dachshund Posted August 16, 2009 Share Posted August 16, 2009 hey, simple question really. i have this log in form <?php if(!$_POST['submit'] == 'login'){ echo "<div id=\"blanket\" style=\"display:none;\"></div>\n"; echo "<div id=\"popUpDiv\" style=\"display:none;\">\n"; echo "<div id=\"close\">\n"; echo "<a href=\"#\" onclick=\"popup('popUpDiv')\">\n"; echo "<img src=\"/phptesting/images/login/close.png\">\n"; echo "</a>\n"; echo "</div>\n"; echo "<div id=\"loginform\">\n"; echo "<form name=\"login\" method=\"post\" action=\"/phptesting/account/index.php?\n"; echo $rows['id']; echo "\">\n"; echo "<span class=\"loginformtext\">Username</span>\n"; echo "<input name=\"username\" type=\"text\" class=\"loginformtext\" id=\"username\" size=\"28\">\n"; echo "<br>\n"; echo "<span class=\"passwordformtext\">Password</span>\n"; echo "<input name=\"password\" type=\"password\" class=\"loginformtext\" id=\"password\" size=\"28\">\n"; echo "<br><br>\n"; echo "<div align=\"left\"\n"; echo "<input type=\"submit\" name=\"submit\" class=\"loginformtext\" value=\"Submit\">\n"; echo "</div>\n"; echo "</div>\n"; echo "</form>\n"; echo "</div>\n"; echo "<span class=\"login_text\"><a href=\"#\" onclick=\"popup('popUpDiv')\">LOG IN</a> |\n"; }else { $user= $_POST['username']; $pass= $_POST['password']; if($user && $pass){ $sql = "SELECT * FROM `users` WHERE `username`='$user'"; $res = mysql_query($sql) or die(mysql_error()); if(mysql_num_rows($res) == 1){ $epass = md5($password); $sql2 = "SELECT * FROM `users` WHERE `username` ='$user' AND `password`='$epass'"; $res2 = mysql_query($sql2) or die(mysql_error()); if(mysql_num_rows($res2) ==1){ //success $row = mysql_fetch_assoc($res2); $_SESSION['uid'] = $row['id']; echo "<span class=\"login_text\"><a href=\"/phptesting/account/\">ACCOUNT</a> | <a href=\"logout.php\">LOGOUT</a></span>\n"; }else { echo "<span class=\"login_text\">USERNAME OR PASSWORD INCORRECT | <a href=\"/phptesting\">HOME</a></span>\n"; } }else { echo "<span class=\"login_text\">USERNAME OR PASSWORD INCORRECT | <a href=\"/phptesting\">HOME</a></span>\n"; } }else { echo "<span class=\"login_text\">PLEASE SUPPLY ALL VALID FIELDS | <a href=\"/phptesting\">HOME</a></span>\n"; } } ?> and then when they go to their accounts page i have <?php if($session_name){ echo "Welcome to your account.\n"; } else { echo "You are not logged in. Please log in to view your account.\n"; } ?> all i want to add is something like echo "Welcome to your account.\n"; echo $username; any help would be much appreciated. Link to comment https://forums.phpfreaks.com/topic/170502-solved-recognise-user-once-logged-in/ Share on other sites More sharing options...
DEVILofDARKNESS Posted August 16, 2009 Share Posted August 16, 2009 if($_SESSION['uid']){ $userid = $_SESSION['uid']; $query = "SELECT username FROM users WHERE userid = '$userid'"; $result = mysql_query($query); list($username) = mysql_fetch_row($result); echo "Welcome. \n"; echo $username; }else{ Change it to your table settings Link to comment https://forums.phpfreaks.com/topic/170502-solved-recognise-user-once-logged-in/#findComment-899413 Share on other sites More sharing options...
dachshund Posted August 16, 2009 Author Share Posted August 16, 2009 cheers! Link to comment https://forums.phpfreaks.com/topic/170502-solved-recognise-user-once-logged-in/#findComment-899443 Share on other sites More sharing options...
dachshund Posted August 16, 2009 Author Share Posted August 16, 2009 oh actually, is there a way that I can change it to $query = "SELECT * FROM users WHERE userid = '$userid'"; so that i can then echo something like $rows['email'] thanks! Link to comment https://forums.phpfreaks.com/topic/170502-solved-recognise-user-once-logged-in/#findComment-899448 Share on other sites More sharing options...
dachshund Posted August 16, 2009 Author Share Posted August 16, 2009 will post as new question Link to comment https://forums.phpfreaks.com/topic/170502-solved-recognise-user-once-logged-in/#findComment-899459 Share on other sites More sharing options...
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