Question

Is there a way to create a ResourceDictionary Locator.
Right now I have in xaml:

 <ResourceDictionary>
            <ResourceDictionary.MergedDictionaries>
                <ResourceDictionary Source="../../Resources/StringResources.xaml"/>
            </ResourceDictionary.MergedDictionaries>
        </ResourceDictionary>

Where the code behind sets the list of dictionaries based on currentculture

I’d like to have

 <ResourceDictionary>
            <ResourceDictionary.MergedDictionaries>
                <Locator "StringResources"/>
            </ResourceDictionary.MergedDictionaries>
        </ResourceDictionary>

Or something like that such that I don’t have to modify each v.xaml.cs files

Answer 1

You could create your own ResourceDictionary:

public class ResourceDictionaryLocator : ResourceDictionary
{
    public ResourceDictionaryLocator()
    {
        switch (CurrentLocalization)
        {
            case "English":
                base.Source = new Uri("pack://application:,,,/Languages/English.xaml");
                break;

            case "French":
                base.Source = new Uri("pack://application:,,,/Languages/French.xaml");
                break;
        }
    }
}

Then consume it like this:

<Application x:Class="TestingWPF.App"
             xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
             xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
             xmlns:local="clr-namespace:TestingWPF"
             StartupUri="MainWindow.xaml">
    <Application.Resources>
        <ResourceDictionary>
            <ResourceDictionary.MergedDictionaries>
                <local:ResourceDictionaryLocator />
            </ResourceDictionary.MergedDictionaries>
        </ResourceDictionary>
    </Application.Resources>
</Application>

This is generally not done, however. You can load different resource dictionaries at runtime. Something like the following:

Application.Current.Resources.MergedDictionaries.Clear();
Application.Current.Resources.MergedDictionaries.Add(dictionary);