DLR Posted February 9, 2010 Share Posted February 9, 2010 Hi, I have spend a number of frustrated hours trying to see where my error is. I have copied this from the security tutorial (by DanielO), but cannot see where I am making my mistake. Any pointers will be appreciated. The error I get is Warning: mysql_real_escape_string() expects parameter 2 to be resource, string given in C:\wamp\www\sfwebsite\admin\SF201.php on line 93 The code for connection to database sfwebsite works elesewhere in the same programme if(!$conn) { die('connection error'); } $db_selected = mysql_select_db("sfwebsite",$conn); // this is line 93 if(! $db_selected) { die('connection error'); } $name = mysql_real_escape_string($_SESSION['username'],'$conn'); $sql = "SELECT level FROM sf_admin WHERE username = '$name' "; $res = mysql_query($sql) or die('Failed to find Admin records'); Quote Link to comment Share on other sites More sharing options...
trq Posted February 9, 2010 Share Posted February 9, 2010 $name = mysql_real_escape_string($_SESSION['username'],$conn); Quote Link to comment Share on other sites More sharing options...
DLR Posted February 9, 2010 Author Share Posted February 9, 2010 Thanks. I could kick myself. Quote Link to comment Share on other sites More sharing options...
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