No results appearing when fetching array - PHP / MYSQL

Question

i have this problem where my query wont pull the data from the database into the website, i've tried many things but i either get an offset error or nothing appears at all. Could someone have a look please?

website with dropdown -

$dbc = @mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect to MySQL: ' . mysqli_connect_error() );
//echo "Connected to MySQL<br>";
$q = "SELECT artistname FROM tbl_artists ORDER BY artistname ASC";      
$r = @mysqli_query ($dbc, $q);

    // Form start
echo '<form action="php/process_dropdown.php" method="post">

<select name="selection">
<option value=""> -- Please select an Artist -- </option>';


// Fetch and print all the records:
    while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
        echo '<option>' . $row['artistname'] . '</option>
        ';
    }

    echo '<p><input type="submit" value="Send"></p>

    </select>

</form>'; // End of form.

?>

The dropdown process

$dbc = @mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect to MySQL: ' . mysqli_connect_error() );

$name = $_POST['selection'];


$q = "SELECT tbl_songs.songtitle, tbl_songs.yearrelease, tbl_artists.genre, tbl_songs.price
FROM tbl_artists INNER JOIN tbl_songs ON tbl_artists.artistID = tbl_songs.artistID
WHERE (((tbl_artists.artistID)='$name'))";

$r = @mysqli_query ($dbc, $q) or die("Error: ".mysqli_error($dbc));

    // Table header.
    echo '<table>
    <tr>
    <th scope="col">Song</th>
    <th scope="col">Year Released</th>
    <th scope="col">Genre</th>
    <th scope="col">Price</th>
    </tr>
';

    // Fetch and print all the records:
    while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
        echo '<tr><td>' . $row['songtitle'] . '</td><td>' . $row['yearrelease'] . '</td><td>' . $row['genre'] . '</td><td>' . $row['price'] . '</td></tr>   ';
    }

echo '</table>'; // Close the table.

?>

Thank you.

Do you get any error report?

Herman Nz
– Herman Nz

2015-02-21 02:52:02 +00:00
Commented Feb 21, 2015 at 2:52 — Herman Nz
– Herman Nz, Commented Feb 21, 2015 at 2:52

Herman Nz · Accepted Answer · 2015-02-21 03:06:32Z

1

EDIT:

$link = mysqli_connect('host', 'user', 'password', 'database');

Change into:

$dbc = mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die 
('Could not connect to MySQL: ' . mysqli_connect_error() );
//echo "Connected to MySQL<br>";
$q = "SELECT artistname FROM tbl_artists ORDER BY artistname ASC";      
$r = mysqli_query ($dbc, $q);
..................

And see this:

WHERE (((tbl_artists.artistID)='$name'))";

You missed: tbl_songs. before $name.

Besides, It's not good to use @ since it hides the error message. (It's a poor scripting)

edited Feb 21, 2015 at 3:06

answered Feb 21, 2015 at 1:58

Herman Nz

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2 Comments

Mo Martin Over a year ago

I don't really know what you changed, but something has fixed it. Also, adding the tbl_songs before $name doesn't make a difference. It works without it, but won't work with it. Also, i realised that i wrote WHERE (((tbl_artists.artistID)='$name'))"; rather than WHERE (((tbl_artists.artistname)='$name'))"; which meant that i defined the wrong field

Herman Nz Over a year ago

that's what makes me confused too. you need to find artistID but grabbing it from name. now you're working with Inner Join. there are three types of it, one of them is here > tutorialspoint.com/sql/sql-inner-joins.htm

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