MadTechie

try this change `count_` = 'count_' +'"1"' WHERE to `count_` = (`count_` + 1) WHERE Note the ` instead of ' might also try without brackets

ok not sure why that didn't work can you post the results of this <?php print_r($row); echo "---<br>" $X = split("/",$row[cover]); print_r($X); echo "$X[0]tb_$X[1]"; ?>

So is $variable infact part of the string ? ie "this is some test data $variable in the db" or a field in the database ? i need some examples really!

first thing with isset, is that if it is null it is still set, so you may want to check if its empty also are you using the session correctly <?php session_start(); // start up your PHP session! ?> Change if ($usertheme == '1'); { // Not Logged or Free $themed = 'theme/platinum/'; } to if ($usertheme == '1') { // Not Logged or Free $themed = 'theme/platinum/'; } note the ");" has been removed

erm.. not sure what your asking but.. heres a basic get file index.php <?php echo "x = ".$_GET[x]; ?> so index.php?x=10 would display x = 10

i found this very useful and easy to setup, phpMyDataGrid http://www.phpclasses.org/browse/package/2943.html either that or a form with tables

OK the reason for that is because $row[cover] = "images/520027.jpg" so try this <?php $X = split("/",$row[cover]); echo "<td><img src=".$X[0]."tb_".$X[1]."></td>"; ?> if this fails can you also post the results of a <?php print_r($row[cover]); ?>

just a quick note if ($usertheme == '1'); can you show your echo points!

I'm not sure what your asking but you have the submit button in its own form!! so that will fail, also the other form refers to $check and the 2nd refers to index.html!!

Personally i could have a function ie function displaypost($foo) { echo "<h1>"; echo $foo; echo "</h1>"; } and use while ($result = mysql_fetch_array($query)) { displaypost($result['row_that_was_fetched']) }

try echo "<td><img src=tb_$row[cover]></td>"; or for testing $X = "tb_".$row[cover]; <td><img src=$X></td>;

i think its here +'"1"' should be +'1'