[英]How can I create a new instance of a class at run time using c#?
I have the following class which I pass Entity Framework Model when making an instance of it.我有以下类,我在创建它的实例时传递了实体框架模型。
public class TableMapper<TSource>
{
protected IQueryExtractor QueryExtractor { get; set; }
protected string TableAliasPrefix { get; private set; }
public TableMapper(IQueryExtractor queryExtractor, string tableAliasPrefix = null)
{
QueryExtractor = queryExtractor;
TableAliasPrefix = tableAliasPrefix;
}
public IReportRelation GetForeignRelation<TProperty>(Expression<Func<TSource, TProperty>> property)
{
return new ReportRelation
{
Column = GetReportColumn(null, Self, property),
TableName = this.Table(),
TableAlias = this.GetSqlAlias(),
ModelType = typeof(TSource),
QueryExtractor = this.QueryExtractor
};
}
...
...
...
}
I call this above class yet from another class like so我把这个班级称为另一个班级,就像这样
var clientMapper = new TableMapper<Client>(QueryExtractor, "Client");
var = clientMapper.GetForeignRelation(x => x.Id);
Please note that Client
is an Entity Framework 6 model.请注意,
Client
是实体框架 6 模型。
From within my GetForeignRelation
method, I set the type of Client
or <TSource>
so I can create a new instance of the TableMapper
class at run time.在我的
GetForeignRelation
方法中,我设置了Client
或<TSource>
的类型,以便我可以在运行时创建TableMapper
类的新实例。
Here is what I tried to do in an attempt to create a new instance of it at run time.这是我尝试在运行时创建它的新实例时尝试做的事情。
var RunTimeModel = Activator.CreateInstance(relationsMapping.ModelType);
var RunTimeMapper = new TableMapper<RunTimeModel>(relationsMapping.QueryExtractor, relationsMapping.TableAlias);
But that is giving me an error.但这给了我一个错误。
The type or namespace name
RunTimeModel
could not be found (are you missing a using directive or an assembly reference?)RunTimeModel
类型或命名空间名称RunTimeModel
(您是否缺少 using 指令或程序集引用?)
How can I correctly create a new instance of the same class at run time?如何在运行时正确创建同一类的新实例?
You will get the class name of TSource with您将获得 TSource 的类名
typeof(TSource).Name
You should be able to create an instance with:您应该能够创建一个实例:
var instance = Activator.CreateInstance(typeof(TSource));
as long as the type is in the same assembly Other properties can be use to obtain the full namespace只要类型在同一个程序集中就可以使用其他属性来获取完整的命名空间
EDIT:编辑:
Do this example apply to your scanario?此示例适用于您的 scanario 吗? Had to strip down your provided code as I don't know what's behind it all :)
不得不剥离您提供的代码,因为我不知道这背后是什么:)
class Program
{
static void Main(string[] args)
{
var test = new TableMapper<A>();
var fr = test.GetForeignRelation();
var type = fr.ModelType;
var newInstance = Activator.CreateInstance(type);
}
}
public class TableMapper<TSource>
{
public TableMapper()
{
}
public ReportRelation GetForeignRelation()
{
return new ReportRelation
{
ModelType = typeof(TSource)
};
}
}
public class ReportRelation
{
public Type ModelType { get; set; }
}
public class A
{
public string Test { get; set; }
public A()
{
Test = "Some string";
}
}
You cannot pass variable to a type parameter您不能将变量传递给类型参数
var RunTimeModel = Activator.CreateInstance(relationsMapping.ModelType);
var RunTimeMapper = new TableMapper<RunTimeModel>(relationsMapping.QueryExtractor, relationsMapping.TableAlias)
Because RunTimeModel is not a type but a variable.因为 RunTimeModel 不是类型而是变量。 I think you are missing purpose of method which you are using.
我认为您缺少正在使用的方法的目的。
You can use the following code to instantiate your generic type :您可以使用以下代码来实例化您的泛型类型:
var RunTimeMapper = Activator.CreateInstance(typeof(TableMapper<>).MakeGenericType(new[] { relationsMapping.ModelType }), new[] { relationsMapping.QueryExtractor, relationsMapping.TableAlias });
However,the return value type for RunTimeMapper is object.I cann't figure out how to cast RunTimeMapper to its actual type.So,it's a little inconvenient to use.但是RunTimeMapper的返回值类型是object。我不知道如何将RunTimeMapper转换为它的实际类型。所以,使用起来有点不方便。
Referer: How to: Examine and Instantiate Generic Types with Reflection Referer: 如何:使用反射检查和实例化泛型类型
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