如何知道 WPF 窗口是否打开

Question

In a WPF window, how do I know if it is opened?

My goal to open only 1 instance of the window at the time.

So, my pseudo code in the parent window is:

if (this.m_myWindow != null)
{
    if (this.m_myWindow.ISOPENED) return;
}

this.m_myWindow = new MyWindow();
this.m_myWindow.Show();

EDIT:

I found a solution that solves my initial problem. window.ShowDialog();

It blocks the user from opening any other window, just like a modal popup. Using this command, it is not necessary to check if the window is already open.

Answer 1

In WPF there is a collection of the open Windows in the Application class, you could make a helper method to check if the window is open.

Here is an example that will check if any Window of a certain Type or if a Window with a certain name is open, or both.

public static bool IsWindowOpen<T>(string name = "") where T : Window
{
    return string.IsNullOrEmpty(name)
       ? Application.Current.Windows.OfType<T>().Any()
       : Application.Current.Windows.OfType<T>().Any(w => w.Name.Equals(name));
}

Usage:

if (Helpers.IsWindowOpen<Window>("MyWindowName"))
{
   // MyWindowName is open
}

if (Helpers.IsWindowOpen<MyCustomWindowType>())
{
    // There is a MyCustomWindowType window open
}

if (Helpers.IsWindowOpen<MyCustomWindowType>("CustomWindowName"))
{
    // There is a MyCustomWindowType window named CustomWindowName open
}

Answer 2

You can check if m_myWindow==null and only then create and show window. When the window closes set the variable back to null.

    if (this.m_myWindow == null)
    {
           this.m_myWindow = new MyWindow();
           this.m_myWindow.Closed += (sender, args) => this.m_myWindow = null;           
           this.m_myWindow.Show();
    }

Answer 3

Here is another way to achieve this using LINQ.

using System.Linq;

...

public static bool IsOpen(this Window window)
{
    return Application.Current.Windows.Cast<Window>().Any(x => x == window);
}

Usage:

bool isOpen = myWindow.IsOpen();

Answer 4

If you need to activate the window if it is found, you can use the code below:

//activate a window found
//T = Window

 Window wnd = Application.Current.Windows.OfType<T>().Where(w => w.Name.Equals(nome)).FirstOrDefault();
 wnd.Activate();

Answer 5

Put a static bool in your class, named _open or something like that. In the constructor then do this:

if (_open)
{
    throw new InvalidOperationException("Window already open");
}
_open = true;

and in the Closed event:

_open = false;

Answer 6

Is that you search ?

if (this.m_myWindow != null)
{
    if (this.m_myWindow.IsActive) return;
}

this.m_myWindow = new MyWindow();
this.m_myWindow.Show();

If you want a singleton, you should read this : How can we create a Singleton Instance for a Window?

Answer 7

    void  pencereac<T> (int Ops) where T : Window , new()
    {
        if (!Application.Current.Windows.OfType<T>().Any()) // Check is Not Open, Open it.
        {
           var wind = new T();
            switch (Ops)
            {
                case 1:
                    wind.ShowDialog();
                    break;
                case 0:
                    wind.Show();
                    break;
            }
        }
        else Application.Current.Windows.OfType<T>().First().Activate(); // Is Open Activate it.
    }

Kullanımı:

void Button_1_Click(object sender, RoutedEventArgs e)
{
  pencereac<YourWindow>(1);
}

Answer 8

This works if you want to Check if the Second Form is already Open and avoidt opening it again on buttong Click.

int formcheck = 0;
private void button_click()
{
   Form2Name myForm2 = new Form2Name();
   if(formcheck == 0)
   {
      myForm2.Show(); //Open Form2 only if its not active and formcheck == 0
      // Do Somethin

      formcheck = 1; //Set it to 1 indicating that Form2 have been opened
   {   
{

Answer 9

public bool IsWindowOpen<T>(string name = "") where T : Window
{
    return Application.Current.Windows.OfType<T>().Any(w => w.GetType().Name.Equals(name));               
}