编程方式打开的属性序列化.NET的EF code首先 [英] Programatically turn on properties serialization on .NET EF Code First

问题描述

我与$ EF上仿照6 C $ cFirst我的数据。
我建立一个Web API来将有机会获得其客户不同类型的，但取决于client's配置他们应该看到与否模型的某些属性。

I have my data modeled with CodeFirst on EF 6. I'm building a Web API to which different type of clients would have access but depending on the client´s configuration they should see or not certain properties of the models.

¿如何打开或关闭 [JsonIgnore] 或 [连载] ？是否有可能设置一组特定的规则，要做到这一点，就像一个验证？

¿How do I turn on or off the [JsonIgnore] or [serialized]? Is it possible to set a certain set of rules to do this, like a validator?

推荐答案

您可以创建自定义的合同解析器和创建响应时使用它：

Option 1: Using a custom ContractResolver

You can create a custom contract resolver and use it when creating the response:

public class TestContractResolver : DefaultContractResolver
{
    public string ExcludeProperties { get; set; }
    protected override IList<JsonProperty> CreateProperties(Type type,
                                           MemberSerialization memberSerialization)
    {
        if (!string.IsNullOrEmpty(ExcludeProperties))
            return base.CreateProperties(type, memberSerialization)
                        .Where(x => !ExcludeProperties.Split(',').Contains(x.PropertyName))
                        .ToList();

        return base.CreateProperties(type, memberSerialization);
    }
}

这里的用法：

[HttpGet]
public HttpResponseMessage Test()
{
    var person = new Person() { Id = 1, FirstName = "x", LastName = "y", Age = 20 };
    string excludeProperties= "FirstName,Age";
    string result = JsonConvert.SerializeObject(person, Formatting.None,
                    new JsonSerializerSettings
                    {
                        ContractResolver = new TestContractResolver() 
                        { 
                            ExcludeProperties = excludeProperties
                        }
                    });
    var response = this.Request.CreateResponse(HttpStatusCode.OK);
    response.Content = new StringContent(result, Encoding.UTF8, "application/json");
    return response;
}

和结果将是：

{"Id":1,"LastName":"y"}

选项2：使用字典

您可以用逗号分隔的属性名的字符串忽略，然后选择属性，并把它们（名称和值），在字典中，并用它们作为结果是：

Option 2: Using Dictionary

You can have a comma separated string of property names to ignore, then select properties and put them (name and value) in a dictionary and use them as result:

[HttpGet]
public Dictionary<string, Object> Test()
{
    var person = new Person() { Id = 1, FirstName = "x", LastName = "y", Age = 20 };

    string excludeProperties = "FirstName,Age";
    var dictionary = new Dictionary<string, Object>();
    person.GetType().GetProperties()
          .Where(x => !excludeProperties.Split(',').Contains(x.Name)).ToList()
          .ForEach(p =>
          {
              var key = p.Name;
              var value = p.GetValue(person);
              dictionary.Add(key, value);
          });

    return dictionary;
}

和结果将是：

{"Id":1,"LastName":"y"}

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