I cannot receive the values of a PHP form via POST

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I’m having a problem in a php form that I don’t know how to fix. I have a file called login.php that contains a form that sends the data via POST to the.php data file. In the.php data I do the necessary processing. It turns out that whenever I try to simulate a login, I get the following message: Notice: Undefined variable: usuario Notice: Undefined variable: pass My login.php code is this:

<!DOCTYPE html>


<form  method="POST" action="dados.php">
    <label>Login</label>
    <input type="text" name="usuario" class="form-control" id="usuario" autocomplete="off" autofocus required>

    <label for="senha">Senha</label>
    <input type="password" class="form-control" id="pass" name="pass" autocomplete="off" required>

    <input type="submit" value="Login" id="entrar" name="entrar">
</form>

    <!-- Scripts -->
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript" src="js/ocultarSenha.js"></script>

My.php data code is this:

<?php
    // Conexão com o banco
    $bdServidor = '127.0.0.1';
    $bdUsuario = 't3carvvo_MatheusEdnei';
    $bdSenha = 't3carvvo_#';
    //$bdUsuario = 'root';
    //$bdSenha = 'root';
    $bdBanco = 't3carvvo_pessoas';

    $conexao = mysqli_connect($bdServidor, $bdUsuario, $bdSenha, $bdBanco);

    if(mysqli_connect_errno($conexao)) {
        echo "Problemas para conectar no banco. Erro:";
        echo mysqli_connect_errno;
        die();
    }

    if (isset($_POST['usuario'])) {
        $usuario = $_POST['usuario'];
        echo "entrei no usuario  ";
        echo $usuario;
    } else {
        echo "não entrei no usuario";
    }
    if (isset($_POST['pass'])) {
        $pass = MD5($_POST['pass']);
        echo "entrei na senha   ";
        echo $pass; 
    } else {
        echo "não entrei na senha";
    }


    $contem = strpos($usuario, '@');
    if($contem === false){
        echo "cpf";

    } else {
        $sqlSelecionarEmail = "SELECT email FROM pessoa WHERE email = '$usuario'";
        $selecaoEmail = mysqli_query($sqlSelecionarEmail,$conexao);
        $arrayEmail = mysqli_fetch_array($selecaoEmail);
        print($arrayEmail);
        $loginarray = $arrayEmail['email'];

        $sqlSelecionarSenha = "SELECT senha FROM pessoa WHERE senha = '$pass'";
        $selecaoSenha = mysqli_query($sqlSelecionarSenha,$conexao);
        $arraySenha = mysqli_fetch_array($selecaoSenha);
        $senhaArray = $arraySenha['senha'];

        if($loginarray === $usuario and $senhaArray === $pass ) {
            echo"
            <script language='javascript' type='text/javascript'>
                alert('Login e senha corretos');
                //window.location.href='dados.php';
            </script>
            ";
        } else {
            echo"
            <script language='javascript' type='text/javascript'>
                alert('Login ou senha incorretos');
                window.location.href='dados.php';
            </script>
            ";
        }

    }

?>

Edit 1:

I modified the code a little bit and now the return is as follows:

Notice: Undefined index: usuario in C: xampp htdocs t3carvvo_tarefa1 dados.php on line 31

Notice: Undefined index: pass in C: xampp htdocs t3carvvo_tarefa1 dados.php on line 32

Notice: Undefined index: usuario in C: xampp htdocs t3carvvo_tarefa1 dados.php on line 33

php web-application form post

  • it does not display the line where is giving the error?

  • It prints the "I didn’t sign in to the user" message indicating that isset($_POST['user']) returns an empty value. If I withdraw this check it informs me that there is an error when I do : $user = $_POST['user'];

  • tried to give a var_dump($_POST) , to see if you’re afraid of something?

  • I did it now and it returned to me: array(0) { } . Looking at the Chrome developer tools in the network tab I noticed that the "Request method" is indicating that the request was of type GET, even putting the form as POST.

1 answer

0

I ran your code and found the following errors:

On the lines where you use the function mysqli_query the function parameters are exchanged, the first parameter should be $conexão and the second $sqlSelecionarEmail.

Wrong:

$selecaoEmail = mysqli_query($sqlSelecionarEmail,$conexao);

Right:

$selecaoEmail = mysqli_query($conexao, $sqlSelecionarEmail);

Another mistake is on the line where you do print($arrayEmail);. Here the error is that you are passing an array but the function printf expects a string as parameter.

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