Webapi - Tryupdatemodel or Updatemodel in an Apicontroller

Question

Webapi - Tryupdatemodel or Updatemodel in an Apicontroller

Asked 8 years, 12 months ago

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1

I have the following situation, a Person class, with properties Id, Name, Birth, I want to implement a page that can update only the Person Name. My implementation should take a person by their id, and change only the Name field.

I made the implementation below in MVC, but how could I do the same with Webapi, since I don’t have Tryupdatemodel or Updatemodel. I implement functionality with this routine to facilitate generalization in the future.

Controller Code in MVC:

public class HomeController : Controller
{
    public ActionResult Index()
    {
        return View();
    }

    [HttpPost]
    public JsonResult SetNome()
    {
        var sample = Pessoa.FindById(12);
        TryUpdateModel<Pessoa>(sample);
        return Json("sucesso");
    }
}

public class Pessoa
{
    [Required]
    public int Id { get; set; }

    [Required]
    public string Nome { get; set; }

    [Required]
    public DateTime? Nascimento { get; set; }

    public static Pessoa FindById(int id)
    {
        return new Pessoa() { Id = 12, Nome = "Meu Nome", Nascimento = Convert.ToDateTime("1989-01-31") };
    }
}

View code:

<!DOCTYPE html>
<html>
<head>
    <script src="https://code.angularjs.org/1.4.3/angular.js"></script>
    <script>

        var app = angular.module("MainModule", []);

        var MainController = function ($scope, $http) {

            var url = 'http://localhost:47107/home/setnome';
            //var url = 'http://localhost:47107/api/values';

            $scope.pessoa = {};
            $scope.pessoa.nome = "xpto";

            $scope.enviar = function () {

                $http({
                    url: url,
                    method: "POST",
                    data: $scope.pessoa
                })
                .then(function (response) {
                    // success
                    console.log(JSON.stringify(response));
                },
                function (response) {
                    // failed
                    console.log(JSON.stringify(response));
                });
            }
        };

        app.controller("MainController", MainController);

    </script>
</head>

<body ng-app="MainModule">
    <div ng-controller="MainController">
        <input id="pessoa.nome" ng-model="pessoa.nome" />
        {{pessoa.nome}}
        <button ng-click="enviar()">Enviar</button>
    </div>
</body>
</html>

Current code in Webapi:

public class ValuesController : ApiController
{
    public void Post([FromBody]Pessoa pessoa)
    {
        var sample = Pessoa.FindById(12);

        sample.Nome = pessoa.Nome;
    }
}

I wanted to have to abstract the Name field, as it was done in MVC, with Updatemodel, it identifies which fields were sent by View and fills only them, the others it keeps original. With Webapi I have to do this manually.

2 answers

1

Ricardo, I solve this problem in a different way.

Initially, my Action or Operation does not receive an Entity/Model, but a DTO.

Receiving a DTO, I would do this treatment, that you want, in my repository.

Example of a Repository method with this feature:

public void Alterar(TDTO DTO)
    {           
            var entidadeSemAlteracao = Context.Set<Compra>().Find(DTO.sID);
            DTO.ToEntity(entidadeSemAlteracao);
            Context.SaveChanges();            
    }

The Toentity is an extension on my DTO, calling the Automapper.

public static TDestino ToEntity<TOrigem, TDestino>(this TOrigem DTO, TDestino entidadeDestino) where TDestino : EntidadeBase
    {
        var result = Mapper.Map(DTO, entidadeDestino);
        return result;
    }

Everything happens when the extension is called, which will leave the properties as Modified and the Savechanges() will change only these properties.

What do you think?

That’s what you wanted?

Very good Henryle solution, although I am trying to avoid using Automapper in this project as it is an overhead too much work.

– Ricardo Carvalho

2015/09/30 at 14:07

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by Ricardo Carvalho • **455** points · Answer 1 · 2015-09-29T21:23:12+00:00

Sharing knowledge with anyone who has the same problem. Recalling that the purpose of this function is not to validate the entity and just fill (complement) an entity coming from the database with the data that came from the view.

    protected virtual void UpdateModel<T>(T original, bool overrideForEmptyList = true)
    {
        var json = ControllerContext.Request.Content.ReadAsStringAsync().Result;
        UpdateModel<T>(json, original, overrideForEmptyList);
    }

    private void UpdateModel<T>(string json, T original, bool overrideForEmptyList = true)
    {
        //faz um comportamento similar ao do UpdateModel do Controller, onde sobreescreve as propriedades Enumarables,
        //caso no json a mesma esteja preenchida, se ela estiver vazia (não nula), sobreescreve também a não ser que,
        //o parametro overrideForEmptyList seja false
        var newValues = JsonConvert.DeserializeObject<T>(json);            
        foreach (var property in original.GetType().GetProperties())
        {
            var isEnumerable = property.PropertyType.GetInterfaces().Any(t => t.IsGenericType && t.GetGenericTypeDefinition() == typeof(IEnumerable<>));

            if (isEnumerable && property.PropertyType != typeof(string))
            {
                var propertyOriginalValue = property.GetValue(original, null);
                if (propertyOriginalValue != null)
                {
                    var propertyNewValue = property.GetValue(newValues, null);

                    if (propertyNewValue != null && (overrideForEmptyList || ((IEnumerable<object>)propertyNewValue).Any()))
                    {
                        property.SetValue(original, null);
                    }
                }
            }
        }

        JsonConvert.PopulateObject(json, original);
    }

    public void Post()
    {           
        var sample = Pessoa.FindById(12);
        UpdateModel(sample);            
    }