Reputation: 39364
I have an Enum and a Class:
public enum Colors { Red = 1, Yellow = 2 }
public class Color {
public int Value { get; set; }
public string Name { get; set; }
}
Can I create a List from the enum Colors values using a lambda expression?
Thank you, Miguel
Upvotes: 1
Views: 7972
Reputation: 74227
Theres at least two ways to do it (though I can't imagine why you'd need to):
static void Main( string[] args )
{
List<Color> colorList1 = Enum.GetValues(typeof(Colors)).Cast<Colors>().Select( c => new Color( c.ToString(), (int)c ) ).ToList() ;
List<Color> colorList2 = Enum.GetNames(typeof(Colors)).Zip( Enum.GetValues(typeof(Colors)).Cast<int>() , (n,v)=> new Color(n,v)).ToList() ;
return ;
}
class Color
{
public Color( string name , int value )
{
Name = name ;
Value = value ;
}
public string Name { get ; set ; }
public int Value { get ; set ; }
public override string ToString()
{
return string.Format( "{0}:{1}" , Name , Value ) ;
}
}
enum Colors
{
Red = 1 ,
Orange = 2 ,
Yellow = 3 ,
Green = 4 ,
Blue = 5 ,
Indigo = 6 ,
Violet = 7 ,
}
Upvotes: 1
Reputation: 16498
Assumes you meant for the Name
property to be a string
instead of an int
( ( IEnumerable<int> )Enum.GetValues(typeof(Colors)) )
.Select(ev =>
new Color()
{
Value = ev,
Name = Enum.GetName(typeof(Colors), ev)
})
.ToList();
Upvotes: 6
Reputation: 66439
This will give you a list of the enum values. You don't need to create a custom Color
class.
var myColorsList = Enum.GetValues(typeof (Colors)).Cast<Colors>().ToList();
If you really want to use your Color
object:
var myColors = Enum.GetValues(typeof (Colors)).Cast<Colors>()
.Select(x => new Color{Value = (int)x, Name = x.ToString()})
.ToList();
Upvotes: 1
Reputation: 15549
First change the Name property to a string and then try this:
public enum Colors { Red = 1, Yellow = 2 }
public class Color
{
public int Value { get; set; }
public string Name { get; set; }
}
static void Main(string[] args)
{
var colorClasses = from name in Enum.GetNames(typeof(Colors))
select
new Color()
{
Value = (int)Enum.Parse(typeof(Colors), name),
Name = name
};
}
Upvotes: 1