Adam

Well when you develop the code just indent as you go along. You generally indent code when it's encased with curly braces: function myfunction() { // encased once so 1 indent if ($this == $that) { // encased twice so 2 indents } } There are a few exceptions which don't follow that, like when using a switch () .. but genrally anything encased just gets an extra indented. You'd be best following a coding standard, like PEAR's: http://pear.php.net/manual/en/standards.php .. can't go far wrong with following that! Indention is purely for readbility of the developer.. Adam

Fair point ILMV .. Anyway, coming back to his original question, your problem lies with your query. You need to look into using "LIMIT" in mysql.. For example: SELECT * from forumtutorial_posts where parentid='0' order by lastrepliedto DESC LIMIT 0, 10 That would display onnly the first 10 rows.. Adam

Hah well getting technical.. ?><table class='maintable'><?php should be.. ?><table class="maintable"><?php I think things like not printing HTML is just nit picking - never going to waste enough server resources to actually matter.. perhaps a microsecond or two, if that? Adam

Oh no sorry - it's early! Not thinking right.. I would hae thought it would work with what you have? Adam

foreach ($images as $imageArr) { foreach ($imageArr as $image) { echo '<img src="$image['name']" alt="$image['description']" title="$image['description']" />'; } } .. should do the trick!

if(isset($_POST['submit'])) { There you are testing to see if the submit button has been pressed, but you haven't named the submit button "submit" .. Change: <input type="submit" value="Post Comment"> to: <input type="submit" value="Post Comment" name="submit"> Basically you're testing to see if an input named "submit'" has been sent to the script via the post method (think <form method="post" ..), yet you hadn't named the submit button "submit". Adam

Fix it in what way?

Yeah show us the formvalidator.js script ..

Because you're not using it, nothing happens in your if ($verify == ..... session_start(); if(isset($_POST['submit'])) { $verify = isset($_POST['verify']) ? strtolower($_POST['verify']) : ""; if ($verify != $_SESSION['verify']) { die('BAD IMAGE VERIFICATION!'); } } You'll want to find a neater way of doing it though.. Adam

Probs because you had name=name without any quotes around "name" .. session_start(); print_r($_POST); print_r($_SESSION); $str = '<form method="post" action="' . $PHP_SELF . '">'; $str .= '<input name="name" type="text">'; $str .= '<input name="cmd" type="submit" value="Send">'; $str .= '</form>'; echo $str; $_SESSION['name'] = $_POST['name'];

You'll need to look into AJAX to load external files. Take a look at these examples (mainly the top one): http://www.w3schools.com/Ajax/ajax_examples.asp Adam

To be honest mate I very much doubt you'll get anything like that for free. You'd either have to pay someone or create it yourself... Freelance forum: http://www.phpfreaks.com/forums/index.php/board,8.0.html Adam

Have you got another form tag opened but not closed anywhere on the page? Can't imagine it would be that but, worth a look! EDIT: ignore the last bit i said! Adam

Doesn't make any sense? I thought file1 had a form filled with data from a database? Not a form to control file2, which also gets data from a database?? ???

Hah sorry, yeah look into "window.open()" ..

Unless the functions print straight to the page, you made need to echo out the return value of "con()" .. Also don't forget you've got qutoes in there which will trigger an error: $db_text = "blah blah <?php echo con(\"123\"); ?> blah"; Adam

Give chmod a go... http://uk.php.net/chmod

To be honest if you're new to javascript this will be a tough learning experience, but I once found this helpful for blending / fading objects in and out .. http://brainerror.net/scripts/javascript/blendtrans/ Adam

ahh okay, it might be trying to divide by a string - try converting it to a float (not 100% sure this would work): if ($total == '' || $total == null || $total == 0) { $total = $total + $row['cost']; } else { $vat = (float) $row['float']; $total = $total + ($row['cost'] + ($row['cost'] / $vat)); } Oh also make sure there's no % on the front of it either!

Oh wait sorry, missed the 'error in bottom left corner' bit.. Errmm only thing that looks a little out of place is: "void(0);" .. try removing that?

Try putting it into a function within <script type="text/javascript"></script> tags .. When you use inline js like that sometimes browsers may see it as malicious code.. By the way this is the PHP forum.. Adam

What data type is set for the vat field in your database?

??? if ($total == '' || $total == null || $total == 0) { $total = $total + $row['cost']; } else { $total = $total + ($row['cost'] + ($row['cost'] / $row['vat'])); } try that?

Well technically it's a statement and should be written: include 'your_file.php';

Also remember you still need PHP tags around the code within your external file...