PHP Prepare method not working when calling it twice?

Question

I am using the prepare method as follows:

$db= new mysqli("localhost","***","***","***");

if ($db->connect_error)          


        

Answer 1

You need read $stmt->error;

To read, you need rewrite code as in the example below.

...
$stmt = new mysqli_stmt($db);
if($stmt->prepare('SELECT name FROM table WHERE id = ? '))
{
  $stmt->bind_param('i', $id);
  $stmt->execute();
  echo "Success
";
}
else {
  var_dump($stmt->error);
}
...