How to check if a windows form already exists, open it if it does, creates a new one if it doesn't?

Question

I basically want something that does this:

if (form.isOpen() == true)
    form.ShowDialog();
else
    form f = new form();

I have a form that displays a list of items added to an order, but when I go to another page and navigate back to make an order form it calls form f = new form();, which I believe resets the form. Any suggestions of how this can be overcome?

Basically, you want to know if there any instance of `form` opened? — The_Black_Smurf, Jan 14 '16 at 14:45
If form is still in scope then you shouldn't have any need to check if already exists or not, you should just be able to always show it — Sayse, Jan 14 '16 at 14:53

The_Black_Smurf · Answer 1 · 2020-09-09T14:33:05.367

2

You can check which forms are open using Application.OpenForms property as follows:

if (Application.OpenForms.OfType<MyForm>().Any())
{
    Application.OpenForms.OfType<MyForm>().First().BringToFront();
}
else
{
    form f = new MyForm();
    f.show();
}

edited Sep 09 '20 at 14:33

answered Jan 14 '16 at 14:51

The_Black_Smurf

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This works well, and you don't need to use `showDialogue` and make the form modal. Of course, you can only have *one* instance of that form. – bgmCoder Feb 27 '20 at 02:12

score 1 · Answer 2 · answered Jan 14 '16 at 14:54

Use the static property Application.OpenForms

YourFormType f = Application.OpenForms.OfType<YourFormType>().FirstOrDefault();
if (f != null)
    f.ShowDialog();
else
{
    using(f = new YourFormType())
    {
        f.ShowDialog();
    }
}

However I don't know if this is going to solve your problem. You might be better off separating your form from it's backing data and using binding to connect the two. When you navigate back you just pass the backing data to the new copy of the form. Look in to concepts like MVVM for more information.

score 0 · Answer 3 · answered Jan 14 '16 at 15:01

If you call ShowDialog() you can't go to another form within same application (until you have multiple UI threads). If you meant Show() instead of ShowDialog i'd say you just have a

     Form _form;

in your class and add some guarding code, like that (generic idea, though):

     void ShowOrCreate()
     {
        if(_form==null)
        {
            _form = new MyForm();
            _form.Closed += OnMyFormClosed();
            _form.Show();
        }
        else
        {
            _form.BringToFront();
        }
      }

      void OnMyFormClosed(...)
      {
         _form = null;
      }

sorry for missing parts and method signatures, I am sure the can be easily checked in MSDN.

score 0 · Answer 4 · answered Jan 14 '16 at 15:20

A kind of singleton assuming you want single instance of given form per application:

public MyForm: Form
{
    private static MyForm _instance;

    public static MyForm Instance
    {
        get
        {
            if(_instance == null)
                _instance = new MyForm();
            return instance;
        }
    }
}

Now you can simply use MyForm.Instance. Note, with this approach you don't want to call MyForm.Instance.Close() ever (or you have to re-create instance). Use either Show/Hide (modeless) or ShowDialog/Hide (modal).

How to check if a windows form already exists, open it if it does, creates a new one if it doesn't?

4 Answers4

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