""Terion Miller"" <webdev.terion@xxxxxxxxx> wrote in message news:37405f850901071354p7abd0aa8i7c96cf69c81fadb5@xxxxxxxxxxxxxxxxx >> > $result=mysql_query($query) or die('Queryproblem: ' . mysql_error() >> > . >> > '<br />Executed query: ' . $query); >> > if (mysql_num_rows($result) >= '1'){ >> > while ($row = mysql_fetch_assoc($result)){ >> > echo $row['AddEditAdmin']; //to print out the value of >> > column >> > 'var1' for each record >> > } >> > }else{ >> > echo 'No records found.'; >> > } >> > ?> >> > >> > anyone have ideas for me, the session user is working, and I need to >> > use >> > it >> > in the query to pull only that users data I also on the login page >> > where >> I >> > set that session all set it to = $UserName but when I try and use that >> > in >> > the query UserName = $UserName I get an undefined variable error... >> > >> > Really trying but not quite getting it... >> > >> >> >> >> -- >> PHP General Mailing List (http://www.php.net/) >> To unsubscribe, visit: http://www.php.net/unsub.php >> >> Well I am def. further along, the query is working, I can echo fields in > the $row and get the results but when I try and use this: > > <?php > if ($row['AddEditAdmin'] == 'YES') { > ?> > to sort out what menu items to load it just doesn't do its job. > I would say do this to see if what is in the return is what you are expecting foreach($row as $key=>$value){ echo $key , ': ' , $value , '<br>'; }; just to make sure that the value is "yes", and not 1 or true or something like that. Frank -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
