Setting DataContext in XAML in WPF

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This code will always fail.

As written, it says: “Look for a property named “Employee” on my DataContext property, and set it to the DataContext property”. Clearly that isn’t right.

To get your code to work, as is, change your window declaration to:

<Window x:Class="SampleApplication.MainWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    xmlns:local="clr-namespace:SampleApplication"
    Title="MainWindow" Height="350" Width="525">
<Window.DataContext>
   <local:Employee/>
</Window.DataContext>

This declares a new XAML namespace (local) and sets the DataContext to an instance of the Employee class. This will cause your bindings to display the default data (from your constructor).

However, it is highly unlikely this is actually what you want. Instead, you should have a new class (call it MainViewModel) with an Employee property that you then bind to, like this:

public class MainViewModel
{
   public Employee MyEmployee { get; set; } //In reality this should utilize INotifyPropertyChanged!
}

Now your XAML becomes:

<Window x:Class="SampleApplication.MainWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        xmlns:local="clr-namespace:SampleApplication"
        Title="MainWindow" Height="350" Width="525">
    <Window.DataContext>
       <local:MainViewModel/>
    </Window.DataContext>
    ...
    <TextBox Grid.Column="1" Grid.Row="0" Margin="3" Text="{Binding MyEmployee.EmpID}" />
    <TextBox Grid.Column="1" Grid.Row="1" Margin="3" Text="{Binding MyEmployee.EmpName}" />

Now you can add other properties (of other types, names), etc. For more information, see Implementing the Model-View-ViewModel Pattern

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