c# - WPF DataType resolution for interface DataContexts -

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• automatic template selection in wpf not working interface 1 answer

i've set generic dialog window takes in viewmodel , title, display according window's xaml.

for example, code sets title , data context of window , shows it:

public void showwindow(object viewmodel, string title) {     var win = new dialogwindow()     {         title = title,         datacontext = viewmodel     };     win.show(); } 

the window's codebehind contains nothing of relevance, xaml content presenter looks like:

<contentpresenter x:name="dialogpresenter" content="{binding}">     <contentpresenter.resources>         <datatemplate datatype="{x:type i:imyinterface}" >             <c:myinterfacescontrol/>         </datatemplate>     </contentpresenter.resources> </contentpresenter> 

the problem datatemplate doesn't fire, viewmodel passed in implementation of interface. however, want code detect whether viewmodel implements that, , use template accordingly.

i've thought of 1 or 2 ways around problem using converters, example checking whether implementation of converter parameter.

but there simpler way here?

edit: not same question suggested in comments (edit 2: that's gone); won't know interface type (many different ones may passed in), , instead have data context of type object work with.

datatemplate doesn't work types interfaces, should use concrete type.

if have interfaces defined

public interface ia {     string { get; set; } } 

and concrete type implementing it:

public class aa : ia {     public string { get; set; } } 

then following data template will not used, when passing new a();

<datatemplate datatype="{x:type local:ia}">     <textblock text="{binding a}" /> </datatemplate> 

instead see like

application.aa

solution 1

you can use concrete type , work:

<datatemplate datatype="{x:type local:aa}">     <textblock text="{binding a}" /> </datatemplate> 

solution 2

use property of interface type , bind (not concrete type), described in this answer.

note: can use same name of property return different interface (the 1 choose data template) in concrete types.

solution 3

you can use template selector. simple, give keys data template:

<datatemplate x:key="ia">     <textblock text="{binding a}" /> </datatemplate> 

and choose 1 use in code:

public class templateselector : datatemplateselector {     public override datatemplate selecttemplate(object item, dependencyobject container)     {         if (item ia)             return (datatemplate)((frameworkelement)container).findresource("ia");          return base.selecttemplate(item, container);     } } 

contentcontrol has contenttemplateselector property, don't use contentpresenter this.