I can not print the value of a $ _Session with an echo

1

I have a login authentication system that is this:

<?php
if(isset($_POST['txtpass'])) 
{
    session_start();
    //variable de conexion: recibe dirección del host , usuario, contraseña y el nombre base de datos
    $mysqli = new mysqli("localhost", "root", "", "bdpersona") or die ("Error de conexion porque: ".$mysqli->connect_errno);
    // comprobar la conexión 
    if (mysqli_connect_errno()) 
    {
        printf("Falló la conexión: %s\n", mysqli_connect_error());
        exit();
    }

    $login = $mysqli->real_escape_string($_POST['txtlogin']);   
    $pass = $mysqli->real_escape_string($_POST['txtpass']);

    $resultado = $mysqli->query("SELECT * FROM tbusuario where login='$login' and pass='$pass' and activo!=0");
    $valida=$resultado->num_rows;
    if($valida != 0)
    {
        $datosUsu = $resultado->fetch_row();
        $_SESSION['nombreusu'] = $datosUsu[3];
        $_SESSION['perfil'] = $datosUsu[4];             
        echo "<META HTTP-EQUIV='Refresh' CONTENT='0; URL=listar.php'>";
    }
    else
    {
        echo 
        "<script> 
            var textnode = document.createTextNode('Usuario ó Password Incorrecto');
            document.getElementById('msg').appendChild(textnode);
        </script>";     
    }   
}

?>

As you can see, if I accept the login and pass correctly, I take two values from the table where the login is and pass it and I save it in a variable session and I go to a page .php

$_SESSION['nombreusu'] = $datosUsu[3];
$_SESSION['perfil'] = $datosUsu[4];         

When I go to the next page, which is list.php, I have this:

<?php
session_start();
if(isset($_SESSION['nombreusu']))
{

?>
<!----Todo el código----> 

<?php
}
 else
{
?>
     <META HTTP-EQUIV="Refresh" CONTENT="0; URL=index.php">
<?php
}
?>

Well, in "All the code", I want to print $_SESSION['perfil']

And I do this:

  <?php 
  $quieneres= $_SESSION['perfil']
  echo $quieneres;
  ?>

And I get the following error:

Parse error: syntax error, unexpected 'echo' (T_ECHO) in C:\xampp\htdocs\fondomarino\listar.php on line 419

Why can not I see the value per screen of the Session variable ???

Thank you very much.

    
asked by Vidal 22.03.2018 в 11:24
source

1 answer

3

You are missing the ; at the end of the first line, try this:

 <?php 
  $quieneres= $_SESSION['perfil'];
  echo $quieneres;
  ?>
    
answered by 22.03.2018 / 11:41
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