maxudaskin

<div style="overflow:scroll;height:400px"><?php echo $text; ?></div> That way the text will scroll in the DIV tag, but not make the page long...

How can I use an array in while... I just totally blanked out...

Try making it <br />

Anyone want to help me?

What on earth are the characters that go in this? <?php preg_match_all('/\[img\](.*?)\[\/img\]/', $message, $image, PREG_PATTERN_ORDER); ?> Well the find part... how do I write something in there?

I got an error with your code... Warning: getimagesize(images/TOPLOGO.jpg) [function.getimagesize]: failed to open stream: No such file or directory in /home/.grable/vzoom/virtualzoom.net/Untitled-1.php(19) : runtime-created function on line 2 This is some text and this: will be converted to proper code

What is with the @? $result = @mysql_query ($query); // Run the query.

What he is trying to say is: <?php if( $categorie == "1" ) {$header = "Algemene Juridische Hulp"}; if( $categorie == "2" ) {$header = "Consumenten"}; if( $categorie == "5" ) {$header = "Vreemdelingen"}; if( $categorie == "8" ) {$header = "Wonen en Huur"}; ?> should be <?php if( $categorie == "1" ) {$header = "Algemene Juridische Hulp";} if( $categorie == "2" ) {$header = "Consumenten";} if( $categorie == "5" ) {$header = "Vreemdelingen";} if( $categorie == "8" ) {$header = "Wonen en Huur";} ?> The semicolon ( should go after the action, not the curly bracket... A good way to remember this is to write the code like this: <?php if( $categorie == "1" ){ $header = "Algemene Juridische Hulp"; }elseif( $categorie == "2" ){ $header = "Consumenten"; }elseif( $categorie == "5" ){ $header = "Vreemdelingen"; }elseif( $categorie == "8" ){ $header = "Wonen en Huur"; } ?>

What will happen if there are two images?

I thought ../ is root?

Try else... I don't think els will work to well in php... but java and turing, yes. Nahh, not an AK-47 but a MK11 / SR 25 Sniper Rifle http://www.globalsecurity.org/military/systems/ground/mk11.htm

Are you trying to link it? <a href="../pic/photo.jpg" target="_blank"><img src="../pic/photo.jpg" /></ a> If not <img src="../pic/photo.jpg" />

Try changing <?php mail( "[email protected]", "Feedback Form Results", $message, "From: $email" ); ?> to <?php mail( "[email protected]", "Feedback Form Results", $comments, "From: $email" ); ?> $message to $comments

Screenshot of the table would be nice and providing what $pID is set to would help a lot...

But the text will also contain other stuff... Here is a real example... GLAD to see we have our new forum!

Is it possible to isolate the URL?

That is why I am gifted

<?php if(!isset($_GET['refresh'])){ echo "<meta http-equiv=\"Refresh\" content=\"0;url=\"page.php?refresh=1\">"; } ?>

Given: $text = "[ img ]http://www.url.com/image.jpg[ /img ]" Wanted: $var = "http://www.url.com/image.jpg"; I want the url Isolated.. but the url is dynamic.

<?php $terms = $_POST['search']; $query = mysql_query("SELECT * FROM `table` WHERE `column` LIKE '$terms'"); while($row = mysql_fetch_array($query)){ echo "$row['column']"; echo "<br />"; } ?> Search Term: egg Results: eggnogg eggs eggmcmuffin egg

Let me help your brain out. Organize your script... <?php $variable = "value"; if($variable != "something"){ // Checks if $variable is not equal to something echo "Variable actually says: " . $variable; // Variable is not equal to something }else{ // Variable is equal to something while($i != 10){ // If $i is not equal to 10 do this echo $i; // Output $i $i++; // Increase $i by one } } ?> See how I indented the code and added comments in the appropriate sections? You can also align the variables... <?php $that = 9; $goo = 65; $apples_and_bananas = "yum"; ?> Please note that in the above code, it does not align because a space in your browser is not the same width as a character. In dreamweaver and notepad they are.

But how can I isolate the URL?

Is it possible? If not, is there another way to do it? What I want to do is if <img src="someurl.com/image.jpg" /> is found (the url would vary) to getimagesize("someurl.com/image.jpg"). To do this, we need to extract the URL... is it possible? <?php $smilies = array("[img=","]"); $locations = array('<img src="','" />'); echo str_replace($smilies,$locations,nl2br($text)); ?> $text might be Hey, I found a cool pic! [img=http://www.url.com/image.jpg]

Any Idea's what so ever? Even a partial idea will help.

Ok, before you think, oh wow, this guy doesn't know anything, I do know how to do it, but, I am stuck on a method for a code I made. <?php /* FORUM ARRAYS */ $find = array("[img]", "[/img]"); $replace = array("<img src=\"","\" />"); ?> <?php echo str_replace($find,$replace,nl2br($textsql['text'])); ?> What I want it to do is after it changes [ img ] and [ /img ] to <img src=\" and \" /> for it to check the image size, probably with <?php $getimg = getimagesize("somepic.jpg"); list($width, $height, $type, $size) = $getimg; ?>