Execute Array stored in a variable

Question

I stored an array in the variable $stringPedidos. I'm trying to use this variable as an array in $resp but it's not working.

I'm trying to display a keyboard that contains the product names depending on a user order ID extracted from the db. The $stringPedidos variable contains the array needed for displaying the keyboard at a Telegram Bot using PHP. I want to use $stringPedidos as an array, the problem is that $stringPedidos is executed as a string.

$query = mysqli_query($conn, "SELECT * FROM pedidos WHERE usuario_id='".$id."'");

      if(mysqli_num_rows($query) > 0){

        $stringPedidos= "array(";

        $query = "SELECT producto_id FROM pedidos WHERE usuario_id='".$id."'";
        $result = mysqli_query($conn, $query) or die(mysqli_error($conn));
        while ($row = mysqli_fetch_array($result, MYSQLI_BOTH)) {
            $producto_id = $row['producto_id'];

                 $queryProductos = "SELECT name FROM productos WHERE id='".$producto_id."'";
                    $resultProductos = mysqli_query($conn, $queryProductos) or die(mysqli_error($conn));
                    while ($rowProductos = mysqli_fetch_array($resultProductos, MYSQLI_BOTH)) {
                        $productoName = $rowProductos["name"];
                          $stringPedidos .= 'array("'.$productoName.'"),';


                                      }

        }

        $stringPedidos .= 'array("Return↩️")';
        $stringPedidos .= ")";


    $resp = array("keyboard" => eval($stringPedidos),"resize_keyboard" => true,"one_time_keyboard" => true,"remove_keyboard" => true);

}

Output: array(array("testProduct"),array("testProduct2"),array("Return↩️"))

The output is right, but it's not executed as an array, because it isn't working.

Answer 1

What you need to do is to add eval("\$newStringPedidos = $stringPedidos;"); before the last line. This will store your array in a new variable $newStringPedidos. Now the variable $newStringPedidos is really an array and not a string, so you can use it in the last line and it shall work.

eval("\$newStringPedidos = $stringPedidos;");
$resp = array("keyboard" => $newStringPedidos,"resize_keyboard" => true,"one_time_keyboard" => true,"remove_keyboard" => true);

The problem is that you are using the eval() function incorrectly. Visit https://www.w3schools.com/pHP/func_misc_eval.asp

Furthermore, it is usually much better to avoid eval() function, since it is quite confusing and in your case I think you really do not need it.