php mysql query ORDER BY $variable

[yandoo]

Hi I was wondering if someone could help me please? I need to add the ORDER BY option of my query to be a $variable so a user can choose (by either using a drop down menu or hitting hyperlink) to determine how the outputted results will be ordered. I don't know how to do this, if someone could help that would be ace.

$sql4 = mysql_query("SELECT media.*, IF (ISNULL(rating.ip), 0, 1) AS HasRated
FROM media LEFT JOIN rating ON media.media_id = rating.media_id AND rating.ip = '".$ipaddress."'
WHERE media.media_type = 'Book' ORDER BY $variable $limit");

Thank you

[yandoo]

Ok so I've got as far as:

if (isset($_POST['order'])) {

$order = mysql_real_escape_string($_POST['order']);

$sql4 = mysql_query("SELECT media.*, IF (ISNULL(rating.ip), 0, 1) AS HasRated
FROM media LEFT JOIN rating ON media.media_id = rating.media_id AND rating.ip = '".$ipaddress."'
WHERE media.media_type = 'Book' ORDER BY $order ASC $limit");	
}

<form action="books.php" enctype="multipart/form-data" name="myOrder" id="myOrder" method="post">		   

        <select name="order">
          <option value="<?php echo $media_title;?>">Title</option>
		   <option value="<?php echo $rating;?>">Popular</option>
        </select>
		<input name="submit" type="button" value="submit"/>

  </form>

I'm not sure where to go from here.. any ideas please?

[cretaceous]

Your sql should be ok but the select values look wrong:

<select name="order">
          <option value="<?php echo $media_title;?>">Title</option>
		   <option value="<?php echo $rating;?>">Popular</option>
        </select>

should probably be

<select name="order">
          <option value="title">Title</option>
		   <option value="rating">Popular</option>
        </select>

so it's the name of the database column that is passed as the 'order' - not the title of a book or a rating

You don't say what is wrong - also note mysql is now past its sell by date and you should look at mysqli or PDO