66
votes

I'm facing some issues with a WPF binding to a static property. In my application there is a class called Globals which contains a static property to ThisAddIn. This object allows me to access a User property which basically is what I want to use in my Binding. So, my XAML looks like this:

<TextBlock Text="{Binding Path=User.Name
                   , Source={x:Static Member=s:Globals.ThisAddIn}}" />

The namespace s is declared at the top by the following line:

xmlns:s="clr-namespace:ConsoleApplication16.Model"

I found a ton of different approaches of how to refer to the static Property in XAML, but none of them worked except for this one, which also seems the most logical for me. Some samples I found, did not use Path or Member at all.

After some testing I found the right way to do this, is using "x:Static Member" to define the class and the static property you want to use, while Path defines the correct Properties inside this object, just like a normal binding would do.

Although this solution compiles without any complains, a XAMLParseException pops up directly telling me that the StaticExtension value cannot be resolved to an enumeration, static field, or static property.

2
How exactly is that the right way if it throws an exception? - H.B.
Also you should be able to drop the Path= and Member= because that will invoke the respective contructors which set the relevant properties. - H.B.
I usually bind to static classes/properties using {Binding Source={x:Static local:MyStaticClass.SomeStaticProperty.PropertyName}. Just be sure your properties have get/set assessors. For example, be sure you use public string PropertyName { get; set; } instead of public string PropertyName; - Rachel
@Rachel: That surely won't work as x:Static has a fixed syntax of ns:Class.Property. - H.B.
@H.B. You're right, I'm not sure why I put the PropertyName in there. {Binding PropertyName, Source={x:Static local:MyStaticClass.SomeStaticProperty} - Rachel

2 Answers

121
votes

Ensure Globals.ThisAddIn is public.

You may also get this if you are using a resource file in which case you also need to ensure the access modifier is set to Public:

enter image description here

36
votes

In output library project, you should ensure that the Resources.resx file's generator has been modified as PublicResXFileCodeGenerator instead of ResXFileGodeGenerator by default.

Replace following with...

Not public resource

This one

correct public resource