Automatically choose images to display in while loop

Johny7126

I display thumbnail image previews for my website using a while loop. Each time the loop goes through 6 thumbnails it starts a new line - basicly giving columns and rows.

The images have naming conventions like this:

category_name_2nd category colour(not all have a 2nd a category).jpg e.g.

mountainbike_handlebar_black.jpg

In the loop I need to display images with and without a 2nd category i.e.

mountainbike_handlebar_black.jpg

mountainbike_handlebar.jpg

For the below code I tried to put a preg_match if statement around this (image names are in the url and so you can use a preg_match on a GET):

$files = "http://localhost/images/small/".$image_name[0].".jpg";

But no luck - this displays all images with a second category only.

So please suggest how I would tackle this and get all images displaying.

	....

echo "<table align='center' cellspacing='7px' cellpadding='7px'>";

$i = 1;

while ($f = mysql_fetch_array($filelist)) 
{	
	echo "<td>";

	$image_name_original = $f['filename'];
	$image_name = explode(".",$image_name_original);

	$files = "http://localhost/images/small/".$image_name[0].".jpg";						

	echo "<img src='$files' style='text-decoration: none; border: none;' alt='".$f['item_name']."' title='".$f['item_name']."' width='117' height='161' />";															
	echo "</td>	";	

	if($i%6 == 0)
	{
		echo "<tr>";
		echo "</tr>";

		$i=0;						
	}

	$i = $i+1; // add 1 for each iteration of while loop
}	

....

Thanks

leatherback

I am utterly confused..

Why the pregmatch? Why can't you just loop through the text file and show the images based on the names? Wouldn't that already show all the images? Also.. Why the explode-and-add of extention?

Johny7126

Your right about the explode-and-add extension - that is code I now dont need to use therefore I changed it:

$files = "http://localhost/images/products/small/".$image_name_original;

I have no textfile - the while loop goes through the database while there are still rows. Hope that explains that question.

I agree, its not easy to get your head around this one so feel free to ask any questions you want about it.

anoopmail

put a debugging line and lets see wat it shows

$files = "http://localhost/images/small/".$image_name[0].".jpg";                        

echo $files . '<br>' ; # debug

    echo "<img src='$files' style='text-decoration: none; border: none;'....

Johny7126

It shows all the image URLs for non black images i.e.

http://localhost/images/mountainbike_breaks.jpg

And so on.

Johny7126

I think that the code below might help you to understand the situation. I do actually need to explode $image_name as I need to put $colour before the .jpg to say that it is black - if it has a black image that is. This code does not work completely as it shows the item image and the black item image also for items like that e.g.

item_1 1 image in item cell
item_2 1 image in item cell
item_3, item_3 black 2 images in item cell
item_4 1 image in item cell

				
while ($f = mysql_fetch_array($filelist)) 
				{
...
					$image_name_original = $f['filename'];
					$image_name = explode(".",$image_name_original);

				// If itemis in the database and is in the GET and has a colour family then show that colour image
				if(preg_match('/icon print black/', $_GET['colourfamily']))
				{
					$colour= "_black";
				}
				else
				{
					$colour= "";
				}

				$files_2 = "http://localhost/images/products/small/".$image_name[0].$colour.".jpg";	
				$files = "http://localhost/images/products/small/".$image_name[0].".jpg";	

				echo "<img src='$files_2' style='text-decoration: none; border: none;' alt='".$f['item_name']."' title='".$f['item_name']."' width='117' height='161' />";															
				echo "<img src='$files' style='text-decoration: none; border: none;' alt='".$f['item_name']."' title='".$f['item_name']."' width='117' height='161' />";																				}