如何在MVC5项目中使用Json.NET进行JSON模型绑定？

Question

我一直在网上寻找答案或例子,但还找不到一个.我只想更改默认的JSON序列化程序,用于在模型绑定到JSON.NET库时反序列化JSON.

我发现这个 SO帖子,但到目前为止无法实现它,我甚至看不到System.Net.Http.Formatters命名空间,也看不到GlobalConfiguration.

我错过了什么？

UPDATE

我有一个ASP.NET MVC项目,它基本上是一个MVC3项目.目前我的目标是.NET 4.5并使用ASP.NET MVC 5和相关的NuGet包.

我没有看到System.Web.Http程序集,也没有看到任何类似的命名空间.在这个上下文中,我想注入JSON.NET作为JSON类型请求的默认模型绑定器.

Answer 1

我终于找到了答案.基本上我不需要MediaTypeFormatter那些设计用于MVC环境的东西,但是在ASP.NET Web API中,这就是为什么我没有看到那些引用和命名空间(顺便说一句,那些包含在Microsoft.AspNet.WeApiNuGet包中) .

解决方案是使用自定义值提供程序工厂.这是所需的代码.

    public class JsonNetValueProviderFactory : ValueProviderFactory
    {
        public override IValueProvider GetValueProvider(ControllerContext controllerContext)
        {
            // first make sure we have a valid context
            if (controllerContext == null)
                throw new ArgumentNullException("controllerContext");

            // now make sure we are dealing with a json request
            if (!controllerContext.HttpContext.Request.ContentType.StartsWith("application/json", StringComparison.OrdinalIgnoreCase))
                return null;

            // get a generic stream reader (get reader for the http stream)
            var streamReader = new StreamReader(controllerContext.HttpContext.Request.InputStream);
            // convert stream reader to a JSON Text Reader
            var JSONReader = new JsonTextReader(streamReader);
            // tell JSON to read
            if (!JSONReader.Read())
                return null;

            // make a new Json serializer
            var JSONSerializer = new JsonSerializer();
            // add the dyamic object converter to our serializer
            JSONSerializer.Converters.Add(new ExpandoObjectConverter());

            // use JSON.NET to deserialize object to a dynamic (expando) object
            Object JSONObject;
            // if we start with a "[", treat this as an array
            if (JSONReader.TokenType == JsonToken.StartArray)
                JSONObject = JSONSerializer.Deserialize<List<ExpandoObject>>(JSONReader);
            else
                JSONObject = JSONSerializer.Deserialize<ExpandoObject>(JSONReader);

            // create a backing store to hold all properties for this deserialization
            var backingStore = new Dictionary<string, object>(StringComparer.OrdinalIgnoreCase);
            // add all properties to this backing store
            AddToBackingStore(backingStore, String.Empty, JSONObject);
            // return the object in a dictionary value provider so the MVC understands it
            return new DictionaryValueProvider<object>(backingStore, CultureInfo.CurrentCulture);
        }

        private static void AddToBackingStore(Dictionary<string, object> backingStore, string prefix, object value)
        {
            var d = value as IDictionary<string, object>;
            if (d != null)
            {
                foreach (var entry in d)
                {
                    AddToBackingStore(backingStore, MakePropertyKey(prefix, entry.Key), entry.Value);
                }
                return;
            }

            var l = value as IList;
            if (l != null)
            {
                for (var i = 0; i < l.Count; i++)
                {
                    AddToBackingStore(backingStore, MakeArrayKey(prefix, i), l[i]);
                }
                return;
            }

            // primitive
            backingStore[prefix] = value;
        }

        private static string MakeArrayKey(string prefix, int index)
        {
            return prefix + "[" + index.ToString(CultureInfo.InvariantCulture) + "]";
        }

        private static string MakePropertyKey(string prefix, string propertyName)
        {
            return (String.IsNullOrEmpty(prefix)) ? propertyName : prefix + "." + propertyName;
        }
    }

你可以在你的Application_Start方法中使用它:

// remove default implementation    
ValueProviderFactories.Factories.Remove(ValueProviderFactories.Factories.OfType<JsonValueProviderFactory>().FirstOrDefault());
// add our custom one
ValueProviderFactories.Factories.Add(new JsonNetValueProviderFactory());

这是一个向我指出正确方向的帖子,这个也给了价值提供者和模型装订者一个很好的解释.

Answer 2

我也有这样的问题.我将JSON发布到一个动作,但我的JsonProperty名称被忽略了.因此,我的模型属性始终为空.

public class MyModel
{
    [JsonProperty(PropertyName = "prop1")]
    public int Property1 { get; set; }

    [JsonProperty(PropertyName = "prop2")]
    public int Property2 { get; set; }

    [JsonProperty(PropertyName = "prop3")]
    public int Property3 { get; set; }

    public int Foo { get; set; }
}

我发布到使用此自定义jquery函数的操作:

(function ($) {
    $.postJSON = function (url, data, dataType) {

        var o = {
            url: url,
            type: 'POST',
            contentType: 'application/json; charset=utf-8'
        };

        if (data !== undefined)
            o.data = JSON.stringify(data);

        if (dataType !== undefined)
            o.dataType = dataType;

        return $.ajax(o);
    };
}(jQuery));

我称之为:

data = {
    prop1: 1,
    prop2: 2,
    prop3: 3,
    foo: 3,
};

$.postJSON('/Controller/MyAction', data, 'json')
            .success(function (response) {
                ...do whatever with the JSON I got back
            });

不幸的是,只有foo被绑定(奇怪,因为情况不一样,但我想默认的模型绑定器不区分大小写)

[HttpPost]
public JsonNetResult MyAction(MyModel model)
{
    ...
}

解决方案最终变得相当简单

我刚刚实现了Dejan的模型绑定器的通用版本,它非常适合我.它可能会使用一些虚拟检查(比如确保请求实际上是application/json),但它现在正在做这个技巧.

internal class JsonNetModelBinder : IModelBinder
{
    public object BindModel(ControllerContext controllerContext, ModelBindingContext bindingContext)
    {
        controllerContext.HttpContext.Request.InputStream.Position = 0;
        var stream = controllerContext.RequestContext.HttpContext.Request.InputStream;
        var readStream = new StreamReader(stream, Encoding.UTF8);
        var json = readStream.ReadToEnd();
        return JsonConvert.DeserializeObject(json, bindingContext.ModelType);
    }
}

当我想在特定的动作上使用它时,我只是告诉它我想要使用我的自定义Json.Net模型绑定器:

[HttpPost]
public JsonNetResult MyAction([ModelBinder(typeof(JsonNetModelBinder))] MyModel model)
{
    ...
}

现在我的[JsonProperty(PropertyName ="")]属性不再在MyModel上被忽略,一切都被正确绑定!