Error message instead of thumbnails

ga86

Hi I am trying to get image thumbnails to display on the page. The page displays one large image at the top (code not included since it works fine), and below it are supposed to be 5 smaller thumbnails of images the visitor may also be interested in viewing.

Below is the code for displaying the thumbnails, but i get this error message:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /public_html/view_photo.php on line 50

Line 50 is shown through the comment.

if(isset($_GET['imageid'])) {

$imageid = (int) $_GET['imageid'];

echo "<h3><font color=#C12267>You may also be interested in these images:</font></h3>";
$query2 = "SELECT right FROM rules WHERE left='$imageid' LIMIT 5";
$result2=mysql_query($query2);
while($row2=mysql_fetch_assoc($result2)) { //THIS IS LINE 50

if (!file_exists('thumbnails/'.$row2['imagepath'])) {
    /* Thumbnail generation */
    $im = @imagecreatefromjpeg($row2['imagepath']);
    if($im) {
        $width = imagesx($im);
        $height = imagesy($im);
        $scale = min( 150 / $width, 150 / $height);
        $newwidth = $width * $scale;
        $newheight =$height * $scale;
        $im2 = imagecreatetruecolor($newwidth, $newheight);

        imagecopyresampled($im2, $im, 0, 0, 0, 0, $newwidth, $newheight, $width, $height);

        imagejpeg($im2, "thumbnails/" . $row2['imagepath']);
    }
}

echo '<a href="view_photo.php?imageid='.$row2['imageid'].'">';
echo '<img src="thumbnails/'.$row2['imagepath'].'"/></a><br/><br />';
} 
}

Please can someone help me? Thank you in advance

laserlight

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /public_html/view_photo.php on line 50

The above warning is due to one or more of these factors:

Failed to connect to database server.
Failed to select database.
Failed to execute corresponding SQL statement.

ga86

laserlight wrote:
The above warning is due to one or more of these factors:

Failed to connect to database server.

Failed to select database.

Failed to execute corresponding SQL statement.

Thanks for your reply. Im quite new to php, and i am not sure how can i check if it is connecting or not. Or if it is the select query not working, how do i tell?

The large image on the page displays so i'm assuming it is connecting to the database.

Thank you.

laserlight

Im quite new to php, and i am not sure how can i check if it is connecting or not. Or if it is the select query not working, how do i tell?

Go to the portion of code where you connect to the database server and select the database and change it to something like this:

mysql_connect(...) or die('Could not connect to database server: ' . mysql_error());
mysql_select_db(...) or die('Could not select database: ' . mysql_error());

Also, change:

$result2=mysql_query($query2);

to:

$result2=mysql_query($query2) or die(mysql_error());

This will help you determine where is the problem, but of course for a production script you should have more graceful error handling.

ga86

laserlight wrote:
Go to the portion of code where you connect to the database server and select the database and change it to something like this:
mysql_connect(...) or die('Could not connect to database server: ' . mysql_error());
mysql_select_db(...) or die('Could not select database: ' . mysql_error());
Also, change:
$result2=mysql_query($query2);
to:
$result2=mysql_query($query2) or die(mysql_error());
This will help you determine where is the problem, but of course for a production script you should have more graceful error handling.

It now says:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM rules WHERE left='64' LIMIT 5' at line 1

Any ideas what the problem may be?

Thank you very much.

laserlight

left is probably recognised as a key word (as in LEFT JOIN), so you need to escape it. In the MySQL dialect of SQL, backticks () are used to escape identifiers, so you would write: [code=php]$query2 = "SELECT right FROM rules WHEREleft`='$imageid' LIMIT 5";[/code]

ga86

laserlight wrote:
left is probably recognised as a key word (as in LEFT JOIN), so you need to escape it. In the MySQL dialect of SQL, backticks () are used to escape identifiers, so you would write: [code=php]$query2 = "SELECT right FROM rules WHEREleft`='$imageid' LIMIT 5";[/code]

I tried that but it still says:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM rules WHERE 'left'=64 LIMIT 5' at line 1.

Could it be the query? Maybe i didnt write it correctly..?

bradgrafelman

The error is actually just before that bit - RIGHT is also a reserved word, so you need to escape it as well.