PHP - Displaying Images Horizontally With Php

I'm realitivly new to PHP and was hoping somebody could help.

I have a mysql database that stores information about books. I am currently using the code below to query the database and extract the 3 most recent entries and showing them in a dynamic list:

Code: [Select]
$sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 3");
$productCount = mysql_num_rows($sql);
if ($productCount > 0) { // ensure a book exists
while($row = mysql_fetch_array($sql)){
             $id = $row["id"];
$title = $row["title"];
$author = $row["author"];
$price = $row["price"];
$date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));
$dynamiclist .= //My table showing the products
}
} else {
$dynamicList = "There are currently no Books listed in this store";
}
This works well when showing the most recent books 1 below the other. However, I would like to show these products side by side, horizontally across the page.

Can somebody please point me in the right direction?

Many thanks

So I am trying to display an image from a database (I know it's not the best idea but it's what I was told to do). Anyway I'm having problems displaying it. Here is the code for displaying the image.
$cat = $_GET['cat'];

include 'includes/openDbConn.php';

$sql = 'SELECT * FROM CushionsCategories WHERE CushionCategory = "'.$cat.'"';

echo $sql;

$result = mysql_query($sql); 

echo '<table>';

while($val = mysql_fetch_assoc($result)){

$cush = 'SELECT * FROM Cushion WHERE SKU = "'.$val['CushionSKU'].'"';

echo $cush;

$cushres = mysql_query($cush);

$cushval = mysql_fetch_assoc;

$img = 'SELECT * FROM Images WHERE SKU = "'.$val['CushionSKU'].'"';

echo $img;

$imgres = mysql_query($img);

$imgval = mysql_fetch_assoc($imgres);

if( $i % 3 == 0 ) {

      echo '</tr><tr>';
  

}

echo '<td><img src="image.php?sku='.$val['CushionSKU'].'" name="'.$imgval['FileName'].'" description="'.$imgval['Description'].'" /></td>';

}

echo '</tr></table>';

and the page that gets the image:
$sql = 'SELECT Image FROM Images WHERE SKU = "'.$sku.'"';
//echo $sql;
 $result = mysql_query($sql) or die("Invalid query: " . mysql_error());
 
        // set the header for the image
        header("Content-type: image/jpeg");
        echo mysql_result($result, 0);

Thanks for any help in advance.

I had this page working fine where it retrieves a set of images from the database depending on what thumbnail was click on the previous page. Now I am trying to add an edit and delete button to each of the pictures and it will not display any of it now!

This is the code I am using:
Code: [Select]
<?php
$id = $_GET[id];
?>


<?php
$max_items = 16; /* max number of news items to show */

require_once('config.php');
$db = mysql_connect (DB_HOST,DB_USER,DB_PASSWORD);
if(!$db) {
die('Failed to connect to server: ' . mysql_error());
}
mysql_select_db (DB_DATABASE,$db);

function displayNews($all = 0) {
    global $db, $max_items;
    
    if ($all2 == 0) {
        /* query for news up to $max_items */
        $query13 = "SELECT image,id,description,projectcode,thumbnail " . 
                 
                 "FROM project WHERE projectcode='$id' ORDER BY id DESC LIMIT $max_items";
    } else {
        /* query for all news */
        $query13 = "SELECT image,id,description,projectcode,thumbnail " . 
                 
                 "FROM project WHERE projectcode='$id' ORDER BY id DESC";
    }
    $result13 = mysql_query($query13) or print("<p>Error fetching entries from the database, error: " . "Statement: " . $query13 . "</p>" . mysql_error());


    while ($row13 = mysql_fetch_array($result13)) {
        
        $title = htmlentities ($row13['description']);
        $news = nl2br (strip_tags ($row13['projectcode'], '<a><b><i><u>'));
$image = $row13['image'];
$thumbnail = $row13['thumbnail'];
$id = $row13['id'];
        
        /* display the data */

        echo "<div class='portfolioproject'>";?>

        <div class='boxgrid captionfull'>
<a id="<?php echo $id ?>" class="example_group" href="images/gallery/<?php echo $image ?>" title="<?php echo $title ?>" rel="gallery"><img alt="" src="images/gallery/<?php echo $thumbnail ?>" width="122px" height="122px"</a>
               
          </a>}
               
        <?php

//Check whether the session variable SESS_MEMBER_ID is present or not

if(isset($_SESSION['SESS_MEMBER_ID']) || (!trim($_SESSION['SESS_MEMBER_ID']) == '')) {
echo "<form class='editbtn' action='editportfolio.php' method='POST'>";
echo "<input type='hidden' name='idf' value='$id' />";
echo "<input src='images/editbtn.png' type='image' value='Edit' />";
echo "</form>";
echo "<form class='editbtn' action='deleteportfolio.php' method='POST'>";
echo "<input type='hidden' name='ide' value='$id' />";
echo "<input src='images/delbtn.png' type='image' value='Delete' />";
echo "</form>";
} else {
echo "";
}


echo "</div>";

    }
    
    /* if not displaying all news, give a link to do so */
    if ($all == 0) {
        echo "<div class='show_all'><a class='show' href=\"{$_SERVER['PHP_SELF']}" .
             "?action=all\">Show all</a></div>\n";
    }
}

/* this is where the script decides what do do */

echo "\n";
switch($_GET['action']) {

    case 'all':
        displayNews(1);
        break;
    default:
        displayNews();
}
echo "\n";
?>

<?php
//Check whether the session variable SESS_MEMBER_ID is present or not
if(isset($_SESSION['SESS_MEMBER_ID']) || (!trim($_SESSION['SESS_MEMBER_ID']) == '')) {
echo "<span class='show_all'>";
echo "<a class='show' href='admincp.php'>&nbsp;&nbsp;Admin</a>";
echo "</span>";
echo "<span class='show_all'>";
echo "<a class='show' href='logout.php'>&nbsp;Logout</a>";
echo "</span>";
} else {
echo "";
}
?>

No error messages are displayed, only the 'show all' button, and the 'admin' and 'logout' buttons if signed in!

Any help on this would be much appreciated!

Thank you

Martin

Hello I am using this script right here
http://www.nearby.org.uk/sphinx/search-example5-withcomments.phps
and am trying to show images from my db but I can only get it to show the first image. Does anyone know how to make it show all images if images even exist for that particular search result?

I changed the query to $CONF['mysql_query'] = '
SELECT l.link_id AS id, l.title AS title, l.fulltxt AS body, l.url AS url, m.media_id AS im_id, m.title AS im_title, m.thumb_link AS im_t_link
FROM search1_links AS l LEFT JOIN search1_media AS m ON (m.link_id = l.link_id)
WHERE l.link_id IN ($ids)
';

and added

if ($row['im_id']) {
echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> ';
}
right here

            //Actully display the Results
            print "<ol class=\"results\" start=\"".($currentOffset+1)."\">";
            foreach ($ids as $c => $id) {
                $row = $rows[$id];
                
                $link = htmlentities(str_replace('$id',$row['id'],$CONF['link_format']));
                print "<li><a href=\"$link\">".htmlentities($row['title'])."</a><br/>";
                
                if ($CONF['body'] == 'excerpt' && !empty($reply[$c]))
                    print ($reply[$c])."</li>";
                else

           if ($row['im_id']) {
echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> ';
}

                    print htmlentities($row['body'])."</li>";
            }
            print "</ol>";
            
            if ($numberOfPages > 1) {
                print "<p class='pages'>Page $currentPage of $numberOfPages. ";
                printf("Result %d..%d of %d. ",($currentOffset)+1,min(($currentOffset)+$CONF['page_size'],$resultCount),$resultCount);
                print pagesString($currentPage,$numberOfPages)."</p>";
            }

Can anyone help me to be able to display all images if they exist? Thanks.

Hi!

I want the image of my site to be clickable so that it shows the next image contained in the result set of the sql query.

Code: [Select]
<?php
require_once('includes/connection.inc.php');
$conn = dbConnect();
$sql = 'SELECT filename FROM images ORDER BY image_id DESC LIMIT 1';
$msg = '';
if(!$sql) { echo "Something went wrong with the query. Please try again."; }
$result = $conn->query($sql) or die(mysqli_error());
if(mysqli_num_rows($result) == 0) {
header('Location: empty_blog.php');
$conn->close();
} else {
$row = $result->fetch_row();
$image = $row[0];
}
?>
<?php include('header.php'); ?>
<div class="content main">
<img id="image" src="images/<?php echo $image; ?>"/>
</div>
<?php include('includes/footer.inc.php'); ?>

All the filenames of the images are contained correctly in the $result variable, all I need to do now is to fetch the next image in the set. How would I go about this? Any help is greatly appreciated!