如何打印功能结果，如果工作打印 - 是的，否则错误

Question

I am writing a small data entry form. Data gets entered in html form by user & then into csv file on server. Consider the following snippet, that i wrote to do this

if(!empty($_POST)) {
   $handler = fopen('database.csv','a') or exit("Error: Unable to open file!");
   fputcsv($handler,$_POST);
   fclose($handler);    
}
?>

Pretty simple!

The only "fancy" task is display a message to confirm that the entry has been added to the database(Entry added), clearing the form ready for the next entry. Or incase of error, display error message(Doh! Write failed ). I can use OR statement but that would only be invoked incase of an error.

Any idea how to do this?

[UPdate] Thanks for excellent replies everyone! I added conditional statement as everyone suggested

if(!empty($_POST)) {
$handler = fopen('database.csv','a') or exit("Error: Unable to open file!");
if(fputcsv($handler,$_POST)) {
    echo 'everything okay, add next record';
}
else {
      exit('Error: Record add failed. Try again or contact admin');
}
fclose($handler);   

}

Answer 1

if(!empty($_POST)) {
   $handler = fopen('database.csv','a');
   if ($handler === FALSE) {
      exit("Error: Unable to open file!");
   } else {
       fputcsv($handler,$_POST);
       fclose($handler);    
       echo "everything A-ok";
   }
}