fiddy

Can you pls try this: (if($_REQUEST['photo']=="" && $_REQUEST['memory']=="" && $_REQUEST['datap']==""){)

i ment this. Sorry about my previous one (if($_REQUEST['photo']=="" && $_REQUEST['memory']=="" && $_REQUEST['datap']==""){)

Sorry , my previous post was wrong

Can you pls try with OR (|). (if($_REQUEST['photo']=="" || $_REQUEST['memory']=="" && $_REQUEST['datap']==""){)

Hi, Do you need two NOT (!) . (if(!$_REQUEST['photo']!="" && !$_REQUEST['memory']!="" && !$_REQUEST['datap']!=""){) I dont know what you are trying to do there but i guess it should be like  if($_REQUEST['photo']!="" .....

Kindly try this: http://www.tanguay.info/wamp/installPhp5.php5?step=2

Hi kindly check the script available here: http://codewalkers.com/tutorials/35/1.html I guess if you make some modifications you can achive what you want. [b]I did not test it [/b] Let me know how it goes...

Pls try like this: if ($_REQUEST[gim] == 1) { $gotten = @mysql_query("SELECT * FROM pix ORDER BY pid DESC"); while($row = @mysql_fetch_array($gotten)) {         $title = htmlspecialchars($row[title]);         $bytes = $row[imgdata];         header("Content-type: image/jpeg");         print $bytes;         } else {         $errmsg = "MySql Error!"; } } Sorry if am wrong some where. Just felt like giviving a hint.

Pls try like this: $query = "INSERT INTO blomform (blomform) VALUES ('".$_POST['blomform']."')";

Why dont you try like this:     if($_REQUEST['photo']!="" || $_REQUEST['memory']!="" || $_REQUEST['datap']!=""){ ..... }

array getimagesize ( string filename [, array &imageinfo] )

Hi Friend, Thanks a lot.

Hi Friends, I am in need of the list of all the countries, states and its cities. I am looking for one which is a mysql table. Kindly help me if you have it or guide me to a URL where i can download it for free. Thanks

Hi, I want to import or export contacts from outlook from/to the website i am developing . I am researching on it, if any one have done it or have any ideas , please help me by sharing. Thanks..

Try putting ob_start(); as your first line of code

Hi, You are getting this error because you use just ", you should use \" becuse you have started with ". i feel its better to do like this.So that you can add HTML as you wish without worrying about the quotes and double quotes $html = <<START <tr> <td valign='top' align='center' bgcolor='#212121'> <a href='#' onmouseover="roll_over('name', 'http://www.mysite.com/image-hover.jpg')" onmouseout="roll_over('name', 'http://www.mysite.com/image.jpg')"> <img border='0' src='http://www.mysite.com/image.jpg' name='name'></a> </td> </tr> START;

Ok. Thanks for your guidence.

Thanks friends ..

Hi, i guess you would have two JS functions one for request to the server and other for getting back the responce from the server. In the JS function which receives the responce you can user location.href='file.php'

Did you check the php.ini, if you have uncommented the extention for GD. And are you modifying the right php.ini file.Kindly check if you have php.ini elsewhere. Just a thought...

Ok.., a thought ... Can you please check running phpinfo() and search for GD and check the version. If GD is not there when running phpinfo(), kindly check the ini if you have uncommented (i know you have done this)  and the dll matches the one in the ext dir.

Do you get error , if so can you past it here

Hi, Kindly check if the path to your  extentions dir is  C:\PHP\extensions  or C:\PHP\ext

Hi, Thanks for your help. Will this work if i close the window and login again instantly. Thanks Fiddy

Hi, I am looking for a way to track users who are logged in to my website. I thought of maintaining it in the DB and reset that DB entry when they logout. But, If the user go's out without logging out (by clicking the close button in the browser), it will show the status of the user as online always. Expecting help related to this.... Thanks, Fiddy.[color=black][/color]