Warning: mysqli_query() expects at least 2 Parameters, 1 Given in C: wamp www Trabalho odesi proceslogin.php on line 22

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<?php

// Busca as variaveis $login e $senha 

$login=$_POST['login']; 
$senha=$_POST['senha']; 

// Conecta ao servidor e selecione a database.
 $conn= new mysqli("localhost","root","","registro");
if($conn->connect_error){
die("Falhou a ligação: ".$conn->connect_error);
}
// Proteção contra  MySQL injection 
//$login = stripslashes($login);
//$senha = stripslashes($senha);
//$login = mysqli_real_escape_string($login);
//$senha = mysqli_real_escape_string($senha);

    $result=mysqli_query("SELECT utilizador, senha FROM utilizadores WHERE utilizador='$login' and senha='$senha'");

        $_num=mysqli_num_rows($result);
        if($_num < 1){
            echo " Não esta registado  <br>  faça login <br>";
        }else{
            header;
        }
    exit;
?
php mysqli
  • 1

    Miguel, taking the heat of his question, has even been answered. Since you are using mysqli, I recommend using Prepared Statement and not concatenating variables directly into SQL. I recommend reading: http://mattbango.com/notebook/code/prepared-statements-in-php-and-mysqli/

1 answer

1

You must change the following line:

$result=mysqli_query("SELECT username, password FROM users WHERE username='$login' and password='$password'");

for:

$result=mysqli_query($conn, "SELECT utilizador, senha FROM utilizadores WHERE utilizador='$login' and senha='$senha'");

Notice that before the sql I added the variable $conn. You must pass the connection used and the sql to execute.

You can see more details Clicking Here

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