How to deal with many possible values to make a query?

Question

I\'m building an MVC app in which the user will have the possibilities of using a lot of filters to get exactly what he wants.

Here\'s an overview of those filters based

Answer 1

Ho do you normally face this in JavaScript ? You would create an option object, like

var op = {
   cardName : "..",
   cardType : "" ,
    ...
}

and pass it to a function. Correct ?

The same do in C# too. Define a, say, Conditions class, like

public class Conditions {
    public string CardName {get;set;}
    public string CardType {get;set;}
    .. 
}

and pass it to your function, which controls the values of the properties of that type to act accordingly after.

Answer 2

Here's a way you could handle your problem.

First, create a ViewModel with your search parameters.

public class FilterViewModel
{
    public string ObjName { get; set; }
    public string ObjType { get; set; }
    public string OtherType { get; set; }
}

In your controller, pass this to the view.

public virtual ActionResult SearchResults()
{
    return View(new FilterViewModel());
}

In your SearchResults.cshtml you need to encode the model to a json and then use an ajax call to filter your results. Assumming there's a button with btnFilterResults id. This is going to load the results from a partial view (that should be your table) using ajax.

<script type="text/javascript">
     var dataViewModel = @Html.Raw(Json.Encode(Model));

    $("#btnFilterResults").click(function () {
        dataViewModel.ObjName = $("#objName").val(); //Make sure the ids corresponde to your html. 
        dataViewModel.ObjType = $("#objType").val();
        dataViewModel.OtherType = $("#otherType").val();

        $.ajax({
           url: "@Url.Action(MVC.ControllerName.GetSearchResults())",
           data: { viewModel: JSON.stringify(dataViewModel) },
           success: function(data) {
               $("#tableDiv").empty(); //
               $("#tableDiv").html(data);
           }
        });
    });
</script>

Finally, in your controller you can get the results:

[HttpGet]
public virtual ActionResult LoadTableReporteGeneral(string viewModel)
{
    var filterOptions = JsonConvert.DeserializeObject<FilterViewModel>(viewModel);
    var query = from t in db.YourTableName
                select t;
     //This returns an IQueryable with all of your data, no filtering yet.

     //Start filtering.
     if(!String.IsNullOrEmpty(filterOptions.ObjName))
     {
         query = query.Where(x => x.ObjName == filterOptions.ObjName);
     }  
     if(!String.IsNullOrEmpty(filterOptions.ObjType ))
     {
         query = query.Where(x => x.ObjType == filterOptions.ObjType );
     }  
     //and so on. 
     //Finally return a partial view with your table. 
     return PartialView(MVC.Shared.Views._YourTableName, query);
}

Note: I'm using T4MVC and Json.Net packages.

Hope this helps.

Answer 3

Why do you not use ModelBinder . I think it's very useful with your case. Please see below:

_ I have a Model:

public class Employee
{
   public string Id { get; set; }
   public string FirstName { get; set; }
   public string LastName { get; set; }
   public DateTime BirthDate { get; set; }
   public Address HomeAddress { get; set; }
}

and ModelBinder for Employee:

public class EmployeeBinder : IModelBinder {
        public object BindModel(ControllerContext controllerContext, ModelBindingContext bindingContext) {
            var emp = new Employee {
                                       Id = controllerContext.HttpContext.Request.Form["Id"],
                                       FirstName = controllerContext.HttpContext.Request.Form["FirstName"],
                                       LastName = controllerContext.HttpContext.Request.Form["LastName"],
                                       BirthDate = new DateTime(int.Parse(controllerContext.HttpContext.Request.Form["year"]),
                                           int.Parse(controllerContext.HttpContext.Request.Form["month"]),
                                           int.Parse(controllerContext.HttpContext.Request.Form["day"]))
                                   };

            return emp;
        }
    }

Then Controller:

public ActionResult Example(Employee emp) {
            return View(emp);
}

URL: http://localhost:1034/Home/Example?Id=1&LastName=lazycatit