I have a WPF C# using FirstFloor MUI Framework app, that on start, checks for settings and will show the specific startup uri as per below;
if(somethings_true) {
Application curApp = Application.Current;
//ModernWindow
curApp.StartupUri = new Uri("MainWindow.xaml", UriKind.RelativeOrAbsolute);
}else{
Application curApp = Application.Current;
//ModernWindow
curApp.StartupUri = new Uri("OtherWindow.xaml", UriKind.RelativeOrAbsolute);
}
which works fine, however when the "OtherWindow.xaml" is active first it has a onclick event that does other checks, and on finishing opens the MainWindow.xaml. But in the Button_Click() which does the opening of MainWindow.xaml, I cant get the OtherWindow.xaml to close and ive tried inside OtherWindow.xaml..
this.Close();
&
var OtherWin = new OtherWindow();
OtherWin.Close();
&
var w = Application.Current.Windows[0];
w.Hide();
//Only hides the OtherWindows.xaml (Still runs hidden in background even after MainWindow.xaml is closed)
I use the below code to check if OtherWindow.xaml is still open inside MainWindow.xaml in which it states that it does;
foreach (var wndOtherWindow in Application.Current.Windows)
{
if (wndOtherWindow is OtherWindow)
{
//Its Open Still...
//How to close() "OtherWindow.xaml" from here?
}
}
Is there another way to close() the OtherWindow.xaml?
You should cast to Window or specific type(in your case OtherWindow), that way you can call Close() method. Try this:
foreach (var wndOtherWindow in Application.Current.Windows)
{
if (wndOtherWindow is OtherWindow)
{
(wndOtherWindow as Window).Close();
}
}
Hope helps.
Related
I programatically created a new window and frame, in order to navigate to a separate page (Authentication). Upon closing the window, I want to do some stuff, but the if statement is never returning false.
Window newWindow = new Window();
Page authentication = new Authentication();
Frame newFrame = new Frame();
newWindow.Title = "Authentication";
newWindow.Content = newFrame;
newFrame.NavigationService.Navigate(authentication);
newWindow.Show();
if (IsWindowOpen<Window>("Authentication") == false)
{
//DO THINGS HERE
}
This is my IsWindowOpen method:
public static bool IsWindowOpen<T>(string name = null) where T : Window
{
return string.IsNullOrEmpty(name)
? Application.Current.Windows.OfType<T>().Any()
: Application.Current.Windows.OfType<T>().Any(w => w.Title.Equals(name));
}
I manage to open the new window and run the separate page, but upon closing nothing happens. I've added a print line in the if statement to check, and it doesn't print.
Calling Show() won't block the main thread.
If you want to do something when the window closes you should handle Closing event and do things there.
If you want to block the current thread until the window is closed, you should use ShowDialog() instead.
I am new to WPF and have been hunting for an answer, surely this is not difficult?
I have created a main window with links to multiple windows, and I want them to run modelessly alongside one another, but I don't want to open multiple instances of the SAME window.
In simple terms, I can have Windows A, B, C open at once, but not Windows, A, A, B, C, C.
I need to implement a check for the window I'm trying to open (in this case, EditSettings).
If open - activate it
if not open, open it.
I have the following code in Main, which is not working.
public partial class MainWindow : Window
{
public MainWindow()
{
InitializeComponent();
}
private void MenuItem_Click(object sender, RoutedEventArgs e)
{
EditSettings winEditSettings = new EditSettings();
string isOpen = null;
if (isOpen == "true")
{
winEditSettings.Activate();
}
else
{
winEditSettings.Show();
isOpen = "true";
}
}
}
Now I know what's wrong with this logic - every time I press the button to open EditSettings, it's setting isOpen to null again. If I don't set a value to isOpen, the If condition breaks.
I could initialise the variable 'isOpen' as a public variable outside the MenuItem_Click method, but then I think I would need an isOpen variable for each window I create!! Surely there is a better way?
The other option I tried is:
private void MenuItem_Click(object sender, RoutedEventArgs e)
{
EditSettings winEditSettings = new EditSettings();
if (winEditSettings.IsLoaded)
{
winEditSettings.Activate();
}
else { winEditSettings.Show(); }
I can't figure out why this isn't working, I tried isVisible, isLoaded, isActive - nothing is stopping the window from opening more than once. Thank you for the help!
There are people who'll perhaps throw a fit at the idea, but whenever I've needed to do this, I made the child window objects part of the application. Then, in your MenuItem_Click(), test if winEditSettings is null, instead.
It's still a member variable for each window (like your provisional isOpen solution), but having the window objects available can have advantages later, if you need to bridge information between the windows. In my cases, we wanted to be able to close all the child windows together, which (most trivially) meant keeping track of those objects in a central location.
Alternatively, if you want the setup completely decoupled, you could take a singleton-like approach and put the logic into your child window classes. Specifically, you could call EditSettings.Activate and let the class keep track of whether a window needs to be created or the existing window merely Show()n.
If I were handed your code to rewrite, I'd move it something like this:
private static EditSettings winEditSettings = null;
public static void WakeUp()
{
if (winEditSettings == null)
{
winEditSettings = new EditSettings();
}
winEditSettings.Activate(); // This may need to be inside the block above
winEditSettings.Show();
}
Both of those are part of the class (static), rather than an instance. Your application object therefore calls EditSettings.WakeUp() inside the original MenuItem_Click(), and never actually sees the child window, itself.
If you change your mind about the decoupled architecture later, by the way, you can add a get accessor to your winEditSettings and keep everybody fairly happy.
if (_adCst == null)
{
_adCst = new AddCustomerPage();
_adCst.WindowStartupLocation = System.Windows.WindowStartupLocation.CenterScreen;
_adCst.WindowState = System.Windows.WindowState.Normal;
_adCst.ResizeMode = System.Windows.ResizeMode.NoResize;
_adCst.Activate(); // This may need to be inside the block above
_adCst.Show();
}
else
{
if (!_adCst.IsLoaded == true)
{
_adCst = new AddCustomerPage();
_adCst.WindowStartupLocation = System.Windows.WindowStartupLocation.CenterScreen;
_adCst.WindowState = System.Windows.WindowState.Normal;
_adCst.ResizeMode = System.Windows.ResizeMode.NoResize;
_adCst.Show();
}
_adCst.Activate();
}
My suggestion would be that you set some form of a counter. This will prevent more than one instance of the window being opened.
int windowOpen = 1;
private void button_Click(object sender, RoutedEventArgs e)
{
if (windowOpen == 1)
{
WindowA winA = new WindowA();
winA.Show();
windowOpen++; //increments windowOpen by 1, windowOpen will now = 2
}
else if (windowOpen > 1)
{
MessageBox.Show("Window is already open");
}
}
I hope this helps.
For anyone else with this question, I have found another solution - which works except that it doesn't manage to bring the open window to the front (Activate). It does, however, prevent opening the same window more than once.
foreach (Window n in Application.Current.Windows)
if (n.Name == "winEditSettings")
{ winEditSettings.Activate(); }
else
{ winEditSettings.Show(); }
Can anyone speculate on why the window is not brought to the front, with Activate()?
EDIT
For others with this question, placing the winEditSettings.Activate() outside of the foreach loop does everything I'm trying to achieve:
foreach (Window n in Application.Current.Windows)
if (n.Name == "winEditSettings")
{ }
else
{ winEditSettings.Show(); }
winEditSettings.Activate();
This will stop multiple instances of the same window from opening, and will bring the window to the front if the user attempts to reopen it.
I've got a small C# (.NET 4.0) Console Application that I'd like the user to be able to interact by showing a menu when they right-click the System Tray icon. I can add an icon to the Tray with no problems, but I just cannot get the menu to appear. I'm using the following code:
NotifyIcon trayIcon = new NotifyIcon();
trayIcon.Text = "TestApp";
trayIcon.Icon = new Icon(SystemIcons.Application, 40, 40);
ContextMenu trayMenu = new ContextMenu();
trayMenu.MenuItems.Add("Blah", item1_Click);
trayMenu.MenuItems.Add("Blah2", item1_Click);
trayMenu.MenuItems.Add("Blah3", item1_Click);
trayIcon.ContextMenu = trayMenu;
trayIcon.Visible = true;
... which puts the icon in the tray. However, right-clicking the icon does nothing. I've tried various permutations of MenuItems.Add, but nothing will make the menu appear. I'm sure I'm missing something simple - any ideas what?
Try adding this after you create the icon:
Application.Run()
Note that this method will not return, so you can't do anything after calling it. This means that you'll have to do all your other work in a separate thread.
What happens is that the OS sends your application a message telling it that the tray icon has been right-clicked, but the tray icon code never sees it (because these messages are processed by Application.Run) and so can't respond by opening the menu.
Concerning Application.Run(), this is an alternative to placing all the other code in another thread would be to create the NotifyIcon, menu, events, etc on a thread other than the main thread.
This should include Application.Run() as this allows the standard application message loop to work on the current thread. Then since the events were created on the same thread, the Application.Exit() can be used to close out the notification messaging but still allow the main thread to continue. Here's a small example for a console app...
class Program
{
public static ContextMenu menu;
public static MenuItem mnuExit;
public static NotifyIcon notificationIcon;
static void Main(string[] args)
{
Thread notifyThread = new Thread(
delegate()
{
menu = new ContextMenu();
mnuExit = new MenuItem("Exit");
menu.MenuItems.Add(0, mnuExit);
notificationIcon = new NotifyIcon()
{
Icon = Properties.Resources.Services,
ContextMenu = menu,
Text = "Main"
};
mnuExit.Click += new EventHandler(mnuExit_Click);
notificationIcon.Visible = true;
Application.Run();
}
);
notifyThread.Start();
Console.ReadLine();
}
static void mnuExit_Click(object sender, EventArgs e)
{
notificationIcon.Dispose();
Application.Exit();
}
}
Here is the solution:
You have to use Application.Run() because events of gui in console mode not working.
But you can use this solution:
var task = System.Threading.Tasks.Task.Factory.StartNew(() => ShowTrayIcon());
void ShowTrayIcon()
{
some code with tray icon ...
}
This will start your setup of try icon in new thread ...
Did you add the event-handler for tray Icon mouse click?
trayIcon .MouseDown += new MouseEventHandler(trayIcon_MouseDown);
create context menu and do as following inside the trayIcon_MouseDown function
private void trayIcon_MouseDown (object sender,MouseEventArgs e)
{
//add you menu items to context menu
contextMenu.Items.Add(item);
contextMenu.IsOpen = true;
}
Try this. Think this will help you.
I am stuck when trying to switch windows using the latest version of webdriver in C#.
I have a base window, when i click a button, it opens a new window.
The target code for this button is as below.
window.open(uri, "DisplayPage", " width=1200, scrollbars=yes , resizable = yes , toolbar = no , menubar = no");
I am using the below mentioned code to target the new window
string BaseWindow = _driver.CurrentWindowHandle;
ReadOnlyCollection<string> handles = _driver.WindowHandles;
foreach (string handle in handles)
{
if (handle != BaseWindow)
{
_driver.SwitchTo().Window(handle).Title.Equals("DisplayPage");
}
}
}
As you can see from above, I am switching to the window using the Target Title from the base window. This does not seem to work.
I then noticed that the Title of the opened window was different, it was "Display - Transaction Page"
I then modified the code to this
string BaseWindow = _driver.CurrentWindowHandle;
ReadOnlyCollection<string> handles = _driver.WindowHandles;
foreach (string handle in handles)
{
if (handle != BaseWindow)
{
_driver.SwitchTo().Window(handle).Title.Equals("Display - Transaction Page");
}
}
}
Still no luck.
Interestingly, I do not get any errors saying "Window not found".
The problem is that When i try to click on elements on the newly opened page, i get a NoSuchElementException which means that the newly opened window has not been targeted.
Any ideas?
Regards,
Hasan
You should break the loop once window changed to your window, otherwise it will always switch to last opened window:
foreach (string handle in handles) {
if (handle != BaseWindow) {
if(_driver.SwitchTo().Window(handle).Title.Equals("Display - Transaction Page"))
break;
}
}
You can try with Contains instead of equal, it will simplify the window search:
_driver.SwitchTo().Window(handle).Title.Contains("Display");
Although you cracked the answer yourself, there is another way to handle switching between the windows in C#.
// initiate webdriver
IWebDriver driver = new FirefoxDriver();
//perform some action to open a new window. Like clicking a link.
driver.FindElement(By.Id("btnId")).Click();
//switch to new window.
driver.SwitchTo().Window(driver.WindowHandles.Last());
//if you want to switch back to your first window
driver.SwitchTo().Window(driver.WindowHandles.First());
I have a WPF window that I am launching from inside of a winform app. I only want to allow once instance of that WPF window to be open at a time, and not warn that user if they try to open it again.
I am having a problem however trying to search for that WPF window being open because the window is being launched from a winform. What I normaly do is when searching for a winform, I search for any instances of that winform existing in the Application.Current.OpenForms, and when in WPF I search for Application.Current.Windows
The problem I have is that System.Windows.Application.Current is null when launched from inside of a winform, so I can't search for the WPF window that way. Is there any better way of searching for an existing instance of an open window?
My Code:
if (System.Windows.Application.Current != null)
{
foreach (System.Windows.Window win in System.Windows.Application.Current.Windows)
{
if (win is frmCaseWpf)
{
MessageBox.Show("You may have only one active case open at a time.", "Open Case",
MessageBoxButtons.OK,
MessageBoxIcon.Stop);
win.WindowState = System.Windows.WindowState.Normal;
win.Focus();
win.Activate();
return;
}
}
}
Instead of searching the static application objects, you could instead just track this within your window, with a single static variable. Just keep a variable in the window:
private static frmCaseWpf openWindow = null; // Assuming your class name is frmCaseWpf
When you create a window, either in the initialize routines, or OnLoaded, depending on how you want it to work..:
partial class frmCaseWpf {
public frmCaseWpf {
this.OnLoaded += frmCaseWpf_OnLoaded;
}
private void frmCaseWpf_OnLoaded(object sender, RoutedEventArgs e)
{
if (this.openWindow != null)
{
// Show message box, active this.openWindow, close this
}
this.openWindow = this;
}
}
If you want this window to be reusable, make sure to set this.openWindow = null; when you close the window, as well.
Here's something that's working for me.
private About aboutWin;
private void AboutOpenClicked(object sender, RoutedEventArgs e)
{
if(aboutWin == null)
{
aboutWin = new About();
aboutWin.Closed += (a, b) => aboutWin = null;
aboutWin.Show();
}
else
{
aboutWin.Show();
}
}
It would be better make the frmCaseWpf class a singleton. That way you can't create another instance
Rather than try to search for a Window instance, many people use a session- (or system-) wide "Mutex" or a Mutual Exclusion lock. I was going to rewrite one for you, but I found a good codeproject article demonstrating the technique. It's not complex and very simple.
http://www.codeproject.com/KB/cs/SingleInstanceAppMutex.aspx?msg=2908697
Sneak peek:
[STAThread]
static void Main()
{
bool onlyInstance = false;
Mutex mutex = new Mutex(true, "UniqueApplicationName", out onlyInstance);
if (!onlyInstance) {
return;
}
Application.Run(new MainForm);
GC.KeepAlive(mutex);
}
Hope this helps.
(edit: of course you'll have to modify this slightly for your particular use-case, but it demos the general idea)
I am not really a 'proper' programmer, however I have achieved this in a WPF application (not from a winforms one) by using the following:
Dim wdwDetails As New detailsNew()
Private Sub openNewDetails(ByVal recordID As String)
wdwDetails.Owner = Me
wdwDetails.recordID = recordID
wdwDetails.WindowStartupLocation = Windows.WindowStartupLocation.CenterOwner
wdwDetails.Show()
End Sub
Essentially because I am creating the window object outside of the sub that opens it, there will only be a single window. Any new call to the window open sub will use the same object. But I guess that is what Thomas is referring to also.
Like I said, not sure if this will help you or not though.
You can use XXXwindown.isLoad to check if window is loaded before you create a new window:
if ( !ChildWindow.IsLoaded)
{
childWindow= new ChildWindow();
childWindow.Show();
}