该请求已被阻止，因为敏感信息可能被泄露给第三方网站

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>

这里是回来就要收码

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 = "[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}); 
     }

但是当我启动AJAX请求时，出现此错误

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？

来源

2017-04-06 Eugene

是否启用到您的Gmail帐户的远程访问？ https://support.google.com/accounts/answer/6010255 –

是的，我将其设置为启用@TiesonT。 – Eugene

电子邮件正在发送，我收到了邮件。但是我有这个错误。@ TiesonT。 – Eugene

变化

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

要

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

来源

2017-04-06 09:18:34 MUT

该请求已被阻止，因为敏感信息可能被泄露给第三方网站

回答

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