denno020

Put it inside if statements? That's the only thing I could think of.. Provided you have a line of code that sill send the ajax request, otherwise I don't think it will work.. Denno

I'm not all that familiar with how ajax requests would work, so this might not be possible, but I'll give it a go anyway. You could store the amount of columns being displayed in a variable, and then everytime the window is resized, check to see if this amount changes, and if it does then fire the ajax request. However it may not be that simple... Denno

Well said.

How are they pulled from the different posts? Denno

do a google search for TinyMCE (one word). Will be the first result. Denno

as raknjak just said, position:absolute; wasn't helping at all. I just disabled that css property, and the footer moved perfectly to the bottom of the right column. The left column however, didn't drop to be the same height.. Denno

I just tested what I suggested and it didn't work at all :S.

Have you tried floating the right column to the left too? So then it's placed directly after the two columns before it.. I'm not sure if floating it right will put it on a different display 'zone' or whatever.. Maybe try that. Denno

you don't have session_start() at the top of your php file.. Denno

Where do you specify the background? Is it in a CSS file? Are you saving that CSS file too and then trying to preview. By default, dreamweaver will only save the page that is currently opened and active, so if you've edited the source file and a CSS file, it will only save the one that is active at the time of pressing save. Alternatively you could press save all which will do just that.. Denno

I really like the scrolling header on the index page. I didn't expect that the whole bar would change colour to suit each image, and was very pleasantly surprised by it. The only thing I can say is that the text appears on the small side.. Naturally I could zoom in to view it properly, but I don't want to have to do that.. Denno

lol. You can. I was going to suggest that, but then I figured you might like to keep it separate so that it was easy to understand. As retrieve.php doesn't really hint at their being a form inside it. But yes, you can echo the form out in retrieve.php most definitely. Denno

I'm pretty sure that it's not indicating an error with your PHP, the error is literally with the server. Have you run PHP successfully on this server before? Is it a local server? It says the configuration file could be wrong, I would start there.. Denno

Oh I see.. Well you can't have form1.html as a HTML file, it will need to be php. You then include retreive.php. If your form code, you will add the value="" property to each form field, and then inside the doube quotes for each value, you will echo the php variable that holds the correct text for that field. Example: You pull a name from the database and display it in the form //connect to mysql //run query to grab the information from the database //loop through the query result set, putting each column into it's own variable //you should have a variable $name = $row["name"]; //start printing all of your form code //the HTML/PHP code for the name text field will be <input type="text" name="name" value="<?php echo $name; ?>"/> //end your form printing. So you can see I've got lots of pseudo code above, but that's because I'm assuming you already know how to complete those parts. Hope that helps Denno

http://php.net/manual/en/function.getcwd.php Google is your friend. Denno

It's exactly the same as INSERT INTO, however you use UPDATE. Example: UPDATE (table) SET (column1) = (value from form) , (column2) = (value from form) WHERE id = (edited row_id) Denno

Do you know SQL? Do you know how to run a query to grab the contents of a table cell in a database? I'm pretty sure that if you grab that code and display it on a html page, then it's not going to be written as HTML code, it's going to be written as normal text. That means, the web page will actually show HTML code.. Sounds very confusing, but for the browser to interpret that as HTML code, and therefore add styling for the paragraph element, then you need to store <p> in the database, and not <p> Denno

<?php if(isset($_POST['product'])) { $product = $_POST['product']; $n = count($product); $i = 0; echo "Your selected Products are \r\n" . "<ol>"; while ($i < $n) { if($product[$i] == ""){ //do nothing $i++; }else{ echo "<li>{$product[$i]}</li> \r\n"; $i++; } } echo "</ol>"; } ?> Added: if($product[$i] == ""){ //do nothing $i++; }else{ echo "<li>{$product[$i]}</li> \r\n"; $i++; } Denno

Just use PHP. If you don't know how to write PHP yourself, then you can download a free script called FormToEmail PHP Form Script. That should get you started. Denno

Oops. Thanks for the correction. Denno

Oh so you have fixed the problem already? You didn't make that clear in the original post. Denno

Check this video out. Came across it just today whilst trying to learn some AJAX. Turns out this isn't using ajax, however it is using iFrames, as you're requesting. http://www.youtube.com/watch?v=TiV5vCar_cI&feature=mfu_in_order&list=UL Denno

All images will have a square border around them, which defines the outer most point of each side of the image. You could try using CSS and give img attributes a negative margin (on the left and on the top), so that they overlap the image previous and above. Just make sure that the area around the outside of the shape is transparent.. No promises that it will work, but I can't think why it wouldn't... Denno

Why not use the SMTP on the server hosting your website? I'm sure you've got email included in your package somehow... Denno

When the form is submitted, the php script that handles the processing of that form will grab the value set in the drop down, and then use it in a query that will insert the value into the 'other' table. You obviously need to code that, but that's the general idea of what to do. Denno