Display image from server in a div (newest first) PHP - php

Question

I want to display images in from server. In the database you can find the description and name of the image. the images should be displayed in a div where always the newest is on top. As the number of images increase so does the number of div's. so, the oldest will be at the bottom. I have read a number of posts and forums but I was unable to get the logic and the code on how to do it. I hope you can help me with my problem. Thanks a lot for your answer.
the only code i have is the formatting of the dive's:
<div class="gallery">
<div class= "group">
<div class="images">
</div>
<div class="details">
</div>
</div>
<div class= "group">
<div class="images">
</div>
<div class="details">
</div>
</div>
<div class= "group">
<div class="images">
</div>
<div class="details">
</div>
</div>
</div>
it's something like this:
newest image | some details here
newer image | some datails here
new image | some details here
old image | some details here
EDIT 1
I combined Lauri Elias' and iamde_coder's answer. I come up with this code which works almost similar to what is wanted. the only problem is that it displays the item (image&details in a div) 4 times. How can I eliminate the three? thanks!
$image_query = mysql_query ("SELECT filename, story FROM tbl_contest ORDER BY time DESC");
while($image_data = mysql_fetch_array($image_query)){
$imageName = stripslashes(mysql_real_escape_string($image_data['filename']));
$imageDetails = stripslashes(mysql_real_escape_string($image_data['story']));
$count = 0;
foreach($image_data as $imageName) {
echo '<div class="group">';
echo '<div class="images"><img src="/Mainfolder/image_entry/'.$imageName.'"></img></div>';
echo '<div class="details">'.$imageDetails.'</div></div>';
$count ++;
}
}

Answer 1

Now that you've given a little more code to help try using this:
$count = 0;
$image_query = mysql_query ("SELECT filename, story FROM tbl_contest ORDER BY time DESC") or die(mysql_error());
while($image_data = mysql_fetch_array($image_query)){
$imageName = stripslashes(mysql_real_escape_string($image_data['filename']));
$imageDetails = stripslashes(mysql_real_escape_string($image_data['story']));
$count++;
echo '<div class="group">';
echo '<div class="images"><img src="/Mainfolder/image_entry/'.$imageName.'" alt="Image '.$count.'" /></div>';
echo '<div class="details">'.$imageDetails.'</div></div>';
}

Answer 2

You could, for example, use a MySQL select somewhat like this:
SELECT name, description FROM my_database.images ORDER BY created_at DESC;
Then bind the result set to a variable in PHP like $images and then iterate on it and generate HTML like this:
foreach($images as $image) {
echo '<div class="group">';
echo '<div class="images"><img src="/images_folder/'.$image['name'].'"></img></div>';
echo '<div class="details">'.$image['description'].'</div></div>';
}