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Greetings,
I am basically trying to iterate over all of the top-level rows in a RadGridView that are not filtered out. I have some code that seems to work using the Records property of the RadGridView, but apparently a couple of the related classes (Record/DataRecord) have been obsoleted (Q2.2009) and I can't seem to find substitutes. Here is a snippet of my code:
Ultimately, I am hoping to get a distinct list of values from a certain property from the list of bound business objects; however, it must be visible in the grid in order to be inserted into this list. Any thoughts on how I could achieve this using classes that are not obsoleted? Or maybe there is a better approach overall...
I should also mention that I am using Silverlight 2.
As always, thanks for your assistance.
I am basically trying to iterate over all of the top-level rows in a RadGridView that are not filtered out. I have some code that seems to work using the Records property of the RadGridView, but apparently a couple of the related classes (Record/DataRecord) have been obsoleted (Q2.2009) and I can't seem to find substitutes. Here is a snippet of my code:
| List<ControllerChannelType> controllerChannelTypesList = new List<ControllerChannelType>(); |
| foreach (Record record in this.GridView.Records) |
| { |
| DataRecord dataRecord = record as DataRecord; |
| System.Diagnostics.Debug.Assert(dataRecord != null); |
| if (dataRecord != null) |
| { |
| ChannelView channelView = dataRecord.Data as ChannelView; |
| System.Diagnostics.Debug.Assert(channelView != null); |
| if(channelView != null) |
| { |
| controllerChannelTypesList.Add(channelView.Channel_ChannelType); |
| } |
| } |
| } |
Ultimately, I am hoping to get a distinct list of values from a certain property from the list of bound business objects; however, it must be visible in the grid in order to be inserted into this list. Any thoughts on how I could achieve this using classes that are not obsoleted? Or maybe there is a better approach overall...
I should also mention that I am using Silverlight 2.
As always, thanks for your assistance.
