How to make a control in XAML public in order to be seen in other classes

Question

I\'m working in wpf application i made a checkbox in the XAML, then my code calls a function in a class and in this function there is an if condition where its checking on w

Answer 1

Using the following XML you can define a control as a public field on the class to be able to access it from other classes:

<CheckBox x:Name="myCheckBox" x:FieldModifier="public" />

Now you can access the field directly in code:

if (win.myCheckBox.IsChecked.Value)
{
    // ...
}

I agree with H.B., though, that using the MVVM pattern is a better way to do it. Other parts of your code shouldn't be aware of your UI or directly access it.

EDIT:

With the MVVM approach you should first define your view model class:

public class ViewModel
{
    public bool IsChecked { get; set; }
}

Then you set an instance of this class as DataContext:

• either in code, e.g. window constructor:

public MyWindow()
{
    InitializeComponent();
    DataContext = new ViewModel();
}

• or in XAML, e.g. App.xaml:

<Application x:Class="WpfApplication2.App"
             xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
             xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
             xmlns:vm="clr-namespace:WpfApplication2"
             StartupUri="MainWindow.xaml">
    <Application.Resources>
        <vm:ViewModel x:Key="ViewModel" />
    </Application.Resources>
</Application>

Now you can bind your CheckBox to a property in ViewModel:

<CheckBox IsChecked="{Binding IsChecked, Mode=TwoWay}" />

All that's left is to pass the ViewModel instance to your OnRender function. It is stored in the DataContext property of your window.

EDIT 2:

BTW: You really should have asked that before you accepted the answer.

I'm not sure what you are trying to attempt with the is_clicked property. To set this flag when the button is clicked, you need a Command:

public class CalibrateCommand : ICommand
{
    private ViewModel viewModel;

    public CalibrateCommand(ViewModel viewModel)
    {
        this.viewModel = viewModel;
    }

    public void Execute(object parameter)
    {
        viewModel.IsClicked = true;
    }

    public bool CanExecute()
    {
        return true;
    }
}

You add an instance of this command to your view model:

public class ViewModel
{
    public bool IsChecked { get; set; }
    public bool IsClicked { get; set; }
    public ICommand CalibrateCommand { get; set; }

    public ViewModel()
    {
        CalibrateCommand = new CalibrateCommand(this);
    }
}

You bind it to the button like this:

<Button Content="Calibrate" Height="24" x:Name="Calibrate" x:FieldModifier="public" Width="90" Click="Calibrate_Click" HorizontalContentAlignment="Center" VerticalContentAlignment="Center" HorizontalAlignment="Left" DockPanel.Dock="Left" Panel.ZIndex="0" Padding="0" VerticalAlignment="Center" Command="{Binding CalibrateCommand}" />

You don't need to handle any events of the CheckBox and the Button, everything is handled by the binding.

If you added a dependency property to KinectSkeleton you should bind it to the view model:

<kt:KinectSkeleton ViewModelH="{Binding}" />

Answer 2

Don't make the checkbox visible to the outside, just pass the current state of the checkbox to the function or class. Also consider binding the checkbox value to a class in the DataContext, directly accessing controls can be avoided most of the time in WPF, see also the MVVM pattern.