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What about sha512/salt not sure how a salt would work

Am I able to make a secure rs4 so it is hard for tanji etc?

final public function hashed($password) { return md5($password); }

I think it should be like $rankInfo = mysql_fetch_array(mysql_query("SELECT * FROM `ranks` WHERE `ladder` = '".$i." AND ORDER ASC'")); I think its something like that I am still learning so

If I wanted to use this script <?php $data = new mysqli('localhost', 'root', '', ''); if ($data->connect_error) { die("Connection failed: " . $data->connect_error); } $username = $data->real_escape_string($_GET['username']); $exe = $data->query("SELECT look FROM users WHERE username =...

It had * because of the other things.. <div><img src="%www%/images/icon_habbo_small.png"> User ID: <b><?php echo $getLook['id']; ?></b></div> <div><img src="%www%/images/credits.png"> Credits: <b><?php echo $getLook['credits']; ?></b></div>...

<div><img src="%www%/images/terms_icon.gif"> Rank: <b><?php echo $rankg; ?></b></div> and $rankg = mysql_fetch_array(mysql_query("SELECT r.title FROM ranks r LEFT JOIN users u on u.rank = r.id WHERE u.id = '".$_SESSION['user']['id']."'")); $getstats = mysql_fetch_assoc(mysql_query("SELECT *...

Yea it worked

<div><img src="%www%/images/terms_icon.gif"> Rank: <b><?php echo $rankg; ?></b></div> And $rankg = mysql_fetch_array(mysql_query("SELECT r.title FROM ranks r LEFT JOIN users u on u.rank = r.id WHERE u.id = '".$_SESSION['user']['id']."'")); it isn't working It shows array

So like $rankg = mysql_fetch_array(mysql_query("SELECT '.$_SESSION['user']['id'].', ranks.name FROM users INNER JOIN ranks ON ranks.id = users.rank"));

How can i display a rank name instead of the id of the rank Like SELECT rank FROM users SELECT name from ranks where id or something like that?

You mind remaking the script than?

OMG thank you @Exalted Close thread please.@Mods Thank you. You two are the best.

what do you mean

I don't have the image locally I think @Joel got what I was trying to do I wanted to grab the look from the db and place it in like http://habbo.nlhabbo-imaging/avatarimage?figure={$look}

<img src="http://xotel.co/avatarimage.php?user=Kyle"> It put that

[26-Nov-2015 03:09:00 Europe/Belgrade] PHP Parse error: syntax error, unexpected '"' in C:\inetpub\wwwroot\avatarimage.php on line 11 @Joel @TesoMayn https://db.tt/RUpYKr1X it deosn't show the avatar...

Yes ?

Yes But I want it like url.com/avatar.php?user=WCKD or url.com/avatar.php?user=Joel and it will show their look you know what I mean?