WPF/MVVM Load an UserControl at Runtime

Your ViewModels should not care about UserControls. Instead, have them hold ViewModels, and let WPF resolve how to draw the ViewModel with a DataTemplate.

For example,

<Window x:Class="test.Views.StartView"        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"        xmlns:views="clr-namespace:test.Views"        xmlns:viewmodels="clr-namespace:test.ViewModel"        Title="Window1" Height="300" Width="516">    <Window.Resources>        <DataTemplate DataType="{x:Type viewmodels:UCStastistikViewModel}">            <views:UCStastistikView />        </DataTemplate>    </Window.Resources>    <Grid>        <Menu IsMainMenu="True" Margin="0,0,404,239">            <MenuItem Header="_Einstellungen">                <MenuItem Header="Server" />            </MenuItem>        </Menu>        <ContentControl Content="{Binding LoadedControl}" Margin="0,28,0,128" />    </Grid></Window>

Also, get rid of the <UserControl.DataContext> in your UserControl. The DataContext should be passed in by whatever is using the control, not defined in the UserControl :)

Edit

Based on your comment to an earlier answer about switching out the content of the StartPage by switching a ViewModel, you may be interested in looking at this post of mine. It's titled Navigation with MVVM, however the same concept applies for switching Views or UserControls

Basically, you'd make the property LoadedControl of type ViewModelBase instead of hard-coding it's type, and then set it to whatever object you want displayed in your ContentControl. WPF's DataTemplates will take care of hooking up the correct View for the ViewModel.

WPF doesn't support automatic resolution of the view for a given view model. The naive solution to your problem would be to directly add the UCStatistikView to your StartView and bind the VM to it

<Window x:Class="test.Views.StartView"    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"    xmlns:views="clr-namespace:test.ViewModel"    Title="Window1" Height="300" Width="516"><Grid>    <Menu IsMainMenu="True" Margin="0,0,404,239">        <MenuItem Header="_Einstellungen">            <MenuItem Header="Server" />        </MenuItem>    </Menu>    <UCStatistikView DataContext="{Binding LoadedControl}" Margin="0,28,0,128" /></Grid>

A more elaborated approach would be to implement a ViewLocator (view model first approach) or a ViewModelLocator (view first approach). The locator automatically locates and binds the view(s) to a view model. There are some MVVM frameworks/toolkits out there which implement such a locator.

• Caliburn.Micro: Offers a flexible ViewLocator and ViewModelLocator based on naming conventions. Here is an article about them
• MVVM Light: Offers a ViewModelLocator. Here is an introduction

With Caliburn.Micro you start view would look like this

<Window x:Class="test.Views.StartView"    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"    xmlns:views="clr-namespace:test.ViewModel"    Title="Window1" Height="300" Width="516"><Grid>    <Menu IsMainMenu="True" Margin="0,0,404,239">        <MenuItem Header="_Einstellungen">            <MenuItem Header="Server" />        </MenuItem>    </Menu>    <ContentControl cm:View.Model="{Binding LoadedControl}" Margin="0,28,0,128" /></Grid>

The cm:View.Model="{Binding LoadedControl}" attached property tells caliburn to locate the view for the view model which is bound and set the Content property of the ContentControl to it.

Here's what you should do:

The StartView:

<Window x:Class="test.Views.StartView"        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"        xmlns:views="clr-namespace:test.ViewModel"        Title="Window1" Height="300" Width="516">    <Grid>        <Menu IsMainMenu="True" Margin="0,0,404,239">            <MenuItem Header="_Einstellungen">                <MenuItem Header="Server" />            </MenuItem>        </Menu>        <UCStatistikView x:Name="myUCStatistikView" Margin="0,28,0,128" />    </Grid></Window>

the StartViewModel:

namespace test.ViewModel{    public class StartViewModel : ViewModelBase    {        private UCStastistikViewModel _myControlViewModel;        public StartViewModel()        {            _myControlViewModel = new UCStastistikViewModel();        }        public UCStastistikViewModel MyControlViewModel        {            get { return _myControlViewModel; }            set            {                if (value == _myControlViewModel)                    return;                _myControlViewModel = value;                OnPropertyChanged("MyControlViewModel");            }        }    }}

UCStatistikView:

<UserControl x:Class="test.Views.UCStatistik"             xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"             xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"             xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"              xmlns:d="http://schemas.microsoft.com/expression/blend/2008"              xmlns:vm="clr-namespace:test.ViewModel"             mc:Ignorable="d"              d:DesignHeight="188" d:DesignWidth="508">    <Grid Background="red">           </Grid></UserControl>

Code behind of your StartView:

this.myUCStatistikView.DataContext = ((StartViewModel)this.DataContext).MyControlViewModel;

After having tested different approaches, my conclusion is that the best way for datacontext bindings is the code-behind of the parent view in case you have userControls.

EDIT: ViewModel locators are fine for trivial examples but if your ViewModel must be instantiated dynamically (it's mostly the case when its constructor requires parameters), the you can't use it.I personally stopped using locators because of this.