PHP/MySQL image placement

rakennedy75

Here is what I am trying to do. I am fairly new at PHP so forgive the ignorance.

I am pulling images from a folder on my server through a path stored in a database I built. In the database it has the the image path. This works great but I would like to place the images in a grid format. 3 across and 2 down. This is so they fit a particualar area on my page.

Here is my code:

for ($i=0; $i<=6; $i++)
{
while ($row = mysql_fetch_array($result)){

         if ( $i == 1 || $i == 4)
        		{
       		 		  echo "<tr><td>";
              echo "<img src='".$row['imagePath']."' ></td>";
				      }else if ($i == 2 || $i == 5){
					            echo "<td>";
              echo "<img src='".$row['imagePath']."'></td>";
				         } else if ($i == 3 || $i == 6){
					           echo "<td>
             echo "<img src='".$row['imagePath']."'></td>
             echo "</tr>";
			    }end of if statements

} // end of while loop

} end of for loop

I left out the query and other code to keep this simple

this is not working
Is this the right way to go about it, if not, what can I do to correct it

Thanks for any help!!!

leatherback

Maybe shorter (I am very bad at incrementing.. so the i and i++ ight wrong, but the idea should work..):


for ($i=0; $i<6; $i++) 

{ 

while ($row = mysql_fetch_array($result)){ 
{
$body.="<td><img src='".$row['imagePath']."' ></td>"; 
if ($i=2)
{
$body .= "</tr><tr>";
}
} // end of while loop 

} end of for loop 

?>

rakennedy75

First of all, thank you for taking your time to answer my question. I played with this and maybe didn't approach it the best way. I am still have problems with this.

Here is my revised code:

for ($i=1; $i<=6; ++$i) {
while ($row = mysql_fetch_array($result)){

		             		if ( $i == 1 || $i == 4){
       		 	    	$body .= "<tr>";	
       		 	   }

				             $body .= "<td><img src='".$row['imagePath']."'>";
              $body .= "<br>test ".$i."</td>";

			              	if ($i == 3 || $i == 6){
					                 $body .= "</tr>";
	               }
			   }				
		}

echo $body;

I think the problem comes in is where when it hits the while loop it loops all of the results and than goes back to the for loop. The reason I say this is that I put the verbage:

test."$i"

and this echos out one and dumps each result on the a new row as if.

Is there a way I could combine to loops in one where it will loop for each specific result.

Or is there a better way to approach this.

Thanks again for your time.

Tim_Brown

Try this:

$body .= "<table>";
$i=1;
while ($row = mysql_fetch_array($result)){
switch ($i % 3) {
case 0: // last image in row
$body .= "<td><img src='".$row['imagePath']."'>";
$body .= "<br>test ".$i."</td>";
$body .= "</tr>";
break;
case 1: // first image in row
$body .= "<tr>";
$body .= "<td><img src='".$row['imagePath']."'>";
$body .= "<br>test ".$i."</td>";
break;
case 2: // middle image in row
$body .= "<td><img src='".$row['imagePath']."'>";
$body .= "<br>test ".$i."</td>";
break;
}
$i++
}
$body .= "</table>";
echo $body;

Tim_Brown

Looks like I left of a semicolon. Should be:
i++;

rakennedy75

Thanks a lot, this seems to work perfectly. I appreciate the help. never thought of doing it like that, obviously.

Thanks again for your time.