110,535
社区成员
发帖
与我相关
我的任务
分享
<?xml version="1.0" encoding="utf-8"?>
<configuration>
<appSettings>
<add key="MenuOne" value="FormOne" />
</appSettings>
</configuration>
public class GetObject
{
/// <summary>
/// GetObject getobj = new GetObject("ConsoleApplication1", "ConsoleApplication1.Bll", "Sendmail");
/// object obj = GetObject.Dinstanceclass["Cacul"];
/// </summary>
public static Dictionary<string, object> Dinstanceclass = new Dictionary<string,object>();
public GetObject(string properties,string namespaces,string classname)
{
Assembly ass = Assembly.Load(properties);//程序集名称
Type ty=ass.GetType(string.Concat(namespaces,".",classname));//命名空间名称.类名
object obj=(object )Activator.CreateInstance(ty);//实例化对象(泛型) 后面可根据classname返回具体的类实例
if(!Dinstanceclass.ContainsKey(classname))
Dinstanceclass[classname] = obj;
}
}
零时写了个方法,你看看Type ty = Assembly.Load("程序集名称/dll名称").GetType(string.Format("命名空间名称.{0}",FormOne");
T obj= (T)Activator.CreateInstance(ty);
return obj;
object obj = Activator.CreateInstance(Type.GetType("FormOne"));