simple php script, unable return json - php

Question

I wrote this simple php script, that should return a json.
I can't understand what is wrong here.
if I comment - remove the part of code regarding the connection to mySQL(PDO) I'm able to get the print out as expected otherwise Alamofire and SwiftyJson return me the error
'"JSON could not be serialized because of error:\nThe data couldn’t be read because it isn’t in the correct format."'
<?php
header('Content-Type: application/json');
// if i remove the pdo to connect to mySQL server the everthing work fine
// $host = "127.0.0.1";
// $user = "test";
// $password = "123456789";
// $database = "nx_database";
// try{
// $pdo = new PDO ("mysql:host=$host;dbname=$database", $user, $password);
// $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
// print('connesso db -- ');
// }catch(PDOException $e){
// echo "DB Connection Fail" . $e->getMessage();
// die();
// }
$staff = $_POST['staff_ID'];
$array = [
'isLoggedIn'=>$staff
];
$js = json_encode($array);
echo $js;
?>
I attach the code also use to post the request:
func trysimple (){
let parm : [String : Any] = ["staff_ID": "3879"]
AF.request("http://127.0.0.1/nx/public/testRegister.php", method: .post,parameters: parm, headers: nil, interceptor: nil, requestModifier: nil)
.responseString { st in
print(st)
}
.responseJSON { js in
switch js.result {
case .success(let value) :
let json = JSON(value)
debugPrint(json)
case .failure(let err) :
debugPrint(err.localizedDescription)
}
}
}
}

Answer 1

I don't know why do you need a database connection here, but I think I know where is the mistake. You're displaying text that is not json here: print('connesso db -- '); If you expect a json format, you should display everything only in json format. Even on the failed connection possibility.
Here is how I would write it:
<?php
$host = "127.0.0.1";
$user = "test";
$password = "123456789";
$database = "nx_database";
header('Content-Type: application/json');
try{
$pdo = new PDO ("mysql:host=$host;dbname=$database", $user, $password);
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}catch(PDOException $e){
echo json_encode(['error' => "DB Connection Fail" . $e->getMessage()]);
exit;
}
$staff = $_POST['staff_ID'];
$array = [
'isLoggedIn'=>$staff
];
$js = json_encode($array);
echo $js;
?>

Answer 2

kindly update your code like this, for simple printing of json
<?php
header('Content-Type: application/json');
$array = array();
if(isset($_POST['staff_ID']))
{
$staff = $_POST['staff_ID'];
$array = array(
'isLoggedIn' => $staff
);
}
echo json_encode($array);