Display login user name

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Display login user name

Asked 7 years, 2 months ago

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Good afternoon, I’m early in php and would need a help.

I have this file (index.php) where I want to show the name of the logged in user on the screen after his login, however this giving an undefined variable error ( $_SESSION).

Note: Disregard the last div=main line

            $usuario   = $_SESSION['usuarioSession'];
            $senha     = $_SESSION['senhaSession'];

                $sql = mysql_query("SELECT * FROM syslogin WHERE usuario = '$usuario' AND senha = '$senha'");

                while($linha = mysql_fetch_array($sql)){
                    $usuario = $linha['usuario'];
                    $senha   = $linha['senha'];
                    }
     ?>
    <div class="bemvindo">Bem Vindo! <strong><?php echo $usuario;?></strong> | Hoje é: <?php echo date('d/m/Y');?></div><!--Bem Vindo-->

    <div id="principal">
                <h3>Bem Vindo(a).</h3><br /><br />
                <img src="css/img/logo1.jpg" alt="logo" />
    </div> <!-- Fim da div#principal -->

    <?php include('includes/fimerodape.php'); ?>

Please where I’m going wrong?

That code you put out is the whole page? Remember that to work with SESSION you need to initialize it: session_start(); at the beginning of the page.

– Bsalvo

2017/07/26 at 19:55
Be careful that the code you are using does not protect you against attacks from mysql Injection. At least pass inputs through the function mysql_real_escape_string. And consider switching to mysqli or Prepared statements

– Isac

2017/07/26 at 20:13

1 answer

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by Francisco • **7,472** points · Answer 1 · 2017-07-26T20:05:26+00:00

That’s why you didn’t start the session, do it with the session_start():

session_start();
$usuario   = $_SESSION['usuarioSession'];
$senha     = $_SESSION['senhaSession'];

I also recommend that you make a if to check if the user has Session.

if(isset($usuario)){
    echo "<div class='bemvindo'>Bem Vindo! <strong>$usuario</strong> | Hoje é: ".date('d/m/Y')."</div>";
}
else{
    echo "Você não está logado!";
}