How can I check if a user is logged-in in php?_问答_开发者

I'm pretty new to PHP and I am trying to figure out how to use sessions to check and see if a user is logged into a website so that they would have a开发者_StackOverflow社区uthorization to access specific pages.

Is this something that is complicated or is it because I am a noob that I can't figure it out?

Logins are not too complicated, but there are some specific pieces that almost all login processes need.

First, make sure you enable the session variable on all pages that require knowledge of logged-in status by putting this at the beginning of those pages:

session_start();

Next, when the user submits their username and password via the login form, you will typically check their username and password by querying a database containing username and password information, such as MySQL. If the database returns a match, you can then set a session variable to contain that fact. You might also want to include other information:

if (match_found_in_database()) {
    $_SESSION['loggedin'] = true;
    $_SESSION['username'] = $username; // $username coming from the form, such as $_POST['username']
                                       // something like this is optional, of course
}

Then, on the page that depends on logged-in status, put the following (don't forget the session_start()):

if (isset($_SESSION['loggedin']) && $_SESSION['loggedin'] == true) {
    echo "Welcome to the member's area, " . htmlspecialchars($_SESSION['username']) . "!";
} else {
    echo "Please log in first to see this page.";
}

Those are the basic components. If you need help with the SQL aspect, there are tutorials-a-plenty around the net.

Almost all of the answers on this page rely on checking a session variable's existence to validate a user login. That is absolutely fine, but it is important to consider that the PHP session state is not unique to your application if there are multiple virtual hosts/sites on the same bare metal.

If you have two PHP applications on a webserver, both checking a user's login status with a boolean flag in a session variable called 'isLoggedIn', then a user could log into one of the applications and then automagically gain access to the second without credentials.

I suspect even the most dinosaur of commercial shared hosting wouldn't let virtual hosts share the same PHP environment in such a way that this could happen across multiple customers site's (anymore), but its something to consider in your own environments.

The very simple solution is to use a session variable that identifies the app rather than a boolean flag. e.g $SESSION["isLoggedInToExample.com"].

Source: I'm a penetration tester, with a lot of experience on how you shouldn't do stuff.

Any page you want to perform session-checks on needs to start with:

session_start();

From there, you check your session array for a variable indicating they are logged in:

if (!$_SESSION["loggedIn"]) {
    redirect_to_login();
    die;
}

Logging them in is nothing more than setting that value:

$_SESSION["loggedIn"] = true;

You need this on all pages before you check for current sessions:

session_start();

Check if $_SESSION["loggedIn"] (is not) true - If not, redirect them to the login page.

if($_SESSION["loggedIn"] != true){
    echo 'not logged in';
    header("Location: login.php");
    exit;
}

You may do a session and place it:

// Start session
session_start();

// Check do the person logged in
if($_SESSION['username']==NULL){
    // Haven't log in
    echo "You haven't log in";
}else{
    // Logged in
    echo "Successfully logged in!";
}

Note: you must make a form which contain $_SESSION['username'] = $login_input_username;

Apart from the splendid answer von @hiburn8 what is missing is a secure why to logoff / logout. I realize this was not asked but should be considered as well.

Loging off should contain two elements:

remove all data from the session variable
reset the ID to prevent from session reuse attempts.

Perhaps 2) is not needed if all session data was destroyed but there might be (future) scenarios where this might be a good thing. Luckily PHP offers a function for both the elements and a complete code would be:

session_start();             // get a session to manipulate
session_unset();             // remove all server data on the session
session_regenerate_id(true); // create a new session id to prevent hijacking

This should typically go in a logoff script.

See this script for registering. It is simple and very easy to understand.

<?php
    define('DB_HOST', 'Your Host[Could be localhost or also a website]');
    define('DB_NAME', 'database name');
    define('DB_USERNAME', 'Username[In many cases root, but some sites offer a MySQL page where the username might be different]');
    define('DB_PASSWORD', 'whatever you keep[if username is root then 99% of the password is blank]');

    $link = mysql_connect(DB_HOST, DB_USERNAME, DB_PASSWORD);

    if (!$link) {
        die('Could not connect line 9');
    }

    $DB_SELECT = mysql_select_db(DB_NAME, $link);

    if (!$DB_SELECT) {
        die('Could not connect line 15');
    }

    $valueone = $_POST['name'];
    $valuetwo = $_POST['last_name'];
    $valuethree = $_POST['email'];
    $valuefour = $_POST['password'];
    $valuefive = $_POST['age'];

    $sqlone = "INSERT INTO user (name, last_name, email, password, age) VALUES ('$valueone','$valuetwo','$valuethree','$valuefour','$valuefive')";

    if (!mysql_query($sqlone)) {
        die('Could not connect name line 33');
    }

    mysql_close();
?>

Make sure you make all the database stuff using phpMyAdmin. It's a very easy tool to work with. You can find it here: phpMyAdmin

else if (isset($_GET['actie']) && $_GET['actie']== "aanmelden"){

    $username= $_POST['username'];
    $password= md5($_POST['password']);
    $query = "SELECT password FROM tbl WHERE username = '$username'";
    $result= mysql_query($query);
    $row= mysql_fetch_array($result);

    if($password == $row['password']){
            session_start();
            $_SESSION['logged in'] = true;
            echo "Logged in";

    }
}