Sign in Register
Go to solution
Copied to clipboard

Flag this post as spam?

This post will be reported to the moderators as potential spam to be looked at


  • K.Garrein 164 posts 629 karma points
    Jun 17, 2013 @ 12:49
    K.Garrein
    0

    Get Distinct() on a List<DynamicNode>

    Hey.

    I create a list with multiple DynamicNode items.
    When I try to get the distinct items in the list, it doesn't remove the double items, I just get the same list...

    Is this a bug or a feature?

    e.g.

    List<DynamicNode> myList = new List<DynamicNode>();

    myList.Add( new DynamicNode( 1 ) );
    myList.Add( new DynamicNode( 2 ) );
    myList.Add( new DynamicNode( 2 ) );
    myList.Add( new DynamicNode( 1 ) );
    myList.Add( new DynamicNode( 3 ) );

    myList = myList.Distinct().ToList();

    Result: the list still contains the 5 items
    myList.Count --> 5

    Thank you.
    K.

    Copy Link
  • Lars-Erik Aabech 350 posts 1102 karma points MVP 8x c-trib
    Jun 21, 2013 @ 17:10
    Lars-Erik Aabech
    0

    A DynamicNode is a reference type. Reference types aren't equal unless you've overridden object.Equals() or it's the same instance.

    In your example, the instances at index 0 and 3 are different instances, and hence not equal.

    Distinct can however use a custom equality comparer.

    Have a look here:
    http://msdn.microsoft.com/en-us/library/bb338049.aspx (plain static comparer for a given case)
    and here:
    http://stackoverflow.com/questions/1071609/iequalitycomparer-for-anonymous-type (utility comparer where you specify a lambda for your current need)

    Copy Link
  • Ali Sheikh Taheri 470 posts 1648 karma points c-trib
    Jun 21, 2013 @ 17:21
    Ali Sheikh Taheri
    100

    how about this! first check if the item is not in your list then add it like below:

    var mylist = new List();

    if (!mylist.Select(x => x.Id).Contains(1))
        {
            mylist.Add(new DynamicNode(1));
        }
    Copy Link
  • Jeroen Breuer 4908 posts 12265 karma points MVP 5x admin c-trib
    Jun 21, 2013 @ 19:55
    Jeroen Breuer
    0

    What version of Umbraco are you using? In the newer version there are some useful extension methods. If you add the Umbraco.Core namespace you'll be able to do this:

    var nodes = myList.DistinctBy(x => x.Id);

    Jeroen

    Copy Link
  • K.Garrein 164 posts 629 karma points
    Jun 24, 2013 @ 08:38
    K.Garrein
    0

    I think it must be version 6.0.5 or maybe even newer.

    I think I tried the Distinct( x => x.Id ) but it was not working.

    I changed the code to use a List<int> with the node id's, so I got it working that way.
    When I need something similar again I will retry your other solutions :)

    Thank you.
    Kris.

    Copy Link
  • Lars-Erik Aabech 350 posts 1102 karma points MVP 8x c-trib
    Jun 24, 2013 @ 10:22
    Lars-Erik Aabech
    1

    Note that @Jeroen showed a method called DistinctBy, not Distinct. It's a custom extension in umbraco, not the BCL.

    Copy Link
Please Sign in or register to post replies

Write your reply to:

Draft