Identify Tool, multiple layers, Hyperlinks

Question

The Identify Tool in the SDK works great, but how would you go about setting up one of the returned field as a hyperlink? In cases where multiple layers are returned from the Identify Tool and each layer has one or more hyperlink "fields", how would you go about creating these hyperlinks in the SL datagrid for the Identify Tool? I am looking at the DataGridTemplateColumn.CellTemplate sample I found but since each layer has different fields name is this possible? (see code below, converter check for null value in a field) Note: the dataGrid will have many fields returned but one or more can be a hyperlink field? ESRI - Ideally if we can have a Datatype of Hyperlink that we can define within the feature class (see attached pic) and have it display as a hyperlink in a SL dataGrid without going thru extra steps would be great. Any guidance will be appreciated. Below is the code from the Identify Tool that pertains to the DataGrid: Hyperlink.png

ErikEngstrom · Answer

Dave,
I was able to get it to work! I've made a few tweaks to your code, but for the most part, it works great! I'm still very new to Silverlight (and programming in general). These snippets that the user community provides makes things much more managable and has taught me greatly!

This code is a perfect example of what should have been included from the beginning!
THANKS AGAIN DAVE!

DaveTimmins · Answer

Hi Erik, assuming you have the previous code as a control you need to amend your xaml to use that to display the value, something like

ErikEngstrom · Answer

Thanks Dave, but how would I bind the TextValue to the IdentifyResults? I created the control and added it to my generic.xaml. I can see the stack panel in my mainpage.xaml. However I believe I need point the control to what is returned in my identify results and that's where I'm running into an issue now. As it is, I have a datagrid that binds the results of the IdentifyResults (Key/Value). public void ShowFeatures(List results) { _dataItems = new List(); if (results != null && results.Count > 0) { IdentifyComboBox.Items.Clear(); foreach (IdentifyResult result in results) { Graphic feature = result.Feature; string title = result.Value.ToString() + " (" + result.LayerName + ")"; _dataItems.Add(new DataItem() { Title = title, Data = feature.Attributes }); IdentifyComboBox.Items.Add(title); } IdentifyComboBox.SelectedIndex = 0; } } ShowFeatures(args.IdentifyResults);