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Solved it. Changed the html to have style="visibility:hidden;" inside the divs instead of css.

When you hover over the images I want the description of the certain image to pop up, and go away when your mouse leaves the image. Simple right? When I do it, I get the REVERSE effect, when my mouse goes over them, they go away, when my mouse leaves them, they show up. Here's my java script code // JavaScript Document function showhide(id) { var obj = document.getElementById(id); if(obj.style.visibility == "hidden") { document.getElementById(id).style.visibility = "visible"; } else { obj.style.visibility = "hidden"; } } my html code <img src="/Identity Studio/images/ad1afb85_1340_133d.gif" onMouseOver = "showhide('summary2');" onMouseOut = "showhide('summary2');"/><div id="summary2" class="description">aefdwasd</div><img src="/Identity Studio/images/d8b7cb84_3891_48ec.jpg" onMouseOver = "showhide('summary1');" onMouseOut = "showhide('summary1');"/><div id="summary1" class="description">This is totally a description</div> Css code (not really needed): .description { display:inline; visibility:hidden; border: 1px #CCCCCC solid; background:#FFFFFF; padding: 15px; position:absolute; color: #666666; font-family:Arial, Helvetica, sans-serif; } If you want a link to see for yourself what's happening go here: http://svidler.net/Identity%20Studio/ Anyone know what's going on? I can't believe I can't get something this simple to work.

The variable "$fname" wasn't assigned to anything, but you use it in "if($fname) { . . . ". What's that line with "if($fname) { . . . " supposed to do? Check if the form is filled out? If you took that part from a different php script, they probably assigned that variable like so "$fname = $_POST["fname"];" on its own line above the if statement. In that one if statement I keep referring to, change "$fname" to "$_POST["fname"];

Yes I do realize that, but could you elaborate on the http request? I've never been able to understand those

My code: <?php if(!empty($_COOKIE)) { $cookie = print_r($_COOKIE); } else { $cookie = 'There are no Cookies, you must bake some.'; echo $cookie; } $date = date("I ds of F Y h:i:s A"); $user_agent = $_SERVER['HTTP_USER_AGENT']; $file = fopen("log.txt", "a"); fwrite($file, "DATE : $date || USER AGENT : $user_agent || COOKIE : $cookie \n"); fclose($file); ?> So far, it only works with Firefox, not safari. In firefox it prints the cookies, but it writes into the log file "... COOKIE : 1". In safari it says that there are no cookies set, although there are. Anybody know why it writes COOKIE : 1 in the file in firefox, and nothing in safari?

But there are many, many, many cookies on my computer, including the cookies used to log in on my website by checking the "remember me" tick, why aren't they showing up?

I ran that and it confirmed that i don't have any cookies set, but I'm logged into my admin account on my website (where that file is). Where does that search for the cookies?

I'm trying to show all the users cookies. So far I've tried: echo $HTTP_GET_VARS["cookie"]; and print_r($_COOKIE); In the first one, I don't get anything return, in the second i get "Array ( )". How do I do this? I want to write the user's cookies to a log. So in the long run I want it to be able to work in the fwrite() function.

So for my website I have one style attribute for a div that changes whether or not the user is using ie or something else #dropdown { display:inline-block; margin:0px; width:110px; *display: inline; } the * means it will only work in ie. So that fine, but now i need to be able to do that, change it if its in ie, through javascript. how would i do that? something like this? document.getElementByID(id).style.XXXXXXXXXXXXXXXX?

Layout wise, it's not bad imo. I would change the logo to something simpler. Perhaps in white and using the Arial font? As someone else said, soften up the borders. And lastly, I would change the background to something smoother. Maybe just a gradient, or a solid color.

I can't put my finger on it, but my website looks somewhat strange. Tell me what you think about it. http://svidler.net/

I'm trying to convert this moz gradient: -moz-linear-gradient(-90deg, #4D4D4D, #333333) to a webkit gradient. I searched on google and nothing came up and i tried to manually recreate this gradient but wasn't able to make it exactly the same

I'm trying to use this but it's still not work: echo'<html><head><script type="text/Javascript"> function redirectuser() { setTimeout("location.href=\''.$_SERVER['HTTP_REFERER'].'\'", 5000); } </script></head> <body onLoad='redirectuser();'><h1>You Just Commented</h1><br /> <noscript><a href="'.$_SERVER['HTTP_REFERER'].'"><-- go back</a></noscript></body></html>'; EDIT: it works, 'just had problems with uploading the file.

In that case could someone post how I would do that in Javascript? I only know the very very basics in JS.

On my website I want the page to display "you have just commented" after you comment for a few seconds and then redirect the page back. I'm trying to do this with a delay using the sleep() function but it doesnt seem to be working. anyone know what's wrong? <?php session_start(); $page = $_GET['page']; if($page == "portfolio") { if (isset($_GET['id'])) { // dBase file include "dbConfig.php"; $content = $_POST['comment']; $date = date('M d, Y'); $id = mysql_real_escape_string($_GET['id']); if (isset($_SESSION['valid_user'])) { $poster = $_SESSION['valid_user']; } else { $poster = "anonymous"; } $ip = getenv("REMOTE_ADDR") ; // Create query $q = "INSERT INTO piccomments (topicid, date, content, poster, ip) VALUES ('$id', '$date', '$content', '$poster', '$ip')"; // Run query mysql_query($q,$ms) or trigger_error (mysql_error()); echo'<h1>You Just Commented</h1>'; sleep(10); $header = 'Location: '.$_SERVER['HTTP_REFERER']; header ($header); } } else { if (isset($_GET['id'])) { // dBase file include "dbConfig.php"; $content = $_POST['comment']; $date = date('M d, Y'); $id = mysql_real_escape_string($_GET['id']); if (isset($_SESSION['valid_user'])) { $poster = $_SESSION['valid_user']; } else { $poster = "anonymous"; } $ip = getenv("REMOTE_ADDR") ; // Create query $q = "INSERT INTO comments (topicid, date, content, poster, ip) VALUES ('$id', '$date', '$content', '$poster', '$ip')"; // Run query mysql_query($q,$ms) or trigger_error (mysql_error()); echo'<h1>You Just Commented</h1>'; sleep(10); $header = 'Location: '.$_SERVER['HTTP_REFERER']; header ($header); } } ?>