Warning: mysqli_query () expects at least 2 parameters, 1 given in C: \ wamp \ www \ ProcessWork \ proceslogin.php on line 22 [closed]

Navigation
-4
<?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;
?
    
asked by anonymous 11.03.2015 / 12:16

1 answer

1

You should change the following line:

  

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

To: 

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

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

You can see more details Clicking Here

    
11.03.2015 / 12:26
How JSON works on a project [closed] Doubt with college work in javascript using DOM [closed]