { $results = $db->query( "SELECT 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; }

With this block of code: 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 retrieved results"; exit; }

I had the same problem and the solution in my case was to replace SELECT media_id with SELECT Media.media_id.