All models have similar properties? Assuming you cannot create a single model. Then move the properties to base class and create the other models deriving from the base model. Then also create a single view having base model as model.
Models
public class MyBaseModel
{
public string Property1 { get; set; }
public string Property2 { get; set; }
}
public class ModelA: MyBaseModel{}
public class ModelB: MyBaseModel{}
View
Create a single view for the MyBaseModel and Put the view in the Shared folder and name it something like MySharedView:
@model MyNamespace.MyBaseModel
@using (Html.BeginForm())
{
@Html.AntiForgeryToken()
<div class="form-horizontal">
<h4>Title which you can decide based on model type or get from ViewBag</h4>
<hr />
@Html.ValidationSummary(true, "", new { @class = "text-danger" })
<div class="form-group">
@Html.LabelFor(model => model.Property1, htmlAttributes: new { @class = "control-label col-md-2" })
<div class="col-md-10">
@Html.EditorFor(model => model.Property1, new { htmlAttributes = new { @class = "form-control" } })
@Html.ValidationMessageFor(model => model.Property1, "", new { @class = "text-danger" })
</div>
</div>
... Rest of properties ...
<div class="form-group">
<div class="col-md-offset-2 col-md-10">
<input type="submit" value="Create" class="btn btn-default" />
</div>
</div>
</div>
}
Controller
Then in controllers, when you want to return view, specify the shared view name. For example for model A and the same for model B:
[HttpGet]
public ActionResult A()
{
var model = new ModelA() { Property1 = "A", Property2 = "A" };
return View("MySharedView", model);
}
[HttpPost]
public ActionResult A(ModelA model)
{
// SaveA(model)
// Redirect to index
}
Thanks for the immediate response. I will try this out. How do i save when i click save button, it should save on separate table?
You have different actions if you need different processing. Each action receives the model that it should process.
I can extend the answer to include more options, but to be honest, the whole point is having a base class for all the models and having a single view for the the base model, not easier or not more Initiative.
You can also redesign your models, to use a single model, having all the mentioned properties + another property
public int Stage { get ;set;}and save all data in a single table. But I assume it's an obvious solution, and you don;t want to change your model and all the tables behind.@VigneshKumarA I believe the post answer the main concern of the question. Let me know if you have any question about it or if you found it useful.