Tuesday, April 16, 2013

Returning a partial view on a MVC Ajax Form submit

In my previous post we talked about returning a Json object on a MVC Ajax form submit. There we had to write several Java-Script methods in order to process the response. In that case it was just about whether are registration (form1) was successful or not. But assume we need to show another form (form2) (without redirection), do some data filling there and submit it and that all these things have to be in Ajax.

One way you could achieve this, is to have both these forms in our html page and do a jquery show  and a hide on those forms. below given is the javascript code if are implementing this in our previous example.
function RegisterSuccess(data) {
    try {   
        if (data.Status == "0") {          
        } else {           
    } catch (e) {

If we need to auto fill up some of the information in form2 from the database, we need to write javascript code to set the values for each of those fields. I remeber Scott Hanselman , saying "People who can read java-scripts are weird people!" (people who write too!) in one of this videos. So what if a js library writes all these javascripts for us and that we only need to return a view. Yes, Ajax.BeginForm takes care of all that. You just need to set the UpdateTargetId and the LoadingElementId parameters.

Ok now. Time for some coding! This is how are razor view look like.
@model AjaxMvc4.Models.RegisterModel
    ViewBag.Title = "Register";


Create a new account.

@using (Ajax.BeginForm("Register", "Account", new AjaxOptions { HttpMethod = "POST", UpdateTargetId = "formContent", LoadingElementId = "loader" })) { @Html.AntiForgeryToken() @Html.ValidationSummary()
Registration Form
  1. @Html.LabelFor(m => m.UserName) @Html.TextBoxFor(m => m.UserName)
  2. @Html.LabelFor(m => m.Password) @Html.PasswordFor(m => m.Password)
  3. @Html.LabelFor(m => m.ConfirmPassword) @Html.PasswordFor(m => m.ConfirmPassword)
@section Scripts { @Scripts.Render("~/bundles/jqueryval") }

In the controllers instead of returning a Json object we return a partial view. We can even pass a model and set if we need to fill some of the information in the new form (either from the database or from the previous model posted).

        public ActionResult Register(RegisterModel model)
            if (ModelState.IsValid)
                // Attempt to register the user
                    WebSecurity.CreateUserAndAccount(model.UserName, model.Password);
                    WebSecurity.Login(model.UserName, model.Password);

                    UserModel user = new UserModel();
                    user.UserName = model.UserName;

                    return PartialView("_UserContactDetailsPartial", user);               
                catch (MembershipCreateUserException e)
                    ModelState.AddModelError("", ErrorCodeToString(e.StatusCode));
            return PartialView("_ErrorPartial");    

User Details Partial View :
@model AjaxMvc4.Models.UserModel

@using (Ajax.BeginForm("Account", "ContactInformation", new AjaxOptions { HttpMethod = "POST", UpdateTargetId = "formContent", LoadingElementId = "loader" }))

Contact Details
User Name
@Html.DisplayFor(model => model.UserName) @Html.HiddenFor(model => model.UserName)
First Name
@Html.EditorFor(model => model.FirstName) @Html.ValidationMessageFor(model => model.FirstName)
Last Name
@Html.EditorFor(model => model.LastName) @Html.ValidationMessageFor(model => model.LastName)
Email Address
@Html.EditorFor(model => model.Email) @Html.ValidationMessageFor(model => model.Email)
Phone Number
@Html.EditorFor(model => model.PhoneNo) @Html.ValidationMessageFor(model => model.PhoneNo)
Mobile Number
@Html.EditorFor(model => model.Mobile) @Html.ValidationMessageFor(model => model.Mobile)
@Html.EditorFor(model => model.Address) @Html.ValidationMessageFor(model => model.Address)

Don't forget to reference the jquery.validate.min.js and jquery.validate.unobtrusive.min.js for your ajax form to work!

Monday, April 15, 2013

Simple asp.net MVC Ajax Form Application

For this purpose, I'm using Microsoft asp.net MVC4 application.

Step1. As always, add a new project (Not an empty web project).

 Step2. Choose Internet Application from the project template.

Step3. As it is not an empty web project, there will be a set of code generated for you. For example the log in and the registration modules (Models, Controllers and Views). We are planning to alter them into Ajax forms.

Step4. Ok now, lets do some coding. We'll try to change the registration module. Go to register.cshtml in the Account folder inside the Views folder. Instead of the normal Html.BeginForm , change that to Ajax.BeginForm.

There are some slight changes you need to make when it comes to Ajax.BeginForm method.  You need to provide the HttpMethod and some Java-script functions as AjaxOptions other than controller and the action name. I have provided Java script methods for OnBegin, OnSuccess, OnComplete and OnFailure events. (Check Step 6 for JS implementaions)

Step5. We also need to change the controller, since we are planning to return Json Objects so that we can process them in are Java scripts. I have made the return type as "JsonResult" instead of "ActionResult". ActionResult Class is parent class of JsonResult Class. However, I have opted for JsonResult so that I can be more specific (Register action will only return Jsons). Although, inside my "Register" action the registration process is the same, instead of doing a return View(), I do a return Json(). I need to process this Json at client side, according to the format I specify here. Here I'm simply returning a Status and a Message in my Json.

Step6. Now that we have written the code to send an ajax request and process it in the server side, we now need to handle the response in the client side. Below shown are my Java Script method implementations mentioned in step4. In my RegisterBegin() and RegisterComplete() methods I'm just showing and hiding a progress bar. RegisterFailure() method (called by OnFailure event) prompts for error with the HttpStatus Code. Note that this method gets fired only  for Http errors. Not for our application errors. We need to handle our application errors within the RegisterSuccess() method as they also get return with  http status success 200 (which triggers the OnSuccess event).

In our RegisterSuccess() method we differentiate the success registration and errors in registration based on the Status codes in our Json Object (Check Controller Code). If success, we are redirecting to the home page and if not we are showing the error message passed from the controller via the json object.

Step7. We are ready to launch now. Run the application and start registering. You will see a progress bar once you click on the submit button, and the success message on a successful registration. Try some exception scenarios as well! Good Luck!