SirChick Posted October 10, 2007 Share Posted October 10, 2007 I have a session which bounce's out the user if the session doesn't contain their UserID. I aint even changed the code in any way the only substantial change i have done is copy n pasted the script to a different pc and now it won't work at this pc. But it still does on the other computer even though they are 100% the same. This is the session check: $GetUserID = mysql_query("SELECT * FROM userregistration WHERE UserID='{$_SESSION['Current_User']}'"); // Fetch the row from the database if (!($row = mysql_fetch_assoc($GetUserID))) { echo "User not found!"; exit; echo mysql_error(); } The result gives: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\include.php on line 41 User not found! Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/ Share on other sites More sharing options...
marcus Posted October 10, 2007 Share Posted October 10, 2007 Is the variable defined, or better yet, have a value? Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/#findComment-366507 Share on other sites More sharing options...
SirChick Posted October 10, 2007 Author Share Posted October 10, 2007 $_SESSION['Current_User'] when echo'd = 1 which is correct Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/#findComment-366509 Share on other sites More sharing options...
marcus Posted October 10, 2007 Share Posted October 10, 2007 or die(mysql_error()); error reporting is good Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/#findComment-366510 Share on other sites More sharing options...
SirChick Posted October 10, 2007 Author Share Posted October 10, 2007 Doesn't give any error for that. I echo'd the query which gives: Resource id #5 If that means anything? Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/#findComment-366513 Share on other sites More sharing options...
darkfreaks Posted October 10, 2007 Share Posted October 10, 2007 if (!isset($row = mysql_fetch_assoc($GetUserID))) { echo "User not found!"; exit; echo mysql_error(); Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/#findComment-366515 Share on other sites More sharing options...
marcus Posted October 10, 2007 Share Posted October 10, 2007 -tries to decipher darkfreaks's post- Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/#findComment-366516 Share on other sites More sharing options...
SirChick Posted October 10, 2007 Author Share Posted October 10, 2007 if (!isset($row = mysql_fetch_assoc($GetUserID))) { echo "User not found!"; exit; echo mysql_error(); =/ whats this do exactly ? never seen something like that before. Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/#findComment-366519 Share on other sites More sharing options...
marcus Posted October 10, 2007 Share Posted October 10, 2007 Use mysql_num_rows Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/#findComment-366521 Share on other sites More sharing options...
darkfreaks Posted October 11, 2007 Share Posted October 11, 2007 if (!isset($row = mysql_fetch_assoc($GetUserID))) { echo "User not found!"; exit; echo mysql_error(); =/ whats this do exactly ? never seen something like that before. if the row isnt set it errors else it displays the userid Link to comment https://forums.phpfreaks.com/topic/72686-session-check-is-not-working/#findComment-366633 Share on other sites More sharing options...
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