Display multiple images from DB on each table row

Johny7126

I have to get multiple images (6 images) on rows and columns on my website.

Basicly I have a while loop that gets the blob stored image from the MySQL DB - at the moment the loop only displays the images in one row.

I need it to go to the next row after six images are displayed.

Would I need an array for this as I do not see how it can be done with loops.

echo "<div id='products_div'>";
				echo "<table>";

			while ($f = mysql_fetch_array($filelist)) 
			{														
				echo "<td>";	
				echo "<div id='product_background'>";	
				echo "<a href='products.php?id=".$f['id']."'><img style='text-decoration: none; border: 0px; padding-left: 8px; padding-top: 8px;' src='image.php?id=".$f['id']."' alt='Error - no image in DB' width='98' height='98'>";
				echo "<div id='product_link'>";
				echo $f['description'];
				echo "</div>";	
				echo "</a>";	
				echo "</div>";		
				echo "</td>	";													
			}	

			echo "</table>";
			echo "</div>";

Cheers anyone

leatherback

Use the % operator:

$i = 1;
while(...)
  {
echo '';
if($i%6 = 0)
  {
  echo "</tr><tr>";
  $i=0;
  }
$i = $i+1;

Johny7126

echo "<div id='products_div'>";
				echo "<div id='products_selection'>";		
				echo "</div>";

			echo "<table>";
			echo "<tr>";

			$i = 1;

			while ($f = mysql_fetch_array($filelist)) 
			{	
				if($i%6 = 0)
				{
					echo "</tr><tr>";
					echo "<td>";	
					echo "<div id='product_background'>";	
					echo "<a href='products.php?id=".$f['id']."'><img style='text-decoration: none; border: 0px; padding-left: 8px; padding-top: 8px;' src='image.php?id=".$f['id']."' alt='Error - no image in DB' width='98' height='98'>";
					echo "<div id='product_link'>";
					echo $f['description'];
					echo "</div>";	
					echo "</a>";	
					echo "</div>";		
					echo "</td>	";	

					$i=0;						
				}

				$i = $i+1;
			}	
			echo "</tr>";
			echo "</table>";
			echo "</div>";

Gives this error:

Parse error: syntax error, unexpected '=' in C:\wamp\www\OGIO\includes\products.php on line 143
which is the:

if($i%6 = 0)

Cheers for reply

Johny7126

oops == it should be

Johny7126

Hmm only getting one image on each line but according to the database they are the right images i.e. thats where they would be if it was working.

Any ideas please?

leatherback

How did yu implement it?

I see in the code you pasted above that you inserted all your code in the if() statement I posted. The idea is that if you have had a certain number of records (In the posted situation 6) that the code outputs a </tr><tr> so that you get a new line. So you insert it in the output routine. Not the otherway around.

Johny7126

Of course 🙂
Cheers mate