Insert value of a radio button into the database

Question

I have:

<label class="">
      <input name="user_perfil" type="radio" value="administrador">  Administrador
    </label>
    <label class="">
      <input name="user_perfil" type="radio" value="cliente">  Cliente
    </label>

And in my database I have a column:
enum('administrador', 'cliente', 'suporte_tecnico')

I’m picking up the POST that contains the value of the radio button:
$perfil = $_POST['user_perfil'];

And I’m calling a function:

function novoUsuario($conexao, $user, $password, $email, $nome, $perfil)
{
   $query = "INSERT INTO usuarios (user_name, user_password_hash, user_email,     user_fullname, rank)
            VALUES ('{$user}', '{$user_password_hash}', '{$email}', '{$nome}', '{$perfil}')";
            return mysqli_query($conexao, $query);
}

But in my bank the rank value is not being saved. I will use this rank to open specific pages for each value (Adm, support, etc).

What I’m doing wrong my friends?

Answer 1

Caesar, it seems that you are doing everything right. I found no mistake. I am understanding that only the value of the variable $perfil is not being inserted into mysql. I suggest doing a debug.

1. Let’s see if you’re getting the variable value correctly.
2. Let’s test the entry in the database with some value.

After these results we can determine where the error is.

Use echo "variável=>".$_POST['user_perfil']; If you see the value in the browser, the radio value recovery is ok. If not, the value is not being recovered correctly.

Right after on the next line use $perfil = "teste"; If you do not see the test value in your database table, it is pq the problem if you find the insertion. If you see test in your database table the problem is in the previous phase.

Another thing. Review the size of sweep(?) in creating the table.

for vc check the error in your insertion advise using mo Mysqli error manager Adapt this code to your function if you haven’t already done so:

if (!mysqli_query($con,"INSERT INTO .... "))
  {
     echo("Erro: " . mysqli_error($con));
  }