此请求已被阻止，因为敏感信息可能会泄露给第三方网站

Question

我有 AJAX 请求

这里是代码

  <script>
        $(document).ready(function () {
        $('#save').click(function () {
            send_email();
        });
    });
        function send_email() {
$.ajax({
    url: '/Questions/SendEmail',
    dataType: "json",
    type: "GET",
    contentType: 'application/json; charset=utf-8',
    async: false,
    processData: false,
    cache: false,
    delay: 15,
    success: function (result) {
                    alert("Email Forwarded")
    },
    error: function (xhr) {
        alert('error');
        }
    });
}
</script>

这是后端代码

 public JsonResult SendEmail() {

            SmartSolutions.Models.InvitationMail _objModelMail = new SmartSolutions.Models.InvitationMail();
            var emailList = db.InvitationMails.Select(model => model.To);
            var emailList2 = db.InvitationMails.Select(model => model.Link);
            var link2 = emailList2.FirstOrDefault();
            var emails = String.Join(",", emailList);
            MailMessage mail = new MailMessage();
            mail.To.Add(emails);
            mail.From = new MailAddress(_objModelMail.From = "suhomlin.eugene93@gmail.com");
            mail.Subject = _objModelMail.Subject = "Видео интервью";

            string Body = _objModelMail.Body = "Предлагаем пройти интервью" + link2;

            mail.Body = Body;
            mail.IsBodyHtml = true;
            SmtpClient smtp = new SmtpClient();
            smtp.Host = "smtp.gmail.com";
            smtp.Port = 587;
            smtp.UseDefaultCredentials = false;
            smtp.Credentials = new System.Net.NetworkCredential
            ("***********", "*********");// Enter seders User name and password
            smtp.EnableSsl = true;
            smtp.Send(mail);
            return Json(new { Result = "Success", Message = "Saved Successfully",JsonRequestBehavior.AllowGet});
        } 

但是当我启动 AJAX 请求时，我得到了这个错误

此请求已被阻止，因为敏感信息可能会泄露给第三方网站

我该如何解决？

Answer 1

改变

return Json(new { Result = "Success", Message = "Saved Successfully",JsonRequestBehavior.AllowGet});

收件人

return Json(new { Result = "Success", Message = "Saved Successfully"},JsonRequestBehavior.AllowGet);