Cloud9247

Please disregard this post... I just realized this output is related to an AVG update I have on my browser... Unable to delete this post though.

Hello! I hope someone can help me with an issue I'm having. Recently, I have had an issue displaying my banner on my website, which is has a .php file source. I have had any known changes to my php version (currently on 5.6, and considering to switch to 7.0); also, I am hosted through Dreamhost. If someone could look at the following code and offer any advice, I'd be greatly appreciative! My link on the design: <img src= "http://hayalkarga.com/theme/10/images/banner/banner.php" alt= "" /> The Image Code: <?php $dir = ''; $file = ''; $num = 0; $count = 0; $array = array(); $allowedExtensions = array('jpeg', 'jpe', 'jpg', 'gif', 'png', 'bmp'); mt_srand((double)microtime()*1000000); $dir = opendir('.'); while (false !== ($file = readdir($dir))) { if(!is_dir($file) && in_array(strtolower(substr(strrchr($file,'.'), 1)), $allowedExtensions)){ $array[] = $file; } } $count = count($array); $num = mt_rand(0, $count-1); if($count > 0) { $size = getimagesize($array[$num]); $stream = fopen($array[$num], 'rb'); if(is_array($size) && $stream !== false){ header('Content-type: '.$size['mime']); fpassthru($stream); exit(); } } else { echo 'Error: No Images'; } ?> I should note that the image works in my browser at http://www.hayalkarga.com/theme/10/images/banner/banner.php, but does not incorporate into the design when utilizing the link in HTML. Thank you in advance!

Can someone help me, I need to make a drop down redirection menu with the <select> and <option> tags, please help. Thanks

I u7sed that, but it still is repeating the same series twice, why? this is my query: [code] $sqll= "SELECT DISTINCT id, active, series FROM graphics WHERE active='1' ORDER BY `series` ASC"; $rss= mysql_query( $sqll, $connection ); while($row= mysql_fetch_array( $rss )) { [/code]

currrently I am retrieving data from a database named 'graphics', I want it to list all the series in ther database, but I don't want it to repeat the same series twice. How do I make it so a series' name doesn't show more than once?

'active' is in ALL of the tables And, I am trying to pull the series name from each table, but I want to make it where the series name will not repeat. I also want the series' to sort with eachother, so there won't be a boundry between the tables. EX: I want it to go like: Series 1 Series 2 Series 3 NOT: Series 1 Series 2 Series 1 Series 2 Series 1 Series 2 Series 1 Series 2 ------------- I still don't understand though, could someone please help me with this specific line?: [code] $sql= "SELECT avatars.id, series, active, wallpapers.id, series, active, pngs.id, series, active, layouts.id, series, active FROM avatars, wallpapers, pngs, layouts WHERE  active='1' ORDER BY `series` ASC"; [/code]

I'm sorry, I don't understand what you are saying exactly, could you show me in code?

this is my query: [code] $sql= "SELECT avatars.id, series, active, wallpapers.id, series, active, pngs.id, series, active, layouts.id, series, active FROM avatars, wallpapers, pngs, layouts WHERE  active='1' ORDER BY `series` ASC"; [/code] But it doesn't seem to work, what is wrong with it?

I am trying to make a 'Sort By Series' page for my site, but in order to do that, I have to call from 4 different MySQL tables with one query, at least to make it simple xD Anyways, does anyone know how I could access multiple(4) DB tables with one query?

Actually, I just figured out to replace it with $i= -1, which works fine, thanks for your help everyone!

[code] <?php echo $top1.'Avatars'.$top2.$body1; echo '<table id= "avatars"><tr>'; $i= 0;         $sqll= "SELECT userid, id, clicks, username, submitted, name, credits, comments, series, imgurl, active FROM avatars WHERE active='1' ORDER BY `id` DESC LIMIT $from, $max_results"; // LIMIT $from, $max_results $rss= mysql_query( $sqll, $connection ); while($row= mysql_fetch_array( $rss )) { $dwn= "return false;"; echo '<td id= "avi"><a href= "http://hayalkarga.com/designs.php?viewavatar='.$row['id'].'"><img src= "'.$row['imgurl'].'" id= "graphic" alt= "'.$row['name'].'" oncontextmenu="return false"></a></td>';   if ($i == 3) {       echo "</tr><tr>";       $i = 0;   } $i++; } echo '</tr></table>'; echo $body2.$foot1.$foot2; [/code] comes out as: http://hayalkarga.com/designs.php?designs=avatars It prints out 4 on the first row, but only 3 on the others. Can someone help me fix this?

no, I haven't. But I don't really know how to implement that either.

I have been developing my graphics site lately, and I have run into a problem. Currently I am setting up a 100x100 avatars page, and I am using mysql_fetch_array. I want it to have in each row of avatars, 4 avatars per row. and I am currently using tables for this, so can someone help me get it to where every 4 rows it inserts a '<tr>' and '</tr>', so I could get more rows?

I am wanting the profiles of my members to be viewed through [a href=\"http://the-knighthood.net/~Membername~\" target=\"_blank\"]http://the-knighthood.net/~Membername~[/a], and want it done automatically through PHP, so I don't have to link through a LoNg hyperlink(http://the-knighthood.net/member.php?member=view&profile=4 (too long)). I don't know how to pull this off though, can someone please help? Thank you, Cloud9247

Can someone please help me then?