How to instantiate DataContext object in XAML

戏子无情 提交于 2019-11-29 01:44:05

问题


I want to be able to create an instance of the DataContext object for my WPF StartupUri window in XAML, as opposed to creating it code and then setting the DataContext property programmaticly.

The main reason is I don't need to access the object created externally and I don't want to have to write code behind just for setting the DataContext.

I'm sure I've read somewhere how to instantiate the DataContext object in XAML but I can't find it in any of the usual places...


回答1:


You add an XML namespace for whatever namespace your DataContext lives in, create an instance of it in the Window Resources and set the DataContext to that resource:

<Window x:Class="WpfApplication4.Window1"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    xmlns:local="clr-namespace:WpfApplication4"
    Title="Window1" Height="300" Width="300">
    <Window.Resources>
        <local:MyViewModel x:Key="MyViewModel"/>
    </Window.Resources>
    <Grid DataContext="{StaticResource MyViewModel}">

    </Grid>
</Window>



回答2:


You can just specify this directly in XAML for the entire Window:

<Window 
    ... xmlns definitions ...
>
   <Window.DataContext>
        <local:CustomViewModel />
   </Window.DataContext>
</Window>

This creates a view model named "CustomViewModel" in the namespace aliased to local, directly as the DataContext for the Window.




回答3:


Assuming this code:

public abstract class BaseView { }
public class RuntimeView : BaseView { }
public class DesigntimeView : BaseView { }

Try this:

<Page.DataContext>
    <local:RuntimeView />
</Page.DataContext>
<d:Page.DataContext>
    <local:DesigntimeView />
</d:Page.DataContext>
<ListBox ItemsSource="{Binding}" />

Best of luck!




回答4:


If you need to set the DataContext as same control class:

    <Window x:Class="TabControl.MainWindow"
            xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
            xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"        
            xmlns:local="clr-namespace:TabControl"
            Title="MainWindow" Height="350" Width="525"
            DataContext="{Binding RelativeSource={RelativeSource Self}}"        
            >
</Window>

use RelativeSource binding.

or just

     <Window x:Class="TabControl.MainWindow"
                xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
                xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"        
                xmlns:local="clr-namespace:TabControl"
                Title="MainWindow" Height="350" Width="525"                        
                >
<Window.DataContext>
< new instance of any viewModel here....>
</Window.DataContext>
    </Window>

If want to assign an instance of different class than itself.



来源:https://stackoverflow.com/questions/1521699/how-to-instantiate-datacontext-object-in-xaml

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