I was working on a project using WPF, I want to modify the MainWindow to have some extended window styles. I have referred some docs and found out that I can set the window styles using SetWindowLong(), but in my case it will not work because I am trying to use WS_EX_NOREDIRECTIONBITMAP as the extended window style. when I use SetWindowLong() it does nothing. I have read somewhere that WS_EX_NOREDIRECTIONBITMAP can only be set while creating the window. So is there any way that i could modify the CreateWindowEx() of the WPF MainWindow.
I have found another alternate method to achieve this but it is by using the Undocumented API function.(SetWindowCompositionAttribute()). I need my project to be stable so is there any other method to achieve this?
If SetWindowCompositionAttribute() can set WS_EX_NOREDIRECTIONBITMAP at runtime, there must be a workaround to do this.
Here is the screenshot of what i want to achieve :
It is possible to do it in C++ but in C# WPF i haven't found a way to do this.
If you want to set WS_EX_NOREDIRECTIONBITMAP during creating MainWindow then you can create MainWindow manually. Just remove StartupUri from App.xaml and create new event handler for Startup event.
<Application x:Class="WpfApp1.App"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
Startup="App_Startup">
App.xaml.cs
void App_Startup(object sender, StartupEventArgs e)
{
// replace with your code
MainWindow window = new MainWindow();
window.Show();
}
But it would be useful if you can shore more code.
Related
So I'm still learning how to do C# applications and I have actually a problem with a window.
I've created a WPF project and I've separated some part of my main window in sub-parts (user-controls) so I can have a cleaner xaml code to work with.
I have a lot of different UserControl such as UserControlMenuStrip. All of them are inside the MainWindow.
Inside the MenuStrip was a MenuItem called Parameters :
<MenuItem Header="_Parameters" x:Name="MenuParameters"/>
I have created a new window called ParametersWindow. My goal was to open a child window centered with the main window when I click on the item.
But I don't really know how to proceed? Should I make a click= event and write down the code inside the linked UserControlMenuItem.xaml.cs linked file? Or in the MainWindow.xaml.cs file? Or maybe a new and clean file?
When I try to put it inside UserControlMenuItem.xaml.cs, I can't properly set the owner of the window I create this method but I can't set the owner:
private void OpenParametersWindow()
{
WindowParameters WinParam = new WindowParameters();
WinParam.Owner = MainWindow();
WinParam.WindowStartupLocation = WindowStartupLocation.CenterOwner;
WinParam.Show();
}
And, when I try via the MainWindow.xaml.cs I can't even get the variable...
So... How can I properly open the Window properly? And should I do it in the xaml.cs file or create a new one for a better understanding?
I've Created a class and added a static field as MainWindow to holding reference
class ReferenceClass
{
public static MainWindow mainWindow = null;//firstly null.we will set it in WindowLoaded event.
}
You can create a class like this for accessing reference of your MainWindow from wherever you want.Give your MainWindow reference to its static field.
MainWindow Loaded Event
private void Window_Loaded(object sender, RoutedEventArgs e)
{
ReferenceClass.mainWindow = this; //setting the reference to static field of ReferenceClass.
}
Menu Click (Event called on MenuStrip UserControl)
private void MenuItem_Click(object sender, RoutedEventArgs e)
{
NewWindow nw = new NewWindow();
nw.Owner = ReferenceClass.mainWindow;//Calling the reference of MainWindow from our class.
nw.WindowStartupLocation = WindowStartupLocation.CenterOwner;
nw.Show();
}
Here we go
Project
This is just a way for solving this issue.We can find more solutions which are better than mine but i use this solution when i need.
A handle to the parent window for any usercontrol can be obtained that way:
Window wndParent = System.Windows.Window.GetWindow(this);
WinParam.Show(wndParent);
But when working with WPF, it is more convenient to use the MVVM pattern
As you writing UI in WPF it is preferred to implement it with MVVM pattern. It allows you to have a clear separation of concerns between code and presentation.
Regarding your question about how to set Owner I suggest reading this series of articles about dialogs in WPF implemented with MVVM in mind
https://www.c-sharpcorner.com/article/dialogs-in-wpf-mvvm/
I am trying to learn WPF. I have done mostly back-end programming, except I did some C++ UI programming in the nineties. So far, I have created a simple maintenance application with a few screens and I can run it fine. I can navigate around, insert records and whatnot. However, I have to set my app.xaml startup location to MainWindow.xaml and then instantiate my actual window inside the C# code of the class linked to it. If I delete the MainWindow.xaml file and set my StartupLocation to wndMyMainWindow.cs, I get an error saying that it can not find the file. Is there any way around this? It seems sort of weird to require a non C# file type in what is supposed to be a C# UI framework.
In your App.xaml remove the StartupUri="MainWindow.xaml". Then add Startup="App_OnStartup" and create the matching method in your App.xaml.cs file like:
public partial class App : Application
{
private void App_OnStartup(object sender, StartupEventArgs e)
{
// do some code stuff like initializing your ViewModel or something else
// Instanciate the view you want to display and show it
MainWindow mainWindow = new MainWindow();
mainWindow.ShowDialog();
}
}
I currently have one window designed in WPF and coded in C#. I want one of my buttons to open another window, which I would also like to design in WPF. What is the best way for me to do this? Can I make multiple xaml files and call them from the same .cs class? Or should I just have one xaml file? I tried to add a new window into my xaml but it won't allow me to do that. I want all the code to be in the same C# class.
Yes, you can have multiple XAML files and call them from the same .cs file.
For exemple, let's say you have Window1.xaml and Window2.xaml. Window1 is your main window, and the code behind will look like this :
public partial class Window1 : Window
{
public MainWindow()
{
InitializeComponent();
}
}
In Window1 you have a button named btnOpenWindow. On click, you may do that to open Window2 :
private void btnOpenWindow_Click(object sender, RoutedEventArgs e)
{
var window = new Window2();
window.Show();
}
Then a new Window2 is opened.
However you won't be able to get events or others things coming from Window2 in Window1.xaml.cs, obviously you will control that in Window2.xaml.cs for exemple.
You should use the MVVM pattern in your project.
So you have different windows and just one ViewModel to handel these views and your data.
Have a look on: MVVM: Tutorial from start to finish?
I'm accustomed to Winforms where you can create a window and display it via:
Window.ShowDialogue();
I'm using the default MainWindow.xaml in a class library project. I had to delete the App.xaml file to complete the conversion. I want to launch the main window in a simple test. E.G.
[TestMethod]
public void RunPd()
{
MainWindow window = new MainWindow();
window.ShowDialogue();
}
Show/ShowDialogue() is not available. All I have is
GetChildren<>, GetParents<>, InitializeComponent and LoadTree<>.
How can I display MainWindow.xaml?
UPDATE
Main window code:
public partial class MainWindow : Window
Try ShowDialog()
public void RunPd()
{
MainWindow window = new MainWindow();
window.ShowDialog();
}
I'm not sure if this could be the issue or not (your question is a bit vague); however, right-click on the xaml file in the VisualStudio project and click on Properties. Then make sure that "Build Action" is set to "Page."
If this is already set, please make a comment as to that as well.
I had to add a reference to WindowsBase.dll, System.Xaml, PrentationationCore and PrentationFramework. Once I added these references the unit test worked.
Curious as to why I did not need to add these assemblies in Winforms...?
I'm trying to run a window, close it, and then run a second window, in a similar way that seems to work with Windows Forms.
namespace WpfApplication1
{
public partial class App : Application
{
[STAThread]
public static void Main()
{
Application app = new Application();
//windowMain.Show();
app.ShutdownMode = ShutdownMode.OnExplicitShutdown;
MainWindow windowMain = new MainWindow();
app.Run(windowMain);
Window1 window1 = new Window1();
window1.Show();
app.Run(window1);
}
}
}
I've set the Build Action in the App.xaml properties from ApplicationDefinition to Page, but the programme throws an exception when window1 is initialised. What am I doing wrong?
Edit: I've modified the xaml in App.xaml as suggested by first answer and edited main as suggested by the comment.
<Application x:Class="WpfApplication1.App"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
StartupUri="MainWindow.xaml"
ShutdownMode="OnExplicitShutdown">
<Application.Resources>
</Application.Resources>
</Application>
but I'm still getting the exception.
OK this is what I've divined so far. The Solution Builder looks for a Main() function. Why its not a WinMain() function I'm still not a hundred per cent clear on. If there is no Main(), you get an error. You can have more than one Main() as long as the Project properties: "Application" page/tab: property: "StartUp Object" is set to point to one of the main()s. This is done from an automatically created drop down list.
When a “WPF Application” project is created, Visual Studio(VS) create an xaml file called “App.xaml”. This is a class declaration where “App” is derived from the “Application” Class. VS also automatically generates hidden files for an xaml file. It creates a “name.g.i.cs” file, when the xaml file is created. It creates a “name.g.cs” file the first time the project is built after the creation of the xaml file. In this case it creates “App.g.cs” and “App.g.i.cs”. These files are hidden by default. To view them, press the “Show all files” button at the top of the Solution Explorer, they can be found in “\ obj\86\Debug” folder. When you delete an xaml file the “name.g.i.cs” and the “name.g.cs” files remain and are not deleted.
The “App.xaml” file’s “build Action” property is set to “Application Definition” when created by VS. When this property is set to “Application Definition” a Main() function is automatically created in “name.g.i.cs”:
[System.STAThreadAttribute()]
[System.Diagnostics.DebuggerNonUserCodeAttribute()]
public static void Main()
{
WpfApplication8.App app = new WpfApplication8.App();
app.InitializeComponent();
app.Run();
}
When this property is set to “Page”, the Main() function is automatically removed by VS. You can create new “Application” derived classes in code or in xaml. I haven’t found a neat way to do it in xaml. There doesn’t seem to be a template for an xaml “Application” derived class. I created a “.cs” code file and then renamed it to an .xaml file. For some reason VS won’t allow you to have more than one xaml “Application” declaration file set to “Application Build”, it doesn’t even give you the option of choosing one in the "Project: Properties: Application": “Startup Object” property.
As you can see in the hidden Main(), an instance of “App” is instantiated and run. If using your own Main() function: an instance of, the base “Application” class, or an “Application” derived class (whether declared in code or in xaml), can be declared and run. The “Application” class should only be instantiated once and should only be run once. If the “Application” derived class is declared in xaml then a simple application can be run by using the StartUpUri property in the xaml file: StartupUri="Windowname.xaml". Alternatively the top level UI programme logic can be placed in a Startup event handler. If “Startup="Application_Startup" is placed in the “App.xaml” file then an event handler can be written:
private void Application_Startup(object sender, StartupEventArgs e)
{
MainWindow windowMain = new MainWindow();
windowMain.ShowDialog();
Window1 window1 = new Window1();
window1.ShowDialog();
Shutdown();
}
You have to use ShowDialog() here, because it blocks until the window is closed. If you used Show() instead, it would show one window, then immediately show the other one and shutdown the application. In this case there's no need to call the Run() method yourself, that's done automatically.
The “Application” class instance can be run in code whether its declared in code or in xaml. You can then perform initialisation code prior to calling Run(). This would be placed in the Application_ Startup() event handler using the other way. However, if the “Application.Run” call is ever made in the programme, then no windows should be opened (using show() or ShowDialog()) in Main() or anywhere outside of the Application Class or within events and functions called from those events, called during “Application.Run()”.
The Application class has a ShutdownMode property (Application.ShutdownMode). The default for this is: “OnMainWindowClose”. This property can also be set to “OnLastWindowClose” or “OnExplicitShutdown” in code or in the xaml. You will need to reset this if you don't want the programme to close down when the MainWindow is closed.
I think for my purposes it is better not to use the Application class at all and just call the windows using Show() and “ShowDialog()”. This way I can use WPF pages but I could also call Windows Forms, or DirectX screens, as long as they are not open at the same time, or have no UI at all, if the programme is running remotely. Is there any reason for not doing it this way?
I think your application is shuting down when you close the first window. You need to set Application.ShutdownMode to OnExplicitShutdown.
If all you want to do is to show one window, when that closes, show another and when that closes, shutdown the whole application, you should keep the Build action as ApplicationDefinition, set ShutdownMode to OnExplicitShutdown (probably in App.xaml, but you can it in code-behind too) and put the following code in an event handler of the Startup event of your application:
private void Application_Startup(object sender, StartupEventArgs e)
{
MainWindow windowMain = new MainWindow();
windowMain.ShowDialog();
Window1 window1 = new Window1();
window1.ShowDialog();
Shutdown();
}
You have to use ShowDialog() here, because it blocks until the window is closed. If you used Show() instead, it would show one window, then immediatelly show the other one and shutdown the application.
There's no need to run the Run() method yourself, that's done automatically.