ataria

[13-Sep-2007 11:53:51] PHP Warning: Unknown(): Unable to load dynamic library '/usr/local/lib/php/extensions/no-debug-non-zts-20020429//etc/ixed/ixed.4.4.lin' - /usr/local/lib/php/extensions/no-debug-non-zts-20020429//etc/ixed/ixed.4.4.lin: cannot open shared object file: No such file or directory in Unknown on line 0 That is the error in the error log that just keeps showing, over and over. In fact, the error file itself is about 690k. Lovely. Anyone know what that is from? o_o .. I don't even have access to that file.

What would I use sessions for, though? :X

Yeah. I want something that would keep them logged in at all times, unless they log out, of course.

Anyone? :X

I want to use cookies, although, I am open to switch over to sessions. What is a good way to ensure their information is safe and no one can gain access to their account? -x- Good / Bad way: Storing PHP session IDs in a cookie, and, a row in the database? and have them match up? X:

$query = mysql_query("SELECT * FROM `table` WHERE `id`>0"); // change ID to whatever you use as the index. the id above, ^ while($row=mysql_fetch_array($query)){ $timestamp = strtotime($row[time]); $cutoff = time()-$timestamp; $limit = 864000; if($limit > $cutoff){ $query = mysql_query("UPDATE `table` SET `close`=1 WHERE `id`=$row[id]"); // Of course, the "`close`=1" part would be replaced by whatever you have that represents it being closed. // Again, the ID would change to whatever your index is. } else{ // nothing. } } *** Spaced out so it's easier on the eyes. I believe that should work, not 100% sure. I'd prefer if someone revises it for me. but, yeah, try it if you want.

Do you have the User ID anywhere else? such as, in a cookie, or session, etcetc.

The process gets stopped and says there is an error, no not a PHP/MySQL error, the error message I made. I echo'd the contents, and, it does not show anything .

Anyone?

Okay. so, on my site, users are allowed to donate. When they donate (via paypal) they are directed back to the processing page. I use this: if($_SERVER[HTTP_REFERER] != 'https://www.paypal.com/cgi-bin/webscr') { echo "<center><b>Error.</b> "; to check to see if they came from paypal, which prevents false donations / receiving the donation item multiple times with one payment. BUT, It does not work . Is there any other way to secure that only one donation item is given?

.... anyone?

Do you echo anything? or is your script just that query... ?

Okay. I'll just make a quick little code snippet. <?php // would be the function. function secure_int($variable){ $variable = intval($variable); return $variable; } // would be the script. secure_int($id); echo "Your ID is ".$id."!"; ?> see how I didn't use "$id = function($id);" is that possible? =/

Post your script, let us see if you made an error or something.

anyone?

Okay. What I am looking to do is run a variable into a function ( function($variable); ) and then, have it at the end, predefine it in the function. so, the script would look likeeee. $variable = $row[variable]; function($variable); and, then I can use it on the rest of the script as $variable. is that possible, If so, how do i do it?

Hmm. Okay. How is the image stored? & Is the actual image file deleted? -x- Side Note: I see you used <? quite often.. They got rid of that in PHP5, so, it might be wise to break the habit. ;x

I got a question. what exactly do these lines do? if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) {

Off Topic, but, question.. echo file_get_contents('script.php'); would that do the same as, include or require? (Sorry, just trying to find a connection between them, if there is one)

Can you place supply us with some code?

if(in("show")) { $id = $_POST['info']; $action = "SELECT * FROM entried WHERE id = '$id'"; $result = mysql_query($action); $row = mysql_fetch_array($result); $name_info = $row['name']; $ip_info = $row['ip']; } I am not sure what the exact field name is, so I just put in 'name' and 'ip'. change if needed.

Nevermind. Didn't see it. -relooking over.

Nothing is wrong with your line, your just not running a query. $update = mysql_query($sql) or die(mysql_error()); Put that line below $sql.

Alright. I want the users to be able to type [item]Item Name[/item] Of course. [item] will change to.. <img src=" and, [/item] will change to "> but the thing is, the URL is not the item name, so, I need to run a query into the database, retrieve the URL. and then I can use it. any way I can break off the name they entered and run it in a query? =/ sorry if this sounded confusing.

how do i tell it to do nothing... when a field is empty? just leave it empty.. or put 0 Just leave it empty .