PHP / jQuery: How to get next row in SQL? - php

Question

Im trying to get this next/previous image thing to work. But i don't know how to get this right. I only get the id 88 and 87, not the ones before that.
www.wallpapers.puffys.net
In my index.php page i got this code to show the image and get the next/previous image.
include 'db_connect.php';
#Last
$sql = "SELECT id FROM images WHERE validated=1 ORDER BY id DESC LIMIT 0,1";
$result = mysql_query($sql) or trigger_error(mysql_error()."".$sql);
$row = mysql_fetch_array($result);
$last = $row['id'];
#Next
$sql = "SELECT id FROM images WHERE validated=1 AND id>'".$last."' ORDER BY id DESC LIMIT 0,1";
$result = mysql_query($sql) or trigger_error(mysql_error()."".$sql);
$row = mysql_fetch_array($result);
$next = $row['id'];
#Previous
$sql = "SELECT id FROM images WHERE validated=1 AND id<'".$last."' ORDER BY id DESC LIMIT 0,1";
$result = mysql_query($sql) or trigger_error(mysql_error()."".$sql);
$row = mysql_fetch_array($result);
$previous = $row['id'];
?>
<!--Scripts-->
<script type="text/javascript" src="resources/jquery-1.7.1.js"></script>
<script>
$(document).ready(function() {
function showLoader() {
$('.loader').fadeIn(200);
}
function hideLoader() {
$('.loader').fadeOut(200);
}
$(".next").click(function() {
showLoader();
$("#wallpaper").load("get.php?id=<? echo $next; ?>", hideLoader);
return false;
});
$(".previous").click(function() {
showLoader();
$("#wallpaper").load("get.php?id=<? echo $previous; ?>", hideLoader);
return false;
});
showLoader();
$("#wallpaper").load("get.php?id=<? echo $last; ?>", hideLoader);
});
</script>
The get.php page:
include 'db_connect.php';
$id = mysql_real_escape_string($_GET['id']);
if ($id) {
$sql = "SELECT COUNT(*) FROM images WHERE id='".$id."' AND validated=1";
$result = mysql_query($sql) or trigger_error(mysql_error()."".$sql);
$num_rows = mysql_fetch_row($result);
if ($num_rows[0] == 1) {
$sql = "SELECT image FROM images WHERE id='".$id."' AND validated=1";
$result = mysql_query($sql) or trigger_error(mysql_error()."".$sql);
$row = mysql_fetch_array($result);
echo '
<img src="'.$row['image'].'" width="1024" height="576" alt="wallpaper" border="0">
';
}
else {
}
}
else {
}
?>
I know i am doing this wrong, but i don't know how to get this working. So i would hope someone could help me out.

Answer 1

Your problem is most likely this;
#Next
$sql = "SELECT id FROM images WHERE validated=1 AND id>'".$last.
"' ORDER BY id DESC LIMIT 0,1";
You get all IDs greater than the last, and order from highest to lowest which will give you the highest id available. You'll want to change it around to;
#Next
$sql = "SELECT id FROM images WHERE validated=1 AND id>'".$last.
"' ORDER BY id ASC LIMIT 0,1";
which will sort in the correct order and give you the lowest ID that is higher than the last.

Answer 2

This is my solution:
$NEW_ID = 0;
$SQL_FUNCTION = "SELECT ID FROM TableName";
$SQL_DATA = MYSQL_QUERY($SQL_FUNCTION);
WHILE($SQLDAT = MYSQL_FETCH_OBJECT($SQL_DATA)){
$ID = "$SQLDAT->ID";
if($NEW_ID < $ID){$NEW_ID = $ID;}
}
//NOW INSERT WITH THIS ID...
But I'm not sure if it works when more people simultaneously do this!?
(I would ask this question but i'm closed)