Window Startup Location from resource dictionary

问题:

I'm trying to set all my windows to open in the center of the screen. All my windows use style file:

    <Window.Resources>
    <ResourceDictionary>
        <ResourceDictionary.MergedDictionaries>
            <ResourceDictionary Source="../Styles/Mystyles.xaml"/>
        </ResourceDictionary.MergedDictionaries>
    </ResourceDictionary>
</Window.Resources>

So I just inserted this property to the resource dictionary:

    <Style x:Key="windowStyle" TargetType="Window">
         <Setter Property="WindowStartupLocation" Value="CenterScreen"/>
    </Style> 

But, it doesn't work. Am I missing something?

回答1:

You cannot use a Style to define WindowStartupLocation, this is due to the fact that it is not a dependency property. You can define a StaticResource in your resource dictionary which you will use in your windows:

<WindowStartupLocation x:Key="StartupLocation">CenterScreen</WindowStartupLocation>

and then use it like so:

WindowStartupLocation="{DynamicResource StartupLocation}"

回答2:

You don't need to use x:Key attribute. Your style must look like this:

<Style TargetType="{x:Type Window}">
    <Setter Property="WindowStartupLocation" Value="CenterScreen"/>
</Style> 

回答3:

if you dont want to use implicit styles (as bniwredyc suggested) you must set style explicitly:

<Window **Style="{StaticResource windowStyle}"**>
<Window.Resources>
    <ResourceDictionary>
        <ResourceDictionary.MergedDictionaries>
            <ResourceDictionary Source="../Styles/Mystyles.xaml"/>
        </ResourceDictionary.MergedDictionaries>
    </ResourceDictionary>
</Window.Resources>

回答4:

To make every start every window at Center Screen Add this line in App.xaml

<Application.Resources>
        <WindowStartupLocation x:Key="StartupLocation">CenterScreen</WindowStartupLocation>
</Application.Resources>

and add this line in Window tag

WindowStartupLocation="{StaticResource StartupLocation}"