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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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