helpmeplease2

Yes, all the words are links. I won't be able to fit all my CSS here.

When I use: <div style="font-size:9px;">words</div> the words are not showing up as 9px, they are showing up as 11px because of what I have set for link sizes in my stylesheet.... a:link, a:visited, a:active { font-family: Arial,Verdana,Sans-serif,serif; font-size: 11px; font-weight: bold; text-decoration: none; color: #bdbdbd; } How can I make it do what I tell it to?

I got it working, thanks.

That is returning 0 results however there should be some. And it doesn't include the part about the range.

I am trying to make it so it selects all videos that are not on a playlist. Videos are entered on a playlist in playlistvideos as either a single videoid or a range. So how can I say WHERE videos.ID != playlistvideos.videoid AND videos.ID NOT BETWEEN playlistvideos.range AND playlistvideos.range2 SELECT playlistvideos.videoid , playlistvideos.range , playlistvideos.range2 , videos.ID , videos.VTitle , videos.IFilename , videos.videoid , videos.GTitle FROM playlistvideos INNER JOIN videos ON videos.ID = playlistvideos.videoid OR videos.ID BETWEEN playlistvideos.range AND playlistvideos.range2 WHERE videos.GTitle = '$game' AND videos.VTitle NOT LIKE '%</a>%' ORDER BY videos.ID DESC LIMIT $offset, $rowsPerPage

I am still stuck trying to get this to work. $50 to anyone who will make it work for me. This is what I have so far so there shouldn't be too much work: http://www.gameanyone.com/?p=video&id=9541 Email me at derek_dude12@yahoo.com if you're interested.

I'm thinking I don't want to count them like that, as yes it will get huge. The first day there would be over a million entries.

So I am using this to get the number of views in a week... but do I set 'date' as a normal 'date' type in phpmyadmin? And I don't see how it would know when the views were viewed unless it logged each view in it's own row. <?php $period = 604800; $query = "SELECT * FROM `games` WHERE `date` < '".(($now = time()) - $period)."'"; $result = mysql_query($query); while($row = mysql_fetch_array($result)){ $views=$row['views']; } ?>

How can I do this? I want to keep track of the most viewed games on my website (www.gameanyone.com) I can guess that you add some $b=$a+1; and then insert the $b.. but that would only count views. How can I count the number of views in the last week or better the last 7 days?

I made it so I can show replies correctly. Now how do I make it to INSERT replies? I want to make it so it shows a new textarea box when a user clicks reply. I have done this, but I am having problems inserting the comment id that it is a reply to. http://www.gameanyone.com/?p=video&game=Crysis&id=9541 There is alot of code, too much to post here. And if you click reply you can see that the new textarea box would only appear below the first comment, not under the comment that the reply is for. *cry* If I emailed one of you all the code would you make it work? *cry*

I think I might be able to program basic replies.... should replies should be in their own table?

I am so confused. And if they are replies will it still limit them on a certain amount per page? This is all way over my head. Is someone brave enough to do it in exchange for a link?

I get that part. But I am not sure how I would say to echo replies in the same query as the normal comments to get everything in the right place.

I already have an ajax commenting system, but I want to add the feature to allow users to reply to comments. Is there someone who knows how to do this? I am a beginner at PHP so I'm pretty sure I can't do this myself. If you can add the ability to reply to comments like on YouTube then you can have a link on my website for free. (site: www.gameanyone.com) Email me if you want to help you: derek_dude12@yahoo.com

where id='$a' (use quotes). Try this, it may help.

I am using this code which I found on about.com and I have changed it to my needs but it is not working. No results appear when a search is made. I do have a database with many rows of data. Now can someone head in me the right direction to making this work? <?php echo "<h3>Search</h3> <form name='search' method='post' action='$PHP_SELF'> Search for: <input type='text' name='find' /> in <Select NAME='field'> <Option VALUE='VTitle'>Videos</option> <Option VALUE='GTitle'>Games</option> </Select> <input type='hidden' name='searching' value='yes' /> <input type='submit' name='search' value='Search' /> </form>"; if ($searching =="yes") { echo "<h1>Search Results</h1><p>"; if ($find == "") { echo "<p>You forgot to enter a search term."; exit; } $find = strtoupper($find); $find = strip_tags($find); $find = trim ($find); echo "<table cellspacing='0' cellpadding='0' class='bordercolor' border='0' style='margin-top: 1px; vertical-align: top;' align='center'>"; $row = mysql_query("SELECT GTitle,VTitle,IFilename FROM videos WHERE upper($field) NOT LIKE '</a>' AND upper($field) LIKE '%$find%'"); while($result = mysql_fetch_array( $row )) { $game=$row['GTitle']; $video=$row['VTitle']; $image1=$row['IFilename']; if ($image1==""){ $image="images/favicon.gif"; }else{ $image=$row['IFilename']; } echo "<tr> <td class='windowbg2' width='150' align='center'><div class='windowbg2' style='width:150; align:center;'><img src='$image' border='0' height='75'></div></td> <td class='windowbg2' width='485' valign='center'> <a href='http://www.gameanyone.com?p=game&game=$game'>$game</a>:<br> <a href='http://www.gameanyone.com?p=video&game=$game&video=$video'>$video</a></td></tr>"; } echo "</table>"; $anymatches=mysql_num_rows($data); if ($anymatches == 0) { echo "Sorry, but we can not find an entry to match your query<br><br>"; } echo "<b>Searched For:</b> " .$find; } ?> Edit: To see the page go here: http://www.gameanyone.com/?p=thesearch

I am still using tables, but I added a div within a td and that fixed it.

View this site in IE7 or FF than view it in Opera http://www.gameanyone.com/ Where images are in a tables TD by themselves it screws up in Opera, even though there the width is set for the TD. It looks fine in IE and FF, but Opera is giving me a headache. How can I fix this and still use tables?

I am trying to make thumbnails from each .flv file using ffmpeg. I want a script to make the thumbnails. I am new to php and I am having trouble. I am using this currently, but its not making any thumbnails. ffmpeg does work by itself, so there is something wrong with my noobish PHP. I would also like to somehow remove the .flv extension and make it .jpg. Some help please? <?php $path = "videos"; $dir_handle = @opendir($path) or die("Unable to open $path"); while ($file = readdir($dir_handle)) { if($file!="." && $file!="..") exec("START c:/ffmpeg/ffmpeg -vcodec jpeg -i $file -ss 00:00:30 -vframes 1 images2/$file.jpg"); } closedir($dir_handle); ?>

Help?

ok.... well i just got this next script from a free script website and its not working either: login.php <?php require('includes/config.php'); require('includes/dbconnect.php'); if(isset($_COOKIE['apexclan.com'])) { $username = $_COOKIE['apexclan.com']; $pass = $_COOKIE['apexclan.com']; $check = mysql_query("SELECT * FROM members WHERE username = '$username'")or die(mysql_error()); while($info = mysql_fetch_array( $check )) { if ($pass != $info['password']) { } else { header("Location: members.php"); } } } if (isset($_POST['submit'])) { if(!$_POST['username'] | !$_POST['pass']) { die('You did not fill in a required field.'); } if (!get_magic_quotes_gpc()) { $_POST['email'] = addslashes($_POST['email']); } $check = mysql_query("SELECT * FROM members WHERE username = '".$_POST['username']."'")or die(mysql_error()); $check2 = mysql_num_rows($check); if ($check2 == 0) { die('That user does not exist in our database.'); } while($info = mysql_fetch_array( $check )) { $_POST['pass'] = stripslashes($_POST['pass']); $info['password'] = stripslashes($info['password']); $_POST['pass'] = md5($_POST['pass']); if ($_POST['pass'] != $info['password']) { die('Incorrect password, please try again.'); } else { $_POST['username'] = stripslashes($_POST['username']); $hour = time() + 3600; setcookie(ID_my_site, $_POST['username'], $hour); setcookie(Key_my_site, $_POST['pass'], $hour); header("Location: members.php"); } } } else { ?> <form action="<?php echo $_SERVER['PHP_SELF']?>" method="post"> <table border="0"> <tr><td colspan=2><h1>Login</h1></td></tr> <tr><td>Username:</td><td> <input type="text" name="username" maxlength="40"> </td></tr> <tr><td>Password:</td><td> <input type="password" name="pass" maxlength="50"> </td></tr> <tr><td colspan="2" align="right"> <input type="submit" name="submit" value="Login"> </td></tr> </table> </form> <?php } ?> members.php <?php require('includes/config.php'); require('includes/dbconnect.php'); if(isset($_COOKIE['apexclan.com'])) { $username = $_COOKIE['apexclan.com']; $pass = $_COOKIE['apexclan.com']; $check = mysql_query("SELECT * FROM members WHERE username = '$username'")or die(mysql_error()); while($info = mysql_fetch_array( $check )) { if ($pass != $info['password']) { header("Location: login.php"); } else { echo "<a href='logout.php'>Logout</a>"; } } } else { header("Location: login.php"); } ?> When I click to login it just shows me the login box still where it should show me a logout link.

index.php <?php session_start(); header("Cache-Control: private"); require('includes/config.php'); require('includes/dbconnect.php'); require('includes/logincheck.php'); if(isset($_SESSION['Username']) && isset($_SESSION['Password'])){ include('includes/rightusercp.php'); }else{ include('includes/rightmain.php'); } $p="includes/" . $_GET['p'] . ".php"; if($_GET['p']==''){ $p="includes/main.php"; } $c="includes/" . $_GET['c'] . ".php"; if($_GET['c']==''){ $c="includes/home.php"; } if(isset($_SESSION['Username']) && isset($_SESSION['Password'])){ include($c); }else{ include($p); } ?> logincheck.php <?php if($_GET['logout']=='y'){ session_unset(); require('config.php'); require('includes/dbconnect.php'); } if(empty($_SESSION['Username'])){ if(($_POST['Username']!='') && ($_POST['Password']!='')){ $tmpusr=$_POST['Username']; $results=mysql_query("select Username,Password from members where Username='$tmpusr'"); $row=mysql_fetch_assoc($results); if (mysql_num_rows($results)==0) { echo "Invalid Username!"; }elseif($row['Password']!=$_POST['Password']){ echo "Invalid Password!"; }else{ $Username=$_POST['Username']; $Password=$_POST['Password']; session_register("Username"); session_register("Password"); $Username=$_SESSION['Username']; $Password=$_SESSION['Password']; } } } ?> main.php <table border="0" cellpadding="10" cellspacing="0" align="center"><tr><td align="center"> <div style="width:243px; height:120px;"> <div class="smplan2" align="left"> Members, please sign-in below: <form action="index.php" method="post" style="padding:0px; margin:0px"><table border="0"><tr><td width="60"><div class="smplan2" align="left"> <b>Username:</b></div></td><td><input class="searchf" type="text" name="Username" size="20"></td></tr><tr><td><div class="smplan2" align="left"> <b>Password:</b></div></td><td><input class="searchf" type="password" name="Password" size="20"></td></tr></table> <button type="submit" class="buttons">Login</button> </div> </div> When I login the page doesn't change. The database is working fine theres just something here thats wrong. Please help.

No one here knows ajax?

Hi! On a my page, http://www.gameanyone.com/?p=game&game=Stronghold I would like when the user places the cursor over a video title for it to display a thumbnail image for that video on the right side of that page. So they can move the mouse down the list of video titles and see different thumbnails. How would I go about doing this?

Does anyone know how?