I'd like to make an application, where I can click a button and a new window appears, where new options/buttons are available. I already managed to create a new window win2 after clicking the button:
private void button1_Click(object sender, RoutedEventArgs e)
{
var win2 = new Window();
win2.Show();
this.Close();
}
Now how do I edit the new window. Let's say I want to make new buttons (named: blue, green....), where the user can chose a color for the background.
When you want to create a new Window, you cannot use the Window class directly because it acts as a template.
To add a new Window to your project:
Right Click on your Project --> Add --> New Element --> Window. Name it as you please, I will use the default (Window1).
Now you can modify this window in the same way as you did for your original window. Add any UI elements you like and code them to your desire.
To show the new window:
Window1 secondWindow = new Window1();
secondWindow.Show();
Related
i have. view and viewmodel in same project.i was opennig a new page on button click on code behind like this
private void button_click(object sender, routedeventargs e)
{
var dashboardwindow = new dashboard();
this.navigationservice.navigate(new uri("../view/dashboard.xaml", urikind.relative));
}
now i have added user authentication on this buton click using MVVM.so how can i open this page from view model.?
if i used window instead of page then i can open new window simply like this
Dashboard dl = new Dashboard();
dl.Show();
thats workd.but i have taken pages because i need bowser type navigation in my application so.. please help .Thanks
I have three windows. FirstWindow, SecondWindow and ThirdWindow. FirstWindow has button and click on this button opens the SecondWindow. Analogously, SecondWindow has button and click on this button opens the ThirdWindow. Owner property of the SecondWindow is set as FirstWindow and Owner property of the ThirdWindow is set as SecondWindow. The scenario discribing problem:
Open all windows in a row. It will be looked like this:
Then minimize all windows by click on corresponding icon at top right corner of ThirdWindow.
If you will try to maximize all windows by clicking on FirstLevelWindow or ThirdLevelWinow in taskbar - all will be ok, three windows will be maximized. But if you will click on SecondWindow you will see this:
How can I fix it, or it is just WPF bug? I can give archived expample project if it helps.
UPDATE
Minimize window - click "_" icon, left icon in iconbar of the window. All windows are modal, i.e it opens with ShowDialog() method, not with Show() method. So if you minimize third window - all the windows will be minimized.
Here the code if you don't want download project by link:
FirstWindow XAML:
<Button Click="OpenChildWindow"
Content="ChildWindow"/>
FirstWindow .cs:
private void OpenChildWindow(Object sender, RoutedEventArgs e)
{
var window = new SecondLevelWindow();
window.Owner = this;
window.ShowDialog();
}
SecondWindow XAML:
<Button Click="OpenChildWindow"
Content="ChildWindow"/>
SecondWindow .cs:
private void OpenChildWindow(Object sender, RoutedEventArgs e)
{
var window = new ThirdLevelWindow();
window.Owner = this;
window.ShowDialog();
}
ThirdWindow is empty window without any content.
Here link to example project
I've just found, that bug is not reproduced if property ResizeMode of ThirdWindow is set to "NoResize". Mb it will be usefull information.
Well, I admit I have no idea what is going on. Did you try to add a fourth window? This become even stranger: the second window bring back the third, but the fourth is still not back.
Anyway, If I had to manage this problem, I would keep a reference of my childWindow in each parent Window. This way on any interesting event (like activate on the second window in your example) I could manage the state of my child as required (WindowState.Normal in your case).
It could be something like that: in xaml of secondWindow:
Activated="SecondLevelWindow_OnActivated"
And then in code behind:
private ThirdLevelWindow _window;
public SecondLevelWindow()
{
InitializeComponent();
}
private void OpenChildWindow(Object sender, RoutedEventArgs e)
{
_window = new ThirdLevelWindow ();
_window.Owner = this;
_window.ShowDialog();
}
public void SecondLevelWindow_OnActivated(object sender, EventArgs e)
{
if (_window != null)
{
_window.WindowState = WindowState.Normal;
}
}
This is a start, but you could also inspect your current state to define the state of your child.
Hope it helps.
hello I'm grabbing a windows form application in c # and have a question about a context menu I have my main form within the main form I have a picturebox and I created an event as the next mouse click
private void pictureBox1_MouseClick(object sender, MouseEventArgs e)
{
if (e.Button == MouseButtons.Right)
{
ContextMenu cm = new ContextMenu();
cm.MenuItems.Add("DISPONIBLE");
cm.MenuItems.Add("RESERVAR");
cm.MenuItems.Add("OCUPADA");
pictureBox1.ContextMenu = cm;
}
}
all going well so far shows me my contextual menu when I right click but when I choose the option to "AVAILABLE" another windows form show me someone who can help me please thanks
I think this is not the way. if you did so you need to call show for the context menu, you can set that menu at the initialization of the form or the control, you create the context menu one time it will be shown automatically, you don't have to create it each time you click the control, Add the following code to the constructor of the form just to check if it works
ContextMenu cm = new ContextMenu();
cm.MenuItems.Add("DISPONIBLE");
cm.MenuItems.Add("RESERVAR");
cm.MenuItems.Add("OCUPADA");
pictureBox1.ContextMenu = cm;
Like in web page using css we can show a div on mouse enter or hover, in the same way i want to show a panel on mouse enter event of a button, but i am unable to do this. I am trying like this.
private void btn2_MouseEnter(object sender, EventArgs e)
{
Button btn = (Button)sender;
btn.BackColor = System.Drawing.Color.MistyRose; //this is executed on mouse enter
Point locationOnForm = btn.FindForm().PointToClient(
btn.Parent.PointToScreen(btn.Location));
Panel pnl = new Panel();
Label lbl = new Label();
lbl.Text = "anything";
pnl.Controls.Add(lbl);
pnl.Location = new Point(locationOnForm.X, locationOnForm.Y);
pnl.Size = new Size(500, 500);
pnl.BackColor = Color.SkyBlue;
pnl.Visible = true;
pnl.Show();
}
I am not getting how to solve this. I want to know that
1) Is this the right approach or there is any other way of doing this?
2) If this is ok then what is the mistake i am doing here ?
Thanks.
Don't create the panel on mouse enter, rather have the panel created already then just show and hide it.
private void button1_MouseEnter(object sender, EventArgs e)
{
panel1.Show();
}
You have to add the panel to the Form controls
Form1.Controls.Add(pnl);
If you plan to have a panel hover over the button like a <div> in Web, you will have to call
BringToFront() to ensure that the panel does not appear behind the button or other controls on the form -
pnl.BringToFront();
Also like the previous answer, it may be better to have a panel placed on the form already and just set visible to true or false as well as the panel location otherwise you may end up adding multiple panels to the Form controls.
If you plan on just showing plain text in the panel, it may be easier to use Tooltip control -
MSDN - Tooltip Control
In my WPF application, my window gets deactivated when selecting a differnt menu from the menubar. I want the current window to be the active window. Is there a way to prevent this from happening?
private void HostWindow_Deactivated(object sender, EventArgs e)
{
//Code for making the window small and move to side using 3d animation
}
HostWindow is the window which holds the usercontrol form at runtime