Get the name of a class as a string in C#
C#Entity FrameworkC# Problem Overview
Is there a way to take a class name and convert it to a string in C#?
As part of the Entity Framework, the .Include method takes in a dot-delimited list of strings to join on when performing a query. I have the class model of what I want to join, and for reasons of refactoring and future code maintenance, I want to be able to have compile-time safety when referencing this class.
Thus, is there a way that I could do this:
class Foo
{
}
tblBar.Include ( Foo.GetType().ToString() );
I don't think I can do GetType() without an instance. Any ideas?
C# Solutions
Solution 1 - C#
You can't use .GetType()
without an instance because GetType
is a method.
You can get the name from the type though like this:
typeof(Foo).Name
And as pointed out by Chris, if you need the assembly qualified name you can use
typeof(Foo).AssemblyQualifiedName
Solution 2 - C#
Include requires a property name, not a class name. Hence, it's the name of the property you want, not the name of its type. You can get that with reflection.
Solution 3 - C#
You could also do something like this:
Type CLASS = typeof(MyClass);
And then you can just access the name, namespace, etc.
string CLASS_NAME = CLASS.Name;
string NAMESPACE = CLASS.Namespace;
Solution 4 - C#
typeof(Foo).ToString()
?
Solution 5 - C#
Alternatively to using typeof(Foo).ToString()
, you could use nameof():
nameof(Foo)
Solution 6 - C#
You can use an DbSet<contact>
instead of ObjectSet<contact>
, so you can use lambda as a parameter, eg tblBar.Include(a => a.foo)
Solution 7 - C#
Another alternative using reflection, is to use the MethodBase class.
In your example, you could add a static property (or method) that provides you with the info you want. Something like:
class Foo
{
public static string ClassName
{
get
{
return MethodBase.GetCurrentMethod().DeclaringType.Name;
}
}
}
Which would allow you to use it without generating an instance of the type:
tblBar.Include(Foo.ClassName);
Which at runtime will give you:
tblBar.Include("Foo");