javascript - 消息提醒不起作用

Question

这是我的代码，问题是当我发送电子邮件调用表单 ID 时，电子邮件不会发送，当我删除表单 ID 时，电子邮件正在发送

html代码，html后面是js代码 这是我的代码，问题是，当我发送电子邮件调用表单 ID 时，电子邮件不会发送，当我删除表单 ID 时，电子邮件正在发送

<form id="main-contact-form" class="contact-form" name="contact-form" method="post" action="sendemail.php">
    <div class="col-sm-5 col-sm-offset-1">
        <div class="form-group">
            <label>Name *</label>
            <input type="text" name="name" class="form-control" required="required">
        </div>
        <div class="form-group">
            <label>Email *</label>
            <input type="email" name="email" class="form-control" required="required">
        </div>
        <div class="form-group">
            <label>Phone</label>
            <input type="number" class="form-control">
        </div>
        <div class="form-group">
            <label>Company Name</label>
            <input type="text" class="form-control">
        </div>
    </div>
    <div class="col-sm-5">
        <div class="form-group">
            <label>Subject *</label>
            <input type="text" name="subject" class="form-control" required="required">
        </div>
        <div class="form-group">
            <label>Message *</label>
            <textarea name="message" id="message" required="required" class="form-control" rows="8"></textarea>
        </div>
        <div class="form-group">
            <button type="submit" name="submit" class="btn btn-primary btn-lg" required="required">Submit Message</button>
        </div>
    </div>
</form>

--

ajax代码

var form = $('#main-contact-form');
form.submit(function(event) {
    event.preventDefault();
    var form_status = $('<div class="form_status"></div>');
    $.ajax({
        url: $(this).attr('action'),
        beforeSend: function() {
            form.prepend(form_status.html('<p><i class="fa fa-spinner fa-spin"></i> Email is sending...</p>').fadeIn());
        }
    }).done(function(data) {
        form_status.html('<p class="text-success">' + data.message + '</p>').delay(3000).fadeOut();
    });
});

php代码

               <?php


               header('Content-type: application/json');


             $status = array(
              'type'=>'success',
          'message'=>'Thank you for contact us. As early as possible  we will contact you '
        );




         $name = @trim(stripslashes($_POST['name']));


        $email = @trim(stripslashes($_POST['email'])); 


      $subject = @trim(stripslashes($_POST['subject'])); 


    $message = @trim(stripslashes($_POST['message'])); 



      $email_from = $email;



      $email_to = 'email@email.com';//replace with your email




                   $body = 'Name: ' . $name . "\n\n" . 'Email: ' . $email .   "\n\n" . 'Subject: ' . $subject . "\n\n" . 'Message: ' . $message;




           $success = @mail($email_to, $subject, $body, 'From: <'.$email_from.'>');





         echo json_encode($status);



        die;

Answer 1

这就是您的 ajax 调用的样子

$('#main-contact-form').on('submit', function(event) {
    event.preventDefault();

    var form_status = $('<div class="form_status"></div>');

    $.ajax({
        url        : $(this).attr('action'),
        data       : $(this).serialize(),
        dataType   : 'json',
        type       : 'POST',
        beforeSend : function() {
            form.prepend(form_status.html('<p><i class="fa fa-spinner fa-spin"></i> Email is sending...</p>').fadeIn());
        }
    }).done(function(data) {
        form_status.html('<p class="text-success">' + data.message + '</p>').delay(3000).fadeOut();
    });
});