在ASP.NET的實時服務器中從AJAX提交數據時出現500個內部服務器錯誤

Question

在使用c＃在asp.net中建立網站的實時服務器上從ajax提交數據時出現錯誤。

錯誤為“ 500（內部服務器錯誤）”

這是我的代碼

<script>
    $(function () {
        $(document).on("click", "#submit_mail", function (e) {
            if (validateform() == false) {
                e.preventDefault();
            }
            else {
                $.ajax({
                    type: "POST",
                    url: "contact.aspx/SendMail",

                    data: JSON.stringify({ name: $('#txt_name').val(), email: $('#txt_emailID').val(), subject: $('#txt_subject').val(), message: $('#txt_content').val() }),
                    contentType: "application/json; charset=utf-8",
                    dataType: "json",
                    success: function () {
                        alert("Email Send Successfully, we will contact you successfully...!!!");
                        location.reload();
                    },
                    failure: function () {
                        alert("There is some problem in server to contact with us....Please contact with contact number...!!!");
                    }
                });
            }
        });
    });
</script>

現在，這是代碼。 該代碼在本地服務器上運行良好。 在實時服務器中，此代碼不起作用。

當我從開發人員工具調試此ajax時，它表明它調用了ajax行，即$.ajax({並結束它)}; 在調試之間沒有進行，因此這就是為什么在服務器端代碼的webserver方法中，它會空數據並顯示500內部服務器錯誤。

這是服務器端代碼

 [WebMethod]
    [ScriptMethod(ResponseFormat = ResponseFormat.Json)]
    public static void SendMail(string name, string email, string subject, string message)
    {
        //Thread.Sleep(10000);
        // Gmail Address from where you send the mail
        var fromAddress = "xxxxxxxxxxxxx@gmail.com";
        // any address where the email will be sending
        var toAddress = email.Trim();
        //Password of your gmail address
        const string fromPassword = "xxxxxxxxxxx";
        // Passing the values and make a email formate to display
        string sub = subject.Trim();
        string body = "From: " + name.Trim() + "\n";
        body += "Email: " + email.Trim() + "\n";
        body += "Subject: " + subject.Trim() + "\n";
        body += "Message: \n" + message.Trim() + "\n";
        // smtp settings
        var smtp = new System.Net.Mail.SmtpClient();
        {
            smtp.Host = "smtp.gmail.com";
            smtp.Port = 587;
            smtp.EnableSsl = true;
            smtp.DeliveryMethod = System.Net.Mail.SmtpDeliveryMethod.Network;
            smtp.Credentials = new NetworkCredential(fromAddress, fromPassword);
            smtp.Timeout = 20000;
        }
        // Passing values to smtp object
        smtp.Send(fromAddress, toAddress, sub, body);
    }

因此，簡而言之，我的問題是，為什么在AJAX方法中調試不使用AJAX的網址，數據和其他內容。 因為我的想法是，當數據從AJAX獲取URL，數據和其他內容時，就會出現此錯誤。

Answer 1

 ////Send Email on button click function SendEmail(url) { obj = {}; obj.Name = $("#txtName").val(); obj.Email = $("#txtEmail").val(); obj.MobileNo = $("#txtMobileNo").val(); obj.Subject = ($("#txtSubject").val() != '') ? $("#txtSubject").val() : ''; obj.Message = ($("#txtMessage").val() != '') ? $("#txtMessage").val() : ''; $.ajax({ type: "POST", url: url, data: JSON.stringify(obj), contentType: "application/json; charset=utf-8", dataType: "json", success: OnSuccess, failure: function (response) { alert(response.d); } }); function OnSuccess(response) { $.each(response.d, function () { var Msg; var Result; Msg = this['Text']; Result = this['Value']; if (Result > 0) { Clear(); $("#AlertClass").attr("style", "display: block") $('#AlertClass').addClass('alert alert-success text-center'); $('#lblResult').append('<span><img src="assets/images/Serve.png" style=" height: 40px;" /> &nbsp;&nbsp;' + Msg + '</span>'); } else { $("#AlertClass").attr("style", "display: block") $('#AlertClass').addClass('alert alert-danger text-center'); $('#lblResult').append('<span>"' + Msg + '"</span>'); } }); } } 

  [WebMethod] public static List<ListItem> SendEmail(string Name, string Email, string MobileNo, string Subject, string Message) { int Return = 0; try { } catch (Exception) { ReturnMessage.Add(new ListItem { Value = "0", Text = "There is an error try again after some time!" }); } return ReturnMessage; } 

試試這個。