使用 .NET 4,C#
假设我有一个类 Info
延伸 CustomTypeDescriptor
.类的实例 Info
有字典 <string, object>
对,在运行时加载。
我希望能够将字典键公开为属性(以便 Info
的每个实例都有不同的属性)。属性的值应该是字典中对应的值。
我开始公开属性:
public override PropertyDescriptorCollection GetProperties()
{
var orig = base.GetProperties();
var newProps = dictionary.Select( kvp =>
TypeDescriptor.CreateProperty(
this.GetType(),
kvp.key,
typeof(string)));
return new PropertyDescriptorCollection(
orig.Cast<PropertyDescriptor>()
.Concat(newProps)
.ToArray());
}
问题是,我如何获得它们的值?
var info = new Info(new Dictionary<string, object>{{"some_property", 5}};
var prop = TypeDescriptor.GetProperties(i_info)["some_property"];
var val = prop.GetValue(i_info); //should return 5
我发现在 prop.GetValue()
时获得控制权的唯一方法被调用是为了覆盖 GetPropertyOwner(PropertyDescriptor pd)
,但按照我的理解,它希望我返回另一个具有匹配的真实(编译)属性的 type 的实例。
我希望能够自己编写属性的实际实现(对于本例,返回字典中键与属性名称匹配的值)。
这可能吗?
最佳答案
您需要自己实现 PropertyDescriptor
类重写 GetValue
方法。所以而不是 TypeDescriptor.CreateProperty
你将使用新的 MyCoolPropertyDescriptor(dictionary, kvp.Key)
或喜欢。
这是一个如何实现它的示例:
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Linq;
namespace ConsoleApplication1
{
internal sealed class MyCoolPropertyDescriptor : PropertyDescriptor
{
private Func<object, object> propertyGetter;
private Action<object, object> propertySetter;
public MyCoolPropertyDescriptor(
string name,
Func<object, object> propertyGetter,
Action<object, object> propertySetter)
: base(name, new Attribute[] {})
{
this.propertyGetter = propertyGetter;
this.propertySetter = propertySetter;
}
public override bool CanResetValue(object component)
{
return true;
}
public override System.Type ComponentType
{
get { return typeof(object); }
}
public override object GetValue(object component)
{
return this.propertyGetter(component);
}
public override bool IsReadOnly
{
get { return false; }
}
public override System.Type PropertyType
{
get { return typeof(object); }
}
public override void ResetValue(object component)
{
this.propertySetter(component, null);
}
public override void SetValue(object component, object value)
{
this.propertySetter(component, value);
}
public override bool ShouldSerializeValue(object component)
{
return false;
}
}
public sealed class Info : CustomTypeDescriptor
{
IDictionary<string, object> properties;
public Info(IDictionary<string, object> properties)
{
this.properties = properties;
}
public override PropertyDescriptorCollection GetProperties()
{
var orig = base.GetProperties();
var newProps = this.properties
.Select(kvp => new MyCoolPropertyDescriptor(
kvp.Key,
o => ((Info)o).properties[kvp.Key],
(o, v) => ((Info)o).properties[kvp.Key] = v));
return new PropertyDescriptorCollection(orig
.Cast<PropertyDescriptor>()
.Concat(newProps)
.ToArray());
}
}
internal class Program
{
private static void Main(string[] args)
{
var info = new Info(new Dictionary<string, object>{{"some_property", 5}});
var prop = TypeDescriptor.GetProperties(info)["some_property"];
var val = prop.GetValue(info); //should return 5
Console.WriteLine(val);
}
}
}
关于c# - 获取对象实例上自定义属性的*值*?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7889721/