Coding Help

[erikhillis]

Hey guys

What I'm trying to do is spit out an image gallery with this code... it dosn't seem to work though. If someone could go through it and fix my mistakes that would be great.

As you can see what I'm trying to do is select a certain ammount of images / page, and display them 4 at a time before skipping to the next <tr>.

<?php 

$db = mysql_connect("localhost","un","pw");
mysql_select_db (u);

$a = '0';

if(!isset($_GET['page'])){ 
    $page = 1; 
} else { 
    $page = $_GET['page']; 
} 

$max_results = 25; 

$from = (($page * $max_results) - $max_results); 

$sql = mysql_query("SELECT * FROM screenshots order by id desc LIMIT $from, $max_results"); 

echo "<center><table border=\"0\" cellpadding=\"5\" cellspacing=\"0\" width=\"98%\">";

while($row = mysql_fetch_array($sql)) {  

$id = $row[id];

if ($a == '0') {echo "<tr>";}
if ($a == '4') {echo "<tr>";}
if ($a == '8') {echo "<tr>";}
if ($a == '12') {echo "<tr>";}
if ($a == '16') {echo "<tr>";}
if ($a == '20') {echo "<tr>";}

echo "<a href=\"screenshot.php?image=".$row["url"]."\"><img src=\"".$row["url"]."\" alt=\"XWG Image Gallery\" width=\"115\" height=\"87\" border=\"0\"></a>";

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

}

$total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM screenshots"),0); 

$total_pages = ceil($total_results / $max_results); 

echo "<font face=Verdana size=1 color=818562><br><B>Select a Page</B><br /><font face=Verdana size=1 color=FFFFFF>"; 

if($page > 1){ 
    $prev = ($page - 1); 
    echo "<a href=\"screenshots.php?page=$prev\"><font face=Verdana size=1 color=818562><B>«</B><Previous</a>&nbsp;"; 
} 

for($i = 1; $i <= $total_pages; $i++){
    if(($page) == $i){ 
        echo "$i&nbsp;"; 
        } else { 
            echo "<a href=\"screenshots.php?page=$i\">$i</a>&nbsp;"; 
    } 
} 

if($page < $total_pages){ 
    $next = ($page + 1); 
    echo "<a href=\"archive.php?page=$next\">Next&nbsp;<font face=Verdana size=1 color=818562><B>»</B></a>"; 
} 
echo ""; 
?>

<?
$a = $a + 1;
}
?>

[tsinka]

Hi,

specify a variable outside the while loop like

$pics_per_row=4;

$max_results should be something linke 24 or 28 instead of 25 if you want to print 4 pics a line.

inside the while loop at the beginning:

<?PHP
  if ($a%$pics_per_row == 0)
    echo ($a!=0 ? "</tr>" : "")."<tr>";
?>

Don't forget to put <td> and </td> around the link

Put the
echo "</tr></table>";

outside the while loop and maybe replace it by:

while ($a%$pics_per_row != 0) {
  echo "<td>&nbsp;</td>";
  $a++;
}

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

to make sure that the current row gets finished.

Put the $a = $a +1 line inside the while loop (at the end of the loop and maybe replace it by $a++).

Thomas

[erikhillis]

Can you please plug that into my current code (make the appropriate changes) because I can't seem to get it to work properly:

http://jops.xwargames.net/screenshots.php