spedman975 Posted February 27, 2008 Share Posted February 27, 2008 Hi, I am building a webcomic site and as such I have developed a database to hold and output the comics. I have the index page of the comic to output the image (with the alt equalling the comic id and title) and display the comic's title and subtitle below (with a next/prev/first/last navigation and such)... It's all working fine but for some reason the subtitle variable wont appear and I have no idea why. Details: MySQL Version - 5.0.24a-standard-log (According to myphpadmin - Sorry, new to all this) No errors are output, just no subtitle variable Table Structure: rss_id: int(5) (Primary Key) id: int(5) title: varchar(250) link: varchar(200) subtitle: text description: text pupdate: varchar(35) image: varchar(300) Most of this is used for my rss feed though (ie. rss_id, link, description, pubdate) The page coding is as follows: (excluding the header and footer stuff) <?php include 'config.php'; include 'opendb.php'; // how many rows we have in database $query = "SELECT COUNT(id) AS numrows FROM comic"; $result = mysql_query($query) or die('Error, query failed'); $row = mysql_fetch_array($result, MYSQL_ASSOC); $numrows = $row['numrows']; // how many rows to show per page $rowsPerPage = 1; // by default we show first page $pageNum = $maxPage; // if $_GET['page'] defined, use it as page number if(isset($_GET['page'])) { $pageNum = $_GET['page']; } // counting the offset $offset = ($pageNum - 1) * $rowsPerPage; $query = "SELECT * FROM comic ORDER BY id LIMIT $offset, $rowsPerPage"; $result = mysql_query($query) or die('Error, query failed'); // print the random numbers while($row = mysql_fetch_array($result)) { echo "<img src='" .$row['image']. "' width='650' height='350' alt='" .$row['id']. " - " .$row['title']. "'>"; echo "<br><br><b>" .$row['title']. "</b><br>" .$row['subtitle']. "<br>"; } echo '<br>'; // how many pages we have when using paging? $maxPage = ceil($numrows/$rowsPerPage); $self = $_SERVER['PHP_SELF']; // creating 'previous' and 'next' link // plus 'first page' and 'last page' link // print 'previous' link only if we're not // on page one if ($pageNum > 1) { $page = $pageNum - 1; $prev = " <a href=\"$self?page=$page\">[Prev]</a> "; $first = " <a href=\"$self?page=1\">[First Page]</a> "; } else { $prev = ' [Prev] '; // we're on page one, don't enable 'previous' link $first = ' [First Page] '; // nor 'first page' link } // print 'next' link only if we're not // on the last page if ($pageNum < $maxPage) { $page = $pageNum + 1; $next = " <a href=\"$self?page=$page\">[Next]</a> "; $last = " <a href=\"$self?page=$maxPage\">[Last Page]</a> "; } else { $next = ' [Next] '; // we're on the last page, don't enable 'next' link $last = ' [Last Page] '; // nor 'last page' link } // print the page navigation link echo $first . $prev . " Showing page <strong>$pageNum</strong> of <strong>$maxPage</strong> pages " . $next . $last; include 'closedb.php'; ?> Thats the general code for the whole page. The main issue is the output section: { echo "<img src='" .$row['image']. "' width='650' height='350' alt='" .$row['id']. " - " .$row['title']. "'>"; echo "<br><br><b>" .$row['title']. "</b><br>" .$row['subtitle']. "<br>"; } As I said, all works apart from the ".$row['subtitle']." variable which chooses not to appear. So far I've tried separating the subtitle variable into a seperate echo command, tried changing it to a varchar field type (because the other one's like that seem to work)... Sorry if this all seems a little scattered but any help in this situation would be greatly appreciated!! Thanks! Quote Link to comment Share on other sites More sharing options...
PHP Monkeh Posted February 27, 2008 Share Posted February 27, 2008 Just to make sure, does the subtitle field actually contain data, you could check either by looking at phpMyAdmin or if you have direct access to the server. Quote Link to comment Share on other sites More sharing options...
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