zagorisback
Advanced Member | Редактировать | Профиль | Сообщение | Цитировать | Сообщить модератору Как просмотреть сообщение об ошибке (mysql_errno()==1062) в разборчивой форме Я пытался с "elseif", но это не работает Просмотреть первое "else" Код: if($result){ echo '<div class="alert alert-success alert-dismissible fade show" role="alert"> <strong>Success!</strong> Your entry has been submitted successfully! <button type="button" class="close" data-dismiss="alert" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div>'; } elseif (mysql_errno()==1062) { echo '<div class="alert alert-danger alert-dismissible fade show" role="alert"> <strong>Error!</strong> Record present in database <button type="button" class="close" data-dismiss="alert" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div>'; } | Оригинальный код Код: // Create a connection $conn = mysqli_connect($servername, $user, $password, $database); // Die if connection was not successful if (!$conn){ die("Sorry we failed to connect: ". mysqli_connect_error()); } else{ // Submit these to a database // Sql query to be executed $sql = "INSERT INTO `contact_us` (`username`, `email`, `desc`) VALUES ('$username', '$email', '$desc')"; $result = mysqli_query($conn, $sql); if($result){ echo '<div class="alert alert-success alert-dismissible fade show" role="alert"> <strong>Success!</strong> Your entry has been submitted successfully! <button type="button" class="close" data-dismiss="alert" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div>'; } else{ // echo "The record was not inserted successfully because of this error ---> ". mysqli_error($conn); echo '<div class="alert alert-danger alert-dismissible fade show" role="alert"> <strong>Error!</strong> We are facing some technical issue and your entry ws not submitted successfully! We regret the inconvinience caused! <button type="button" class="close" data-dismiss="alert" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div>'; } | Добавлено: Решено Код: elseif (mysqli_errno($conn)==1062){ echo '<div class="alert alert-danger alert-dismissible fade show" role="alert"> <strong>Error!</strong> Record present in database <button type="button" class="close" data-dismiss="alert" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div>'; } | Я хотел бы просмотреть сообщение об ошибке: Просмотреть, какие дубликаты записей присутствуют в базе данных Добавлено: Решено Код: elseif (mysqli_errno($conn)==1062){ echo '<div class="alert alert-danger alert-dismissible fade text-center show" role="alert"> <strong>Error!</strong> Record present in database <button type="button" class="close" data-dismiss="alert" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div>'; $select = mysqli_query($conn, "SELECT * FROM contact_us WHERE username = '".$_POST['username']."'"); mysqli_num_rows($select); echo'<div class="alert alert-warning alert-dismissible fade text-center show" role="alert"> <strong>Error!</strong> This username <b style="color:blue;">' . $username . '</b> is already used! <button type="button" class="close" data-dismiss="alert" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div>'; $select2 = mysqli_query($conn, "SELECT * FROM contact_us WHERE email = '".$_POST['email']."'"); mysqli_num_rows($select2); echo'<div class="alert alert-warning alert-dismissible fade text-center show" role="alert"> <strong>Error!</strong> This email <b style="color:blue;">' . $email . '</b> is already used! <button type="button" class="close" data-dismiss="alert" aria-label="Close"> <span aria-hidden="true">×</span> </button> </div>'; } |
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