Question

$data = $mysqli->prepare("SELECT amount FROM items WHERE id=:id");
echo 'forward1';
if(!$data->execute(array(':id' => $id)))
    die("error executing".$data->error);
echo '2';
$row = $data->fetch_object();
die('Losing my mind'.$row->amount);

This will only echo “forward1”, not “error executing…” or “2”. It works with *$mysqli->query”. If I add quotes ” to :id in the query, it will echo “forward1error executing”.

Answer 1

First, make sure you understand the prepared statements syntax and working model.

As in:

$data = $mysqli->prepare("SELECT amount FROM items WHERE id=(?)");
            // THIS ^^ actually "prepares" an object to be used in the statement
$data->bind_param("i",$id)
            // ...then you "bind" the parameter for your statement as "i"(nteger)
echo 'forward1';
if(!$data->execute())  // And now you simply run it, with no other args
    die("error executing".$data->error);
echo '2';
$row = $data->fetch_object();
die('Loosing my mind'.$row->amount);

I suggest though using something more like

$data->execute() or die("error executing".$data->error);

The main steps of a prepared statement are:
1. Prepare the query with some placeholder values;
2. “Bind” the required number of values to the query;
3. Execute it!

I fail to see why this is relevant in your case, with such a simple query. I also assume you actually need it for something bigger.
Please let me know if I misunderstood your point or code sample.

Oh, and.. have fun! 🙂