My url is
http://localhost:54684/MyTest/Register
My controller Name is
MyTest
My view(ActionMethod name)is
Register
What I have tried:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using System.Configuration;
using MySql.Data.MySqlClient;
using ModelFields;
namespace WebApplicationPractise.Controllers
{
public class MyTestController : Controller
{
public ActionResult Home()
{
return View();
}
[HttpPost]
public ActionResult Register(RegisterFields rf)
{
string connection = ConfigurationManager.ConnectionStrings["ApplicationServices"].ConnectionString;
using(MySqlConnection mysqlcon =new MySqlConnection())
{
string query = "insert into 3tier.jagadeesh(FirstName,LastName,Age,Address) values(@FirstName,@LastName,@Age,@Address)";
using(MySqlCommand cmd =new MySqlCommand(query))
{
cmd.Connection = mysqlcon;
mysqlcon.Open();
cmd.Parameters.AddWithValue("@FirstName", rf.FirstName);
cmd.Parameters.AddWithValue("@LastName", rf.LastName);
cmd.Parameters.AddWithValue("@Age", rf.Age);
cmd.Parameters.AddWithValue("@Address", rf.Address);
mysqlcon.Close();
}
}
return View(rf);
}
}
}
Above one is my controller and below one is my view
@model ModelFields.RegisterFields
@{
ViewBag.Title = "Register";
Layout = "~/Views/Shared/MyTestLayoutPage.cshtml";
}
@using (Html.BeginForm("Register", "MyTest", FormMethod.Post))
{
@Html.TextBoxFor(model => model.FirstName, new { placeholder = "FirstName" });
@Html.TextBoxFor(model => model.LastName, new { placeholder = "FirstName" });
@Html.TextBoxFor(model => model.Age, new { placeholder = "FirstName" });
@Html.TextBoxFor(model => model.Address, new { placeholder = "FirstName" });
<input type="submit" value="submit" />
}
@if (Model != null)
{
<script type="text/javascript">
$(function () {
alert("data inserted successfully");
});
</script>
}