Php using a variable on other pages through sessions

Question

Hello guys i have a problem with php sessions. I try to send a variable from my second page to my third page but i keep getting undefined variable i will include code from all three pages i am sorry in advance for the pretty large amount of code but i belive it will be easier for you guys to help then. Thanks in advance

PAGE 1

<form method="POST" action="authorize.php">
            <input type="text" name="username" class="username"><br>
            <input type="password" name="password" class="password"><br>
            <label for ="remeberMe">Remember me!</label>
            <input type="checkbox" name="remeberMe" class="rememberMe"><br>
            <input type="submit" name="submitLogin" class="submitLogin" value="Login!">
        </form>

PAGE 2

    session_start();
    include 'connection.php';

    $user = $_POST['username'];
    $pass = $_POST['password'];

    $sql = mysqli_query($con, "SELECT * FROM users WHERE screename = '$user' AND ...");
    $count = mysqli_num_rows($sql);

    if ($count == 1)
    {
        $login = true;
    }
    else
    {
        $login = false;
    }   

    if ($login == true)
    {
        $_SESSION['access'] = true;
        $_SESSION['username'] = $_POST['username']; 
        header("Location:"."mainpage.php?".SID);
        exit;
    }   
    else 
    {
        header("Location:"."index.php?");
        exit;
    }

PAGE 3

    session_start();

    if($_SESSION['access'] != true)
    {
        include("index.php");
        exit;
    }

    if(isset($_POST['username']))
    {
        $user = $_POST['username'];
    }

ERROR: $user on page 3 is undefined

Sidenote: This line is unfinished $sql = mysqli_query($con,... - bad paste? — Funk Forty Niner
– Funk Forty Niner, Commented Nov 11, 2014 at 17:06
Which file is authorize.php? And which values/ variables you want to pass torught which files? — Rizier123
– Rizier123, Commented Nov 11, 2014 at 17:06
$sql = mysqli_query($con, "SELECT * FROM users WHERE screename = '$user' AND .... is that your code? — itachi
– itachi, Commented Nov 11, 2014 at 17:06
@AeroX I saw that. Sometimes when there's info missing from a query, could have some bearing on the outcome. For instance, if the missing query code is like WHERE screename = '$user' AND column = '".$_SESSION['var']."' (wink). I just like to see all the parts. — Funk Forty Niner
– Funk Forty Niner, Commented Nov 11, 2014 at 17:18

danmullen · Accepted Answer · 2014-11-11 17:10:32Z

2

You're accessing a POST variable rather than the one you saved in your session. Change to:

if(isset($_SESSION['username']))
{
    $user = $_SESSION['username'];
}

answered Nov 11, 2014 at 17:10

danmullen

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Rizier123 · Accepted Answer · 2014-11-11 17:16:31Z

1

So first you have to limit your SQL Statement with LIMIT 1 or change this condition to >=:

if ($count == 1) //'>='
{
    $login = true;
}
else
{
    $login = false;
}

After that you have to change on page 3 this to the $_SESSION array:

if(isset($_POST['username'])) //to `$_SESSION['username']`
{
    $user = $_POST['username']; //to `$_SESSION['username']`
}

answered Nov 11, 2014 at 17:16

Rizier123

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2 Comments

Anoxy Over a year ago

Worked perfect thanks! and thanks to everyone else that contributed! if you dont mind me asking why did i made a difference that i changed it to >=1 ? :)

Rizier123 Over a year ago

If your not limiting your query the count could be 2 or 3 so the condition to == 1 wouldn't be true! So if you change it to >= your sure that the query has at min. 1 row or more!

A bit new · Accepted Answer · 2014-11-11 17:13:41Z

1

The undefined variable, are you referring to the $_POST['username'] or the session variable in page 3? The post variable would not be retained in page 3, you need to refer to the session variable $_SESSION['username'].

answered Nov 11, 2014 at 17:13

A bit new

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