enter data to mysql with form

$link = mysqli_connect("localhost", "Root", "MyRootPassword", "MyDbName");

        #Usa esta Verificacion
        if (isset($_POST['username']) && isset($_POST['password']) && isset($_POST['name']) && isset($_POST['email'])) {

            #Asignalas a una variable asi
            $usuario_login =$_POST['username'];
            $usuario_passwd =$_POST['password'];
            $usuario_nombre =$_POST['name'];
            $usuario_email =$_POST['email'];

            $sql = "INSERT INTO members (username, password, name, email) VALUES    ('$usuario_login', '$usuario_passwd', '$usuario_nombre', ' $usuario_email')";

            $query = mysqli_query($link, $sql);
            if (!$query) {
                echo "Error";
            } else {
                echo "Success";
            }
        } else {
            echo "Parametros vacios. Verifique POST del Form";
        }

It's not mysql error, it's php error where you use $ _ POST ['username'] nothing is coming to you.

suggestions,

1) to debug use the GET method and you will see in the address bar what is coming to you. 2) use $ _REQUEST instead of $ _POST works with GET and POST 3) Do not use the direct values because they can make you a sql injection

You can use a function to retrieve the $ _RQUEST data something like

function traerParametro( $nombre ){
  if ( array_key_exists( $nombre, $_REQUEST ) )
  {
    return funcionQueEviteSQLInjection( $_REQUEST[ $nombre ] );
  }
  return '';
}

I can not find your error, but I pass a code I did a while ago, it's php Oriented to objects I think it's easier and more efficient. I hope you serve

Class responsible for managing connections to the database

class DB {

private $servidor; 
private $usuario; 
private $password; 
private $bd; 
private $link; 



//Constructor
function __construct(){
    $this->servidor = 'localhost';
    $this->usuario = 'root';
    $this->password = '';
    $this->bd = 'libreria';
    $this->conectar(); /*Al inicializar la clase ya estará conectada*/ 
} 

//Destructor   
function __destruct() {
    if (isset($this->link) && is_resource($this->link)) {
        mysql_close($this->link);
    }
}

/*Realiza la conexión a la base de datos.*/ 
private function conectar(){ 
  $this->link=mysql_connect($this->servidor, $this->usuario, $this->password);
  if (!$this->link){
       die('Error de coneccion' . mysql_error());  
  }
  else{
        mysql_select_db($this->bd,$this->link); 
        mysql_query("set names 'utf8'");
  }
}

Execute a query returns result

private function ExecuteQuery($query){
    $resultado = mysql_query($query,$this->link);
    if ($resultado)
        return $resultado;
    else
        return false;
} 

Execute a query and return no result

private function ExecuteNonQuery($query){
    if ( (mysql_query($query,$this->link)) == true ){
        return true;
    }
    else{
        return false;
    }
}

Simple insert with parameters

public function InsertarLibro($titulo, $paginas,$precio,$foto,$genero,$autor)
{
    $query="INSERT INTO libros(titulo,paginas,precio,foto,idGenero,idAutor) VALUES('$titulo', $paginas,$precio,'$foto',$genero,$autor)";
    $this->ExecuteQuery($query);
    $id = mysql_insert_id();
    return $id;
}

I hope you serve

The error is telling you that the variables are not defined and in the php code the problem is visible, all the values are being recorded in the same variable $nombre

$nombre = isset($_REQUEST['user_nombre']) ? $_REQUEST['user_nombre'] : '';
$nombre = isset($_REQUEST['user_apellido']) ? $_REQUEST['user_apellido'] : '';
$nombre = isset($_REQUEST['user_email']) ? $_REQUEST['user_email'] : '';
$nombre = isset($_REQUEST['user_telefono']) ? $_REQUEST['user_telefono'] : '';
$nombre = isset($_REQUEST['user_usuario']) ? $_REQUEST['user_usuario'] : '';
$nombre = isset($_REQUEST['user_password']) ? $_REQUEST['user_password'] : '';
$nombre = isset($_REQUEST['user_token']) ? $_REQUEST['user_token'] : '';

You are not defining the other variables. On the other hand I recommend that if you receive the data by POST then you get them through $_POST

----------------------------------------------- -------- EDIT: ---------------------------------------- ---------------

@jolunavi, the problem is that you are using some variable names to get the POST data and other variable names in the VALUES at the time of insertion.

If you use these variable names:

        $nombre =$_POST['user_nombre'];
        $apellido =$_POST['user_apellido'];
        $email =$_POST['user_email'];
        $telefono =$_POST['user_telefono'];
        $usuario =$_POST['user_usuario'];
        $password =$_POST['user_password'];

Then you must correct by changing this line:

$sql = "INSERT INTO usuarios (user_nombre,user_apellido,user_email,user_telefono,user_usuario,user_password) VALUES ('$username','$apellido','$email','$telefono','$usuario','$password')";

For this:

$sql = "INSERT INTO usuarios (user_nombre,user_apellido,user_email,user_telefono,user_usuario,user_password) VALUES ('$nombre','$apellido','$email','$telefono','$usuario','$password')";

You should check the names of the fields very well, the order, and avoid changing the variables.

----------------------------------------------- -------- EDIT 2 ---------------------------------------- --------------

Thanks Adres,

you were right, it was that variable that gave problems

solved like this:

      <?php include_once('header.php'); ?>
      <h1>Control Register Usuario</h1> 
      <?php 
       if (isset($_POST['user_nombre']) && isset($_POST['user_apellido']) && isset($_POST['user_email']) && isset($_POST['user_telefono']) && isset($_POST['user_usuario']) && isset($_POST['user_password']) && isset($_POST['user_token']))
       {      
       # Asigno a una variable asi:
    if (!empty($nombre =$_POST['user_nombre']));
      if (!empty($apellido =$_POST['user_apellido']));
      if (!empty($email =$_POST['user_email']));
      if (!empty($telefono =$_POST['user_telefono']));
      if (!empty($usuario =$_POST['user_usuario']));
      if (!empty($password =$_POST['user_password']));
      if (!empty($password =$_POST['user_token']));

In this I have problems

How do I set up a control?

So if the user already exists ... that allows me to change my username or something like that

        # creí poder solucionar asi, pero no funciona
        if ($_GET['user_usuario']) == $_POST['user_usuario'])
        {
          echo "usuario ya existente"
         }

continuation ......

    //imprimo las variables para saber si llegan o no
    $nombre =   $_POST['user_nombre'];     echo "nombre: ";   echo $_POST ['user_nombre'];   echo "<br />";
    $apellido = $_POST['user_apellido'];   echo "apellido: "; echo $_POST ['user_apellido']; echo "<br />"; 
    $email =    $_POST['user_email'];      echo "email: ";    echo $_POST ['user_email'];    echo "<br />";
    $telefono = $_POST['user_telefono'];   echo "telefono: "; echo $_POST ['user_telefono']; echo "<br />"; 
    $usuario =  $_POST['user_usuario'];    echo "usuario: ";  echo $_POST ['user_usuario'];  echo "<br />"; 
    $password = $_POST['user_password'];   echo "password: "; echo $_POST ['user_password']; echo "<br />"; 
    $token = $_POST['user_token'];         echo "token: ";    echo $_POST ['user_token'];    echo "<br />";
    echo "Los Archivos se reciben perfectamente. <br />";

    $sql = "INSERT INTO usuarios (user_nombre,user_apellido,user_email,user_telefono,user_usuario,user_password) VALUES ('$nombre','$apellido','$email','$telefono','$usuario','$password')";
      $query = mysqli_query($coneccion, $sql);
       if (!$query) {
       echo "Error</br>";
       echo $sql;
       }
       else {
       echo "Success";
       }
       }
       mysqli_close($coneccion);
       ?>
       <?php include_once('footer.php'); ?>