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.