2017-04-06 65 views
1

我有AJAX請求該請求已被阻止,因爲敏感信息可能被泄露給第三方網站

這裏是代碼

<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> 

這裏是回來就要收碼

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 = "[email protected]"); 
      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}); 
     } 

但是當我啓動AJAX請求時,出現此錯誤

此請求已被阻止,因爲敏感信息可能會泄露給第三方網站

我該如何解決?

+0

是否啓用到您的Gmail帳戶的遠程訪問? https://support.google.com/accounts/answer/6010255 –

+0

是的,我將其設置爲啓用@TiesonT。 – Eugene

+0

電子郵件正在發送,我收到了郵件。但是我有這個錯誤。@ TiesonT。 – Eugene

回答

2

變化

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

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