102

我知道有一些问题可以解决这个问题,但答案通常遵循推荐字典或参数集合的思路,这在我的情况下不起作用。

我正在使用一个通过反射工作的库,对具有属性的对象做很多聪明的事情。这适用于已定义的类以及动态类。我需要更进一步,并按照以下方式做一些事情:

public static object GetDynamicObject(Dictionary<string,object> properties) {
    var myObject = new object();
    foreach (var property in properties) {
        //This next line obviously doesn't work... 
        myObject.AddProperty(property.Key,property.Value);
    }
    return myObject;
}

public void Main() {
    var properties = new Dictionary<string,object>();
    properties.Add("Property1",aCustomClassInstance);
    properties.Add("Property2","TestString2");

    var myObject = GetDynamicObject(properties);

    //Then use them like this (or rather the plug in uses them through reflection)
    var customClass = myObject.Property1;
    var myString = myObject.Property2;

}

该库适用于动态变量类型,并手动分配属性。但是我不知道会预先添加多少或哪些属性。

4

3 回答 3

124

你看过 ExpandoObject 吗?

见:http: //blogs.msdn.com/b/csharpfaq/archive/2009/10/01/dynamic-in-c-4-0-introducing-the-expandoobject.aspx

来自 MSDN:

ExpandoObject 类使您能够在运行时添加和删除其实例的成员,还可以设置和获取这些成员的值。此类支持动态绑定,这使您能够使用像 sampleObject.sampleMember 这样的标准语法,而不是像 sampleObject.GetAttribute("sampleMember") 这样的更复杂的语法。

允许您做一些很酷的事情,例如:

dynamic dynObject = new ExpandoObject();
dynObject.SomeDynamicProperty = "Hello!";
dynObject.SomeDynamicAction = (msg) =>
    {
        Console.WriteLine(msg);
    };

dynObject.SomeDynamicAction(dynObject.SomeDynamicProperty);

根据您的实际代码,您可能对以下内容更感兴趣:

public static dynamic GetDynamicObject(Dictionary<string, object> properties)
{
    return new MyDynObject(properties);
}

public sealed class MyDynObject : DynamicObject
{
    private readonly Dictionary<string, object> _properties;

    public MyDynObject(Dictionary<string, object> properties)
    {
        _properties = properties;
    }

    public override IEnumerable<string> GetDynamicMemberNames()
    {
        return _properties.Keys;
    }

    public override bool TryGetMember(GetMemberBinder binder, out object result)
    {
        if (_properties.ContainsKey(binder.Name))
        {
            result = _properties[binder.Name];
            return true;
        }
        else
        {
            result = null;
            return false;
        }
    }

    public override bool TrySetMember(SetMemberBinder binder, object value)
    {
        if (_properties.ContainsKey(binder.Name))
        {
            _properties[binder.Name] = value;
            return true;
        }
        else
        {
            return false;
        }
    }
}

这样你只需要:

var dyn = GetDynamicObject(new Dictionary<string, object>()
    {
        {"prop1", 12},
    });

Console.WriteLine(dyn.prop1);
dyn.prop1 = 150;

从 DynamicObject 派生可以让您提出自己的策略来处理这些动态成员请求,请注意这里有怪物:编译器将无法验证您的大量动态调用并且您不会获得智能感知,所以只需保持记住这一点。

于 2013-04-04T19:12:10.450 回答
51

感谢@Clint 的精彩回答:

只是想强调使用 Expando 对象解决这个问题是多么容易:

var dynamicObject = new ExpandoObject() as IDictionary<string, Object>;
foreach (var property in properties) {
    dynamicObject.Add(property.Key,property.Value);
}      
于 2013-04-05T13:24:23.993 回答
8

您可以将您的 json 字符串反序列化为字典,然后添加新属性然后对其进行序列化。

var jsonString = @"{}";

var jsonDoc = JsonSerializer.Deserialize<Dictionary<string, object>>(jsonString);

jsonDoc.Add("Name", "Khurshid Ali");

Console.WriteLine(JsonSerializer.Serialize(jsonDoc));
于 2019-10-23T19:59:55.523 回答