[upload file help]

check the tutorials on this site, there is one about file uploading

[open url]

ok

i read the url folder( for example: [a href=\"http://www.site.com/fotodirectory/\" target=\"_blank\"]http://www.site.com/fotodirectory/[/a] ):
$directory = $_POST['fotodir'];

then i'm using a script i found on the internet to test it

[code]
$imgdir = "$directory";
        $allowed_types = array('png','jpg','jpeg','gif');
        $dimg = opendir($imgdir);
        while($imgfile = readdir($dimg))
        {
            if(in_array(strtolower(substr($imgfile,-3)),$allowed_types))
            {
                $a_img[] = $imgfile;
                sort($a_img);
                reset ($a_img);
            }
        }
        $totimg = count($a_img); // total image number
        for($x=0; $x < $totimg; $x++)
        {
            $size = getimagesize($imgdir.'/'.$a_img[$x]);
            // do whatever
            $halfwidth = ceil($size[0]/2);
            $halfheight = ceil($size[1]/2);
            echo 'name: '.$a_img[$x].' width: '.$size[0].' height: '.$size[1].'<br />';
        }
[/code]

but the error i get
failed to open dir: not implemented in ....php on line ...

and

readdir() not valid argument

so i know that i am supposed to give a local folder, and that's why i asked if there is a method that opens the url instead of just a dir

[open url]

anyone that can send me in the good direction?

[open url]

i have a form that let's the user give in an url with pictures, so what i want to do is read that url and display the pictures that were found, but is there a way to do this? because with opendir/readdir i get errors, and also when i change my htaccess file with the url_open stuff, it also doesn't work

[simple error, can't figure it out]

[!--quoteo(post=350810:date=Mar 1 2006, 11:19 PM:name=curtis_b)--][div class=\'quotetop\']QUOTE(curtis_b @ Mar 1 2006, 11:19 PM) [snapback]350810[/snapback][/div][div class=\'quotemain\'][!--quotec--]
Here's the error I'm getting:

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'Group = 'KK'' at line 1

The variable $group is being stored in the url as a link from a previous page:
[a href=\"http://ecolab.7cpco.com/pricegroups.php?group=KK\" target=\"_blank\"]http://ecolab.7cpco.com/pricegroups.php?group=KK[/a]

Here's the php code that's causing the error:
$query = "SELECT * FROM item_list_price_groups where Group = '$group'";
I have other pages that work just the same as this is intended to (result based on url-stored variable). I pretty much copied the code verbatim from another application I wrote that works fine. I don't understand what could be wrong with this query. Thanks for your help.
[/quote]

$sql = mysql_query("SELECT * FROM item_list_price_groups WHERE Group = '$group'");

[Help With Mysql data insert]

[!--quoteo(post=350826:date=Mar 2 2006, 12:51 AM:name=PcGamerz13)--][div class=\'quotetop\']QUOTE(PcGamerz13 @ Mar 2 2006, 12:51 AM) [snapback]350826[/snapback][/div][div class=\'quotemain\'][!--quotec--]
It only post blank stuff in the mysql table

[code]$User = $POST['User'];
$Password = $POST['Password'];
$E_Mail = $Post['E_Mail'];
$query = "INSERT INTO Users
VALUES('$User','$Password','$E_Mail')";
mysql_query($query);[/code]
[/quote]

$sql = mysql_query("INSERT INTO Users(user, pass, email)
VALUES('$User','$Password','$E_mail') or die (mysql_error());

[simple sql question]

if i put a very important word in a database from a site, is there anyway that someone could access, or retrieve the important word in any way via the site or via anything else?