PHP - User Is Online?
i got my user login and register working with my sql but now if a non logged in user tries to access the shoutbox i want it to redirect them to the register page.
<?PHP session_start(); if (!(isset($_SESSION['login']) && $_SESSION['login'] != '')) { header ("Location: /login/main.php"); } ?> im using that code but even if im logged in, it redirects me to the register page? my main site is on the root of the site. the login page and the logged in page is in "/login/ Similar Tutorialsok i want to make it to where when you look at there profile it will tell you if they are online or not! how would i do that ? here is my php script that has some modifications on the profile page but wont show if other users are online! here is the login script where i put the session and the profile page. login.php Code: [Select] <?php // Start Session to enable creating the session variables below when they log in session_start(); // Force script errors and warnings to show on page in case php.ini file is set to not display them error_reporting(E_ALL); ini_set('display_errors', '1'); //----------------------------------------------------------------------------------------------------------------------------------- // Initialize some vars $errorMsg = ''; $email = ''; $pass = ''; $remember = ''; if (isset($_POST['email'])) { $email = $_POST['email']; $pass = $_POST['pass']; if (isset($_POST['remember'])) { $remember = $_POST['remember']; } $email = stripslashes($email); $pass = stripslashes($pass); $email = strip_tags($email); $pass = strip_tags($pass); // error handling conditional checks go here if ((!$email) || (!$pass)) { $errorMsg = '<font color="red">Please fill in both fields</font>'; } else { // Error handling is complete so process the info if no errors include 'scripts/connect_to_mysql.php'; // Connect to the database $email = mysql_real_escape_string($email); // After we connect, we secure the string before adding to query //$pass = mysql_real_escape_string($pass); // After we connect, we secure the string before adding to query $pass = md5($pass); // Add MD5 Hash to the password variable they supplied after filtering it // Make the SQL query $sql = mysql_query("SELECT * FROM myMembers WHERE email='$email' AND password='$pass' AND email_activated='1'"); $login_check = mysql_num_rows($sql); // If login check number is greater than 0 (meaning they do exist and are activated) if($login_check > 0){ while($row = mysql_fetch_array($sql)){ // Pleae note: Adam removed all of the session_register() functions cuz they were deprecated and // he made the scripts to where they operate universally the same on all modern PHP versions(PHP 4.0 thru 5.3+) // Create session var for their raw id $id = $row["id"]; $_SESSION['id'] = $id; // Create the idx session var $_SESSION['idx'] = base64_encode("g4p3h9xfn8sq03hs2234$id"); // Create session var for their username $username = $row["username"]; $_SESSION['username'] = $username; //THIS IS WHERE I EDITED THE SESSION TO SAY IF THERE LOGGED IN OR NOT $logedin = $row['id']; $_SESSION['islogedin']=$logedin; mysql_query("UPDATE myMembers SET last_log_date=now() WHERE id='$id' LIMIT 1"); // THIS WAS JUST A TEST BUT WONT UPDATE UNTILL THEY LOGOUT mysql_query("UPDATE myMembers SET online='online' WHERE id='$id' LIMIT 1"); } // close while // Remember Me Section if($remember == "yes"){ $encryptedID = base64_encode("g4enm2c0c4y3dn3727553$id"); setcookie("idCookie", $encryptedID, time()+60*60*24*100, "/"); // Cookie set to expire in about 30 days setcookie("passCookie", $pass, time()+60*60*24*100, "/"); // Cookie set to expire in about 30 days } // All good they are logged in, send them to homepage then exit script header("location: home.php?test=$id"); exit(); } else { // Run this code if login_check is equal to 0 meaning they do not exist $errorMsg = "<h3><font color='red'>Email/Password invalid<br /></font></h3><a href='forgot_pass.php'>Forgot password?</a><div align='right'> <br> Forget to activate you account?</div>"; } } // Close else after error checks } //Close if (isset ($_POST['uname'])){ ?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <link rel="icon" href="favicon.ico" type="image/x-icon" /> <link rel="shortcut icon" href="favicon.ico" type="image/x-icon" /> <link href="style/main.css" rel="stylesheet" type="text/css" /> <script src="js/jquery-1.4.2.js" type="text/javascript"></script> <title>Log In</title> <title>Login Page</title> <style type="text/css"> #stage { top: 0px; left: 0px; z-index: 100; } .stage { position: absolute; top: 0; left: 0; width: 100%; min-width: 900px; height: 1359px; overflow: hidden; } #bg { background: #aedfe5 url(images/sky1.png) 0 0 repeat-x; } #clouds { background: transparent url(images/cloud.png) 305px 10px repeat-x; } #sun { background: url(images/land_sun.gif)0 0 no-repeat; } #hillbottom { background: url(images/hill2.png)0 1270px repeat-x; } </style> <link rel="stylesheet" type="text/css" href="css/loginstyle.css" /></head> <body> <!-- IE6 fixes are found in styles/ie6.css --> <!--[if lte IE 6]><link rel="stylesheet" type="text/css" href="css/ie6.css" /><![endif]--> <script src="js/jquery-1.3.2.min.js" type="text/javascript"></script> <script src="js/jquery-ui-1.7.2.spritely.custom.min.js" type="text/javascript"></script> <script src="js/jquery.spritely-0.5.js" type="text/javascript"></script> <script type="text/javascript"> (function($) { $(document).ready(function() { var direction = 'left'; $('#clouds').pan({fps: 40, speed: 0.5, dir: direction, depth: 10}); }); })(jQuery); </script><div id="bg" class="stage"></div> <div id="container"> <div id="sun" class="stage"></div> <div id="clouds" class="stage"> <div id="stage" class="stage"> <body> <div id="behindform"> <form id="signinform" action="login.php" method="post" enctype="multipart/form-data" name="signinform"> <fieldset> <legend>Log in</legend> <label for="login">Email</label> <input type="text" id="email" name="email" /> <div class="clear"></div> <label for="password">Password</label> <input type="password" id="password" name="pass" /> <div class="clear"></div> <label for="remember_me" style="padding: 0;">Remember me?</label> <input type="checkbox" id="remember" style="position: relative; top: 3px; margin: 0; " name="remember"/ value="yes" checked="checked"> <div class="clear"></div> <br /> <input type="submit" style="margin: -20px 0 0 287px;" class="button" name="commit" value="Sign In"/> </fieldset><?php print "$errorMsg"; ?> </form> </div> </div> </div><div id="hillbottom" class="stage"> </div> </body> </html> profile.php This is only a part where i try. but when i putt it on , it wont echo the other peoples on , like it doesnt get the other sessions or somethig Code: [Select] //HERE IS WHERE I STARTED , BUT dONT KNOW WHAT TO DO ! if (isset($_SESSION['islogedin']) && $logOptions_id != $id) { $isonline = "<font color='green'>online</font>"; } else{ $isonline = "<font color='red'>offline</font>"; } // This is to Check if user is online or not! needs editing //$isonline = mysql_query("SELECT online FROM myMembers WHERE id='$logOptions_id'AND online='online'"); //$isonlinecheck=mysql_query($isonline); //if ($isonlinecheck ="online"){ //$online = "is <font color='green'>online!</font>";} //else { // $online = "is<font color='red'> offline!</font>"; //} // End to Check if user is online or not! ?> it displays is online no matter what! please help me! here is the login.php where the session is started . login.php Code: [Select] <?php // Start Session to enable creating the session variables below when they log in session_start(); // Force script errors and warnings to show on page in case php.ini file is set to not display them error_reporting(E_ALL); ini_set('display_errors', '1'); //----------------------------------------------------------------------------------------------------------------------------------- include 'scripts/connect_to_mysql.php'; // Connect to the database // Initialize some vars $errorMsg = ''; $email = ''; $pass = ''; $remember = ''; if (isset($_POST['email'])) { $email = $_POST['email']; $pass = $_POST['pass']; if (isset($_POST['remember'])) { $remember = $_POST['remember']; } $email = stripslashes($email); $pass = stripslashes($pass); $email = strip_tags($email); $pass = strip_tags($pass); // error handling conditional checks go here if ((!$email) || (!$pass)) { $errorMsg = '<font color="red">Please fill in both fields</font>'; } else { // Error handling is complete so process the info if no errors $email = mysql_real_escape_string($email); // After we connect, we secure the string before adding to query //$pass = mysql_real_escape_string($pass); // After we connect, we secure the string before adding to query $pass = md5($pass); // Add MD5 Hash to the password variable they supplied after filtering it // Make the SQL query $sql = mysql_query("SELECT * FROM myMembers WHERE email='$email' AND password='$pass' AND email_activated='1'"); $login_check = mysql_num_rows($sql); // If login check number is greater than 0 (meaning they do exist and are activated) if($login_check > 0){ while($row = mysql_fetch_array($sql)){ // Pleae note: Adam removed all of the session_register() functions cuz they were deprecated and // he made the scripts to where they operate universally the same on all modern PHP versions(PHP 4.0 thru 5.3+) // Create session var for their raw id $id = $row["id"]; $_SESSION['id'] = $id; // Create the idx session var $_SESSION['idx'] = base64_encode("g4p3h9xfn8sq03hs2234$id"); // Create session var for their username $username = $row["username"]; $_SESSION['username'] = $username; //THIS IS WHERE I EDITED THE SESSION TO SAY IF THERE LOGGED IN OR NOT $_SESSION['logedin'] = $_POST['email']; mysql_query("UPDATE myMembers SET last_activity=now() WHERE id='$id'"); mysql_query("UPDATE myMembers SET last_log_date=now() WHERE id='$id' LIMIT 1"); // THIS WAS JUST A TEST BUT WONT UPDATE UNTILL THEY LOGOUT } // close while // Remember Me Section if($remember == "yes"){ $encryptedID = base64_encode("g4enm2c0c4y3dn3727553$id"); setcookie("idCookie", $encryptedID, time()+60*60*24*100, "/"); // Cookie set to expire in about 30 days setcookie("passCookie", $pass, time()+60*60*24*100, "/"); // Cookie set to expire in about 30 days } // All good they are logged in, send them to homepage then exit script header("location: /socialtscripts/home.php?test=$id"); exit(); } else { // Run this code if login_check is equal to 0 meaning they do not exist $errorMsg = "<h3><font color='red'>Email/Password invalid<br /></font></h3><a href='forgot_pass.php'>Forgot password?</a><div align='right'> <br> Forget to activate you account?</div>"; } } // Close else after error checks } //Close if (isset ($_POST['uname'])){ ?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <link rel="icon" href="favicon.ico" type="image/x-icon" /> <link rel="shortcut icon" href="favicon.ico" type="image/x-icon" /> <link href="style/main.css" rel="stylesheet" type="text/css" /> <script src="js/jquery-1.4.2.js" type="text/javascript"></script> <title>Log In</title> <title>Login Page</title> <style type="text/css"> #stage { top: 0px; left: 0px; z-index: 100; } .stage { position: absolute; top: 0; left: 0; width: 100%; min-width: 900px; height: 1359px; overflow: hidden; } #bg { background: #aedfe5 url(images/sky1.png) 0 0 repeat-x; } #clouds { background: transparent url(images/cloud.png) 305px 10px repeat-x; } #sun { background: url(images/land_sun.gif)0 0 no-repeat; } #hillbottom { background: url(images/hill2.png)0 1270px repeat-x; } </style> <link rel="stylesheet" type="text/css" href="css/loginstyle.css" /></head> <body> <!-- IE6 fixes are found in styles/ie6.css --> <!--[if lte IE 6]><link rel="stylesheet" type="text/css" href="css/ie6.css" /><![endif]--> <script src="js/jquery-1.3.2.min.js" type="text/javascript"></script> <script src="js/jquery-ui-1.7.2.spritely.custom.min.js" type="text/javascript"></script> <script src="js/jquery.spritely-0.5.js" type="text/javascript"></script> <script type="text/javascript"> (function($) { $(document).ready(function() { var direction = 'left'; $('#clouds').pan({fps: 40, speed: 0.5, dir: direction, depth: 10}); }); })(jQuery); </script><div id="bg" class="stage"></div> <div id="container"> <div id="sun" class="stage"></div> <div id="clouds" class="stage"> <div id="stage" class="stage"> <body> <div id="behindform"> <form id="signinform" action="login.php" method="post" enctype="multipart/form-data" name="signinform"> <fieldset> <legend>Log in</legend> <label for="login">Email</label> <input type="text" id="email" name="email" /> <div class="clear"></div> <label for="password">Password</label> <input type="password" id="password" name="pass" /> <div class="clear"></div> <label for="remember_me" style="padding: 0;">Remember me?</label> <input type="checkbox" id="remember" style="position: relative; top: 3px; margin: 0; " name="remember"/ value="yes" checked="checked"> <div class="clear"></div> <br /> <input type="submit" style="margin: -20px 0 0 287px;" class="button" name="commit" value="Sign In"/> </fieldset><?php print "$errorMsg"; ?> </form> </div> </div> </div><div id="hillbottom" class="stage"> </div> </body> </html>and then this is the profile.php where i see if there online , i could only put sertian areas , script is too big profile.php Code: [Select] <?php //on top of the page where it checks the session and updates the time // This updates the database correctly if( isset($_SESSION['logedin']) ) { mysql_query("UPDATE myMembers SET last_activity=now() WHERE id='$logOptions_id'"); // this is where it selects the users id but it wont work , it says online for every user! $age= 60; if( isset($_SESSION['logedin']) ) { $q = mysql_query('SELECT id=`$logOptions_id`, DATE_FORMAT(`last_activity`,"%a, %b %e %T") as `last_activity`,UNIX_TIMESTAMP(`last_activity`) as `last_activity_stamp`FROM `mymembers`WHERE `$logOptions_id` <> "'.($_SESSION['logedin']).'"'); $isonlinecheck = mysql_query($q); if ($isonlinecheck ="last_activity_stamp" + $age < time()){ $isonline = "is <font color='green'>online!</font>";} else { $online = "is<font color='red'> offline!</font>"; } } ?> PLEASE HELP ME im trying to count and display the number on users on my site this is the coding im using cant see where im going wrong, its inserted into the data base correctly but wont delete after 60 seconds, cheers matt $session=session_id(); $time=time(); $time_check=$time-60; $sql="SELECT * FROM onlineusers WHERE session='$session'"; $result=mysql_query($sql); $count=mysql_num_rows($result); if($count=="0"){ $sql1="INSERT INTO onlineusers(session, time, username)VALUES('$session', '$time', '$username')"; $result1=mysql_query($sql1); } else { "$sql2=UPDATE onlineusers SET time='$time' WHERE session = '$session'"; $result2=mysql_query($sql2); } $sql3="SELECT * FROM onlineusers"; $result3=mysql_query($sql3); $count_user_online=mysql_num_rows($result3); $sql4="DELETE FROM onlineusers WHERE time<$time_check"; $result4=mysql_query($sql4); Can somebody put a code that would show all the people that are online on the website? ok well i have this is online script that starts at the login page where it sets a session. well it echos that that person is online even if they are not, i will have all of the code only for the online script so it will be in peaces. ok so here is the login page where the session is started login.php Code: [Select] <?php //ok so if they submit the page and its all right and they login , here is the session that is set for the person, again its just the piece of the script . $_SESSION['logedin'] = $_POST['email']; mysql_query("UPDATE myMembers SET last_activity=now() WHERE id='$id'"); ?> now here is where i call on the session and see if there online , the profile.php is set up to where it sees if it is your profile or not so $logoptions_id is the id that they are logged in as. profile.php Code: [Select] <?php // this is on top of the page , where the session is called and if they are logged in it updates the database where there id is. if( isset($_SESSION['logedin']) ) { mysql_query("UPDATE myMembers SET last_activity=now() WHERE id='$logOptions_id'");// there is where $logOptions_id comes in. } // now this is further down the page(script) where we see if they are logged in or not. $age= 60; //set a variable called age, assign an integer of 60 to it. if( isset($_SESSION['logedin']) ) { $q = 'SELECT id=`$id`, DATE_FORMAT(`last_activity`,"%a, %b %e %T") as `last_activity`,UNIX_TIMESTAMP(`last_activity`) as `last_activity_stamp`FROM `mymembers`WHERE `$id` <> \''.($_SESSION['logedin']).'\''; $isonlinecheck = mysql_query($q); $row = mysql_fetch_assoc($isonlinecheck); if (($row['last_activity_stamp'] + $age)< time()){ $isonline = "is <font color='green'>online!</font>";} else { $isonline = "is<font color='red'> offline!</font>"; } } ?> i wana thank all who helps! your all greatly appreciated Hey Guys/Girls, Thanks for offering to help! I'm currently setting up a small social network for school and I just basically want to know whether the way I'm dealing with in-active users is appropriate and not going to SLOW down my code A LOT, My method is : 3 Functions: Code: [Select] function update_active_user(){ global $connection; global $id; $time = time(); $result3 = mysql_query("UPDATE users5 SET last_update = '$time' WHERE id = '$id'", $connection); } function update_inactive_users(){ global $connection; $time_to_expire = time() - 300; // 300 seconds off the current time $result2 = mysql_query("UPDATE users5 SET online='0' WHERE last_update < $time_to_expire AND online='1'", $connection); } function update_active_users(){ global $connection; $time_to_expire = time() - 300; // 300 seconds off the current time $result2 = mysql_query("UPDATE users5 SET online='1' WHERE last_update > $time_to_expire AND online='0'", $connection); } First function updates the user's field called last_update to the current time (This will only occer when the script loads and the user has done something on my website, it will update their last_update field to current time) Second function sets ALL users that haven't loaded any pages in the last 300 seconds to offline or field online to 0, which obviously means the user is offline Third function sets ALL users that have loaded pages in the last 300 seconds to online if they're offline The reason I'm worried about this is not because it doesn't work, it's working fine at the moment with 2 users, I'm worried when their might be A LOT of users and there are people who are active. My question is : will it make my scripts really slow or be bad for my database in ANY way if I do use these functions for ALL users because each time it searches the table, it is searching ALL users to see if they're active or inactive. Hey, guys. I'm a programmer and i've designed an application to assist in an online game. When users visit my website for whatever reason i want to display the amount of users currently using my application. As you may have guessed i'm not to crash hot at php or any web related languages at that . What i think will work, is when my application starts up, i can make it visit a webpage in the background and just idle there while they are using the program, when they close the program obviously it is no longer viewing that webpage. So if there some sort of php counter i can make for this? I was reading some sources to php counters and they count the amount of "sessions" and display that, i was just curious if my program navigates to a webpage upon opening and it doesn't do anything will this 'time it out'? Anyway, any help is appreciated. Thanks. hi, i have made a website where people resgister their details of them and products. they have to enter the following details in form Name of company name of the product company address email id password mobile number contact and brief details about their company
user can then login with email id and pwd. now after login ..user will get a page where he can upload the photos of products images and their price, so now my question is that when he finishes uploading (|by clicking on upload button) the product images and price text box ..then on final uploaded webspage it should show all other things which he registerd before (company name , mobile number etc) along with images and price...hence the main question that user does not need to enter mobile and address while uploading images and filling proce ..but on the final page it should show mobile and address along with price and images..as user is not going to enter mobile and address again and again as he will have multiple products to upload.
I would appreciate your assistance, there are tons of login scripts and they work just fine. However I need my operators to login and then list their activities for the other operators who are logged in to see and if desired send their clients on the desired activity. I have the login working like a charm and the activities are listed just beautifully. How do I combine the two tables in the MySQL with PHP so the operator Logged in can only make changes to his listing but see the others. FIRST THE ONE script the member logges in here to the one table in MSQL: <?php session_start(); require_once('config.php'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } $login = clean($_POST['login']); $password = clean($_POST['password']); if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'"; $result=mysql_query($qry); if($result) { if(mysql_num_rows($result) == 1) { session_regenerate_id(); $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['member_id']; $_SESSION['SESS_FIRST_NAME'] = $member['firstname']; $_SESSION['SESS_LAST_NAME'] = $member['lastname']; session_write_close(); header("location: member-index.php"); exit(); }else { header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?> ................................................. ................................ Now I need the person who logged in to the table above to be able to make multiple entries to the table below <? $ID=$_POST['ID']; $title=$_POST['title']; $cost=$_POST['cost']; $activity=$_POST['activity']; $ayear=$_POST['aday']; $aday=$_POST['ayear']; $seats=$_POST['special']; $special=$_POST['seats']; mysql_connect("xxxxxx", "xxx350234427", "========") or die(mysql_error()); mysql_select_db("xxxx") or die(mysql_error()); mysql_query("INSERT INTO `activity` VALUES ('ID','$title', '$cost','$activity', '$aday', '$ayear', '$special', '$seats')"); Print "Your information has been successfully added to the database!" ?> Click <a href="member-profile.php">HERE</a> to return to the main menu <?php ?> Actually, what i want to do is to use the email to fetch the $email,$password and $randomnumber from database after Hi, so far I have managed to set up a somewhat basic login website with a mysql database backend. Once they have logged on they go to a "main menu" page. What I need to define is that user A sees button A but only that button, etc. (Then of course that same rule would have to apply if they tried to directly go to the page, but I am guessing I can do that in the same way that I currently do to force a login). If anyone has any tutorials or sample code I would much appreciate it. Thanks, Hi, I am getting frustrated beyond belief at the moment with trying to get a very simple script to run, I am using PHP 5.3.3 and MySQL 5.1 on a Win2k8 server with IIS7.5. Basically my script is connecting to a local database, running a single select query, returning those rows and building up a string from them. The problem is that I am receiving complete BS responses from PHP that the access is denied for the user being specified. This is complete rubbish since the user can connect via mysql, sqlyog, ASP.NET MVC without issue but for some bizarre reason it is not working via PHP. The code for the script is here : Code: [Select] <?php $mysql = mysql_connect('127.0.0.1:3306', 'myuser', 'mypass', 'mydatabase'); if (!$mysql) { die(mysql_error()); $content = "<nobr></nobr>"; } else { $result = mysql_query('SELECT * FROM tblEventGroup'); $content = "<nobr>"; if ($result) { while($row = mysql_fetch_assoc($result)) { $content .= "<span>"; $content .= $row['GroupName']; $content .= "</span>"; $content .= "<a href=\"../Event/EventSearch?groupid="; $content .= $row['GroupId']; $content .= "\" target=\"_blank\">Book here</a> "; } } mysql_close($mysql); $content .= "</nobr>"; } ?> I cannot for the life of me understand what the problem is, the return error is Access denied for user 'myuser'@'localhost' (using password: YES) Hi guys, I am trying to put together a little system that allows users to log onto my website and access there own personal page. I am creating each page myself and uploading content specific to them which cannot be viewed by anyone else. I have got the system to work up as far as: 1/ The user logs in 2/ Once logged in they are re-directed to their own page using 'theirusername.php' Thats all good and working how I need it too. The problem I have is this. If I log onto the website using USER A details - I get taken to USER A's page like I should but - If I then go to my browser and type in USERBdetails.php I can then access USER B's page. This cannot happen!! I need for USER A not to be able to access USER B profile - there is obviously no point in the login otherwise! If you are not logged in you obviously cannot access any secure page. That much is working! Please find below the code I am using: LOGIN <?php session_start(); function dbconnect() { $link = mysql_connect("localhost", "username", "password") or die ("Error: ".mysql_error()); } ?> <?php if(isset($_SESSION['loggedin'])) { header("Location:" . strtolower($username) . ".php"); if(isset($_POST['submit'])) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $mysql = mysql_query("SELECT * FROM clients WHERE username = '{$username}' AND password = '{$password}'"); if(mysql_num_rows($mysql) < 1) { die("Password or Username incorrect! Please <a href='login.php'>click here</a> to try again"); } $_SESSION['loggedin'] = "YES"; $_SESSION['username'] = $username; $_SESSION['name'] header("Location:" . strtolower($username) . ".php"); } ?> HEADER ON EACH PHP PAGE <?php session_start(); if(!isset($_SESSION['loggedin'])) { die(Access to this page is restricted without a valid username and password); ?> --------------------------------------------------- Am I right in thinking it is something to do with the "loggedin" part? The system I have here is adapted from a normal login system I have been using for years. The original just checks the details and then does a 'session start'. This one obviously has to re-direct to a user specific page. To do this I used the <<header("Location:" . strtolower($username) . ".php");>> line to redirect to a page such as "usera.php" or "userb.php" Any help would be greatly appreciated! Ta im creating a members website and i want to show how many people are logging, in my database i have a col named online every time somone logs in there online goes from 0 to 1 and when thay log out it goes back to 0. i need to know how to show the total people that have 1 in there online part of the data base iv tryed this code and its not working <?php $result = mysql_query("SELECT online FROM `members` WHERE online='1'"); $row = mysql_fetch_row($result); echo $row; ?> this works in the sql console in phpmyadmin Hallo everybody,
the user is in the table, but i get error (user not found!).
thank you very much for your help
Rafal
<!DOCTYPE html> <html> <head> <title>index</title> <meta http-EQUIV="CONTENT-LANGUAGE" content="en"> <?php SESSION_START(); include("abc.php"); $link2 = mysqli_connect("$hoster", "$nameuser", "$password", "$basedata") or die ("connection error" . mysqli_error($link2)); $email = $_POST["inp_email"]; $pwd = $_POST["inp_pwd"]; if($email && $pwd) { $chkuser = mysqli_query("SELECT email FROM $table2 WHERE email = '$email' "); $chkuserare = mysqli_num_rows($chkuser); if ($chkuserare !=0) { $chkpwd = mysqli_query("SELECT pwd FROM $table2 WHERE email = '$email'"); $pwddb = mysqli_fetch_assoc($chkpwd); if (md5($pwd) != $pwddb["pwd"]) { echo "Password is wrong!"; } else { $_SESSION['username'] = $email; header ('Location:list.php'); } } else { echo "user not found!"; } } else { echo "enter your Email and Password!"; } mysqli_close($link2); ?> </head> <body style="font-family: arial;margin: 10; padding: 0" bgcolor="silver"> <font color="black"> <br> <form action="index.php" method="post"> <b>Login</b><br><br> <table width="100%"> <tr><td> Email:<br><input type="text" name="inp_email" style="width:98%; padding: 4px;"><br> Password:<br><input type="password" name="inp_pwd" style="width:98%; padding: 4px;"><br> <br> <input type="submit" name="submit" value="Login" style="width:100%; padding: 4px;"> </td></tr> </table> </form> </font> </body> </html> Hello, i've got some shop script which has 2 payment modules which i'd like to use for something else, the payment modules only work if the user is logged in though, i tried to make them standalone scripts but that didn't work out too well. So now i decided to go another way and just let everyone have the same session so everyone will be using the same username&password automatically. the index file looks like this: Code: [Select] <?php include('./inc/config.php'); include('./inc/functions.php'); include('./lang/'.$language.'.lng'); $id = addslashes($_REQUEST["id"]); $user = addslashes($_REQUEST["username"]); $pass = addslashes($_REQUEST["password"]); $language = strtolower($language); if(empty($id)) $id =1; $file = mysql_query('SELECT * FROM navi_'.$language.' WHERE id="'.$id.'"'); if(mysql_num_rows($file)>0) $file = mysql_fetch_array($file); else $file = mysql_fetch_array(mysql_query('SELECT * FROM navi_'.$language.' WHERE id="404"')); if(!empty($user) AND !empty($pass)) {$query = mysql_query('SELECT * FROM users WHERE username="'.$user.'" AND pass="'.md6($pass).'"'); if(mysql_num_rows($query) == 1) {$_SESSION[$session_prefix."user"] = ucfirst($user); echo'<meta http-equiv="refresh" content="0; url=index.php?id=8">';} else $error = 'Username oder Passwort ist falsch.';} include('./designe/'.$designe.'/head.tpl'); include('./designe/'.$designe.'/navi.php'); include('./designe/'.$designe.'/middle.tpl'); if(file_exists('./pages/'.$file["file"])) {echo'<h1>'.ucfirst($file["title"]).'</h1>'; include('./pages/'.$file["file"]);} if(!empty($error)) echo '<font color="red">'.$error.'</font>'; include('./designe/'.$designe.'/foot.tpl'); ?> Now i tried alot of things including adding: Code: [Select] session_start(); $_SESSION["username"] = "peter"; $_SESSION["user"] = "peter"; $_SESSION["id"] = "1"; $_SESSION["pass"] = "peter"; $_SESSION["password"] = "peter"; or Code: [Select] $id = "1"; $user = "peter"; $username = "peter"; $pass = "peter"; $password = "peter"; also a combination of both, nothing works, but i don't understand why ? Any help is appreciated. /Edit, i tried adding it to the paymentmodule .php aswell, but no luck. Hallo everybody,
i have the following code.
but i get allways this error while the user exist in the database.
User not found!
what do i do wrong?
thank you very much for your help
Rafal
<html> <head> <?php $connection = mysql_connect("db.xyz.com", "username", "password") or die ("connection fehler"); mysql_select_db("db0123456789") or die ("database fehler"); $email = $_POST["inp_email"]; $pwd = $_POST["inp_pwd"]; if($email && $pwd) { $chkuser = mysql_query("SELECT email FROM gbook WHERE email = '($email)' "); $chkuserare = mysql_num_rows($chkuser); echo $email; echo $pwd; if ($chkuserare !=0) { $chkpwd = mysql_query("SELECT pwd FROM gbook WHERE email = '($email)' "); $pwddb = mysql_fetch_assoc($chkpwd); if ($pwd != $pwddb["pwd"]) { echo "password is wrong!"; } else { echo "login successed"; } } else { echo "User not found!"; } } else { echo "Pleas enter your email and password!"; } mysql_close($connection); ?> </head> <body> <form action="login.php" method="post"> Email <input type="text" name="inp_email"><br> Password <input type="text" name="inp_pwd"><br> <input type="submit" name="submit" value="login"> </form> </body> </html> Edited by rafal, 21 September 2014 - 04:33 PM. i want online quiz in php mysql Hi, i'm trying to create some detailed statistics about customer activity, i have entries in my mysql db when the customer has been active for the last time and want to create some statistics about that, basically a "online last 24 hours" but from specific countries. Now i've tried this: Code: [Select] $time = date('Y-m-d H:i:s'); $time24 = date("Y-m-d H:i:s", time()-((60*60)*24)); $query = "SELECT COUNT(*) as Anzahl FROM customers WHERE country = 'de' AND time BETWEEN '$time' AND '$time24' "; to get the current date and select all customers that have been available from the current date minus 24 hours, what's my mistake here, as this doesn't seem to work! Thanks "vijdev and 0 Guests are viewing this topic." can someone help me with a pointer on how to get started with something like the above message when viewing different pages in the site...it must be page/topic specific. is this using sessions/cookies/databases?...? |