dannyone Posted March 23, 2009 Share Posted March 23, 2009 hello, im wondering if there is a way to do this. i have a page where users can select from a few drop downs some values and save them into a db. is there a way to create a query were if the user exists in the db then they can not use that page? im storing the user_id in a session. any help would be greatly appreciated. thanks Link to comment https://forums.phpfreaks.com/topic/150745-if-exists-kick-out/ Share on other sites More sharing options...
Maq Posted March 23, 2009 Share Posted March 23, 2009 is there a way to create a query were if the user exists in the db then they can not use that page? Yeah just do a query to return some amount of rows with that user and if it returns rows don't let me insert... session_start(); $user = $_SESSION['user']; $result = myql_query("SELECT * FROM table WHERE user = '$user'") or die(mysql_error()); if(mysql_num_rows($result) > 0) { echo "sorry, you're already in the DB"; } else { echo "not in the DB, do whatever you want in this block"; } ?> Link to comment https://forums.phpfreaks.com/topic/150745-if-exists-kick-out/#findComment-791961 Share on other sites More sharing options...
dannyone Posted March 23, 2009 Author Share Posted March 23, 2009 thanks for the reply Maq, i have tried using your code and modifying to match my data but it doesn't seem to work it just gives me a blank page. do i need to add more detail into it to get it to work? <?php $user = $_SESSION['login']; mysql_connect("localhost", "xx", "xx") or die(mysql_error()); mysql_select_db("xx") or die(mysql_error()); $result = myql_query("SELECT * FROM Marker_Rooms WHERE Marker_ID = '$user'") or die(mysql_error()); if(mysql_num_rows($result) > 0) { echo "sorry, you're already in the DB"; } else { echo "not in the DB, do whatever you want in this block"; } ?> im already calling session further up so not included it here thanks Link to comment https://forums.phpfreaks.com/topic/150745-if-exists-kick-out/#findComment-792005 Share on other sites More sharing options...
Maq Posted March 23, 2009 Share Posted March 23, 2009 $result = myql_query Should be (missing the 's' in sql): $result = mysql_query -You need session_start(); at the top of your page, also add in error reporting and tell me if there is any output. session_start(); ini_set ("display_errors", "1"); error_reporting(E_ALL); Link to comment https://forums.phpfreaks.com/topic/150745-if-exists-kick-out/#findComment-792008 Share on other sites More sharing options...
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