C#实例化实现通用接口的类

时间:2010-05-18 15:18:21

标签: c# reflection

我有一些实现IBusinessRequest<T>的业务类,例如:

public class PersonBusiness : IBusinessRequest<Person>
{ }

除此之外我还有一个功能:

 TypeHelper.CreateBusinessInstance(Type businessType, Type businessRequestType)

业务类的要求是它们必须具有无参数构造函数,我在TypeHelper.CreateBusinessInstance函数中检查它。

我想创建一个businessType类型的实例(PersonBusiness),其businessRequestType的通用值为IBusinessRequest<>

我怎样才能完成这项工作?

EDIT1:

感谢所有答案,它让我走上正轨。我写下的情况不是我正在处理的真实情况。我希望它足以解决我的问题: - )

我现在想出了以下内容,它就像魅力一样。

public interface IBusinessRequest<T> where T : class
{
    T Request { get; set; }
}

public interface IBusiness
{        
    /// <summary>
    /// Validates the request against custom rules
    /// </summary>        
    /// <param name="meldingen">Return a list of validation messages</param>
    /// <returns>returns true is validation went succesfull, false when something is wrong</returns>
    bool Validate(out List<string> meldingen);

    /// <summary>
    /// Executes business logic and returns a response object.
    /// </summary>        
    /// <returns>The strongly typed response object</returns>
    object Execute(object request);
}


public class PersonBusiness :IBusiness, IBusinessRequest<Person>
{ }


public static IBusiness CreateBusinessInstance(Type type, object requestMessage)
        {            
            //some checks on the type, like: is of IBusiness & IBusinessRequest<>
            ...

            //get al constructors of type
            ConstructorInfo[] constructors = type.GetConstructors();

            //if we have more then one constructor throw error
            if (constructors.Length > 1)
            {
                throw new BusinessCreateException(String.Format(PrivateErrors.ToManyConstructorsInTypeError, type, constructors.Length, 1));
            }

            //constructor parameters
            ParameterInfo[] parameterInfo = constructors[0].GetParameters();

            //we do not allow a constructor with more then one parameters.
            if (parameterInfo.Length > 0)
            {
                throw new BusinessCreateException(String.Format(PrivateErrors.ConstructorHasToManyParametersError, type, 0));
            }

            IBusiness instance = null;            
            try
            {                
                //create an instance, invoke constructor with zero parameters
                object invokeResult = constructors[0].Invoke(new object[0]);

                //set property "Request"
                PropertyInfo pi = type.GetProperty("Request");

                //do we have found a property
                if (pi != null)
                {
                    pi.SetValue(invokeResult, requestMessage ,null);
                }

                instance = invokeResult as IBusiness;
            }
            catch (Exception ex)
            {
                throw new BusinessCreateException(String.Format(PrivateErrors.BusinessCreateError, type.FullName), ex);
            }

            //return result
            return instance;
        }

现在,在软件的另一部分运行时,业务类类型被提供给我(即PersonBusiness)。另一个事实是我知道requestMessage的部分与IBusinessRequest相同。我需要这个requestMessage来设置PersonBusiness类的属性Request(从IRequestMessage实现)

我给静态IBusiness CreateBusinessInstance(Type type,object requestMessage)这两个变量,它给了我IBusiness,我进一步用它来执行一些业务功能。

全部谢谢!

的Gr

马亭

7 个答案:

答案 0 :(得分:4)

你的问题有点不清楚。通常对于要使用无参数构造函数创建新实例的泛型,使用new()约束:

public interface IBusinessRequest<T> where T : new()

然后,您可以在new T()的实施中使用IBusinessRequest<T>。无需自己检查 - 编译器会这样做。

然而,目前尚不清楚这是否真的是你在这里所追求的。您对“businessRequestType”的通用值IBusiness<>是什么意思?

答案 1 :(得分:3)

您可以使用两个泛型类型参数创建一个泛型方法 - 一个用于业务请求类型,另一个用于实现类型:

public static IBusinessRequest<T> CreateBusinessInstance<T, TImpl>() where TImpl : IBusinessRequest<T>, new()
{
    return new TImpl();
}

你的例子会像这样使用它:

IBusinessRequest<Person> request = CreateBusinessInstance<Person, PersonBusiness>();

答案 2 :(得分:2)

试试这个:

 Type[] typeArgs = { typeof(businessRequestType) };
return businessType.GetType.MakeGenericType(typeArgs);

答案 3 :(得分:1)

在给定Person类型时,您是否尝试创建PersonBusiness实例?

var typeMap = new Dictionary<Type, Func<object>>
{
    { typeof(Person), () => new PersonBusiness() }
};

var businessInstance = (IBusinessRequest<Person>)typeMap[typeof(Person)]();

答案 4 :(得分:1)

请参阅:How to dynamically create generic C# object using reflection?

编辑2:实际上不需要businessRequestType中的参数。根据定义,类型businessType将实现单一形式的IBusinessRequest&lt;&gt;所以没有必要通过businessRequestType。相应地修改了解决方案。

编辑1:问题仍然有点令人困惑。我认为问题比你想要的更受限制。需要定义businessType和businessRequestType的所有组合。您可以使用分层对象工厂的某些变体,例如:

// Creates an instance of T that implements IBusinessRequest<R>
public static IBusinessRequest<R> CreateBusinessInstance<T, R>() where T :
    IBusinessRequest<R>
{
    return Activator.CreateInstance<T>();
}


// Creates an instance of businessType that implements 
// IBusinessRequest<businessRequestType>
public static object CreateBusinessInstance(Type businessType)
{
    object biz = null;

    if (typeof(PersonBusiness) == businessType)
    {
        biz = CreateBusinessInstance<PersonBusiness, Person>();
    }
    //else if ... other business types


    if (null == biz)
    {
        throw new ApplicationException("Unknown type");
    }

    return biz;
}

答案 5 :(得分:0)

或者你可以使用另一种通用方法:

IBusinessRequest<T> CreateBusinessInstance<T>(T businessType) where T : new() {
}

您需要在设计时知道类型,但这可能没问题(完全不了解您的要求)......

干杯 的Matthias

答案 6 :(得分:0)

public class GridController<TModel> : Web.Controllers.ControllerBase where TModel : new()
    {
        public ActionResult List()
        {
            // Get the model (setup) of the grid defined in the /Models folder.
            JQGridBase gridModel = new TModel() as JQGridBase;

此方式您可以从基类继承并调用T模型类型

希望这个帮助

JU。