Drumminxx

I always just use a hidden flash animation and then a little DHTML to make it appear when the user clicks the upload button. it's not a true progress bar but at least it gives the user some animation to look at while it uploads and its quick and easy. an animated gif won't work by the way

if ($info[118] => 2500) { $active = 'ACTIVE'; } else { $active = ''; } change the operator to >= if ($info[118] >= 2500) { $active = 'ACTIVE'; } else { $active = ''; }

if your trying to display the color in the drop down you need to set the style property take a look at his http://pietschsoft.com/blog/Post.aspx?PostID=204

No problem Glad it worked for you!!

if you go to the link I provided above, I would think that any of the first four would work. It's just a matter of what would work best. for example if this is similiar to a donation thing then use the donations or if the admin will be making recurring payments then use that

When you load the player just set the src to file name using $_GET I'm not sure of the syntax of loading quicktime, you may already have it or you can find it on the web I'm sure but it would go something like this provided that id is what holds the file name [code] <EMBED SRC="<? echo $_GET['id'] ?>" WIDTH="262" HEIGHT="15" AUTOPLAY="false"></EMBED> [/code]

another problem with that is if the user has javascript turned off... here's an example using eregi which is case insensitive [code]<? $s = "Ssertfrt633"; if (eregi('^[[:alnum:]]{2,}$',$s)) { echo 'ok'; }else{ echo 'bad'; } ?>[/code]

Change this: [code] $sql = mysql_query("INSERT INTO users (username, password)         VALUES('$username', '$passwd')") or die (mysql_error()); $result = mysql_query($sql,$connection) or die (mysql_error()); [/code] to [code] $sql = "INSERT INTO users (username, password)         VALUES('$username', '$passwd')"; $result = mysql_query($sql,$connection) or die (mysql_error()); [/code] your using the result of the query instead of the query itself or you could just remove this line altogether if you prefer [code] $result = mysql_query($sql,$connection) or die (mysql_error()); [/code]

syntax error define('PN8', ; could be others but this is what I found for starters

yea here's the link I just found it [url=https://www.paypal.com/cgi-bin/webscr?cmd=p/mer/directory_intro-outside]https://www.paypal.com/cgi-bin/webscr?cmd=p/mer/directory_intro-outside[/url]

its been awhile but I recall paypal having some scripts available for developers to use for this purpose, they also had a tutorial on the site. are you already using them? if not take a look at paypal

I don't think a double submit would be a problem on a select (drop down) box but in case it is you could more easily put it into a function. oh yea, this is javascript and really doesn't belong here, sorry mods but its an easy one [code] <script language="javascript" type="text/javascript"> function senddata(e) { e.disabled=true; document.filter.submit(); } </script> [/code] then in your select use this [code] <select name='select' onchange="senddata(this)"> [/code] the post value is what is being sent but its name is the name of the select box, in your case its [code] $_POST['select'] [/code]

yea, the above example should work provided that you created a loadVars object in your action script but I figured we should get the php right first and go from there

Let me know if this works for you or not [code] echo "&answer=$send_answer&"; [/code]

in your php.ini file you show session.save_path no value no value you need to set the save path or the session does not get created