Comments by Member 14024497 (Top 3 by date)

yeah php code already fix

but still getting same error it's so frustrating can't find the answer any place

<?php
session_start();
error_reporting(0);
echo '<meta charset="UTF-8">';
$host_db = "hehe";
$user_db = ":D";
$pass_db = "D:";
$db_name = "=DDD";
$conexion = new mysqli($host_db, $user_db, $pass_db, $db_name);
if ($conexion->connect_error) {
die("La conexion falló: " . $conexion->connect_error);
}

$sql = "SELECT * FROM Usuarios where Correo = '".$_POST['Email']."' and Password = '".$_POST['Password']."'";
$result = $conexion->query($sql);
if ($result->num_rows > 0) {
$row = mysqli_fetch_array($result);

$response['Email'] = $row['Correo'];
$response['Password'] = $row['Password'];
}
else
{
$response['Email'] = 'Error';
$response['Password'] = 'Error';
}
echo json_encode($response);
mysqli_close($conexion);
?>

can you be more specific?

the data is being sent to a URL which is linked to my PHP code
i'm consulting the data from xcode using json
let myURL = URL(string: "hehe")