为foreach()警告提供的无效参数

我想在mysql表中插入json数组数据。我已经写了这段代码。

if (mysqli_connect_errno()){
    $response["success"] = 0;
    $response["message"] = "Database Error!";   
    die(json_encode($response));
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Check connection
if ($con->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
echo "Connected successfully";
if(isset($_GET['doctorJson'])){
    $json = $_GET['doctorJson'];
    $array = json_decode($json, true);
    foreach($array as $item){
        $result = mysqli_query($con, "INSERT IGNORE INTO doctor_visit_track (id, doctor_name, doctor_email, date, time) VALUES 
            ('".$item['id']."', '".$item['doctorName']."', '".$item['doctorEmail']."', '".$item['date']."', '".$item['time']."')");
            }
    if($result){
        $response["message"] = "Success";
        echo json_encode($response);
    } else{
        $response["message"] = "Failure";
        echo json_encode($response);
        }
    }
mysqli_close($con);

当我使用xampp时，上面的代码工作得很好。但是，当我将此代码上传到服务器时，相同的代码会发出警告"为foreach()提供无效参数"，而不是插入表。但是在xampp中使用，代码工作良好，并成功插入数据。

$sql = <<<EOF
INSERT IGNORE INTO 
  doctor_visit_track (id, doctor_name, doctor_email, date, time) VALUES 
  ('?', '?', '?', '?"', '?')")
EOF;
$stmt = mysqli_prepare( $con, $sql);
foreach ($array as $item){
    mysqli_stmt_bind_param( $stmt, "sssss",
      $item['id'], $item['doctorName'], $item['doctorEmail'],
      $item['date'],$item['time']
      );
    $result = mysqli_stmt_execute($stmt);
    // rest of your code
};    

{
"id" : "hackerid",
"doctorName" : "I am a Hacker",
"doctorEmail": "hacker@hacker.com",
"date": "1999-12-31",
"time": "'"); drop table doctor_visit_track; -- Muhahahaha "
}