How to check if a windows form is already open, an

Question

I have a form "fm" that is a simple info window that opens every 10 mins (fm.Show();).

How I can make that every 10 mins it will check if the form "fm" is open and if it is open it closes it and open it again!

Now the form fm is always created with form fm = new form();
so when I try to check if the form is open it will always be false and open a new window even if there is one form before!

I need to have a tool to give it a unique identity and then check if this form with unique identity is opened or not!

I do not want to just update the data on the form (fm), because I have a complicated info with buttons.

The form name is "UpdateWindow"

Solved using the following:

Form fc = Application.OpenForms["UpdateWindow"]; 

if (fc != null) 
   fc.Close(); 

fc.Show();

Answer 1

if( ((Form1)Application.OpenForms["Form1"]).Visible == true)
    //form is visible
else
    //form is invisible

where Form1 is the name of your form.

Answer 2

Funny, I had to add to this thread.

1) Add a global var on form.show() and clear out the var on form.close()

2) On the parent form add a timer. Keep the child form open and update your data every 10 min.

3) put timer on the child form to go update data on itself.

Answer 3

Try this, it will work :

//inside main class
Form1 Fm1 = new Form1();<br>

//in button click
if (Fm1.IsDisposed)
{
    Fm1 = new Form();
}
Fm1.Show();
Fm1.BringToFront();
Fm1.Activate();

Answer 4

In my app I had a mainmenu form that had buttons to navigate to an assortment of other forms (aka sub-forms). I wanted only one instance of each sub-form to be running at a time. Plus I wanted to ensure if a user attempted to launch a sub-form already in existence, that the sub-form would be forced to show "front&center" if minimized or behind other app windows. Using the currently most upvoted answers, I refactored their answers into this:

private void btnOpenSubForm_Click(object sender, EventArgs e)
    {

        Form fsf = Application.OpenForms["formSubForm"];

        if (fsf != null)
        {
            fsf.WindowState = FormWindowState.Normal;
            fsf.Show();
            fsf.TopMost = true;
        }
        else
        {
            Form formSubForm = new FormSubForm();
            formSubForm.Show();
            formSubForm.TopMost = true;
        }
    }

Answer 5

I know I am late... But for those who are curious... This is another way

if (Application.OpenForms.OfType<UpdateWindow>().Count() == 1)
    Application.OpenForms.OfType<UpdateWindow>().First().Close();

UpdateWindow frm = new UpdateWindow()
frm.Show();

Answer 6

In addition, may be this will help

class Helper
    {
        public void disableMultiWindow(Form MdiParent, string formName)
        {
            FormCollection fc = Application.OpenForms;
            try
            {
                foreach (Form form in Application.OpenForms)
                {
                    if (form.Name == formName)
                    {
                        form.BringToFront();
                        return;
                    }
                }

                Assembly thisAssembly = Assembly.GetExecutingAssembly();
                Type typeToCreate = thisAssembly.GetTypes().Where(t => t.Name == formName).First();
                Form myProgram = (Form)Activator.CreateInstance(typeToCreate);
                myProgram.MdiParent = MdiParent;
                myProgram.Show();
            }
            catch (Exception ex) { MessageBox.Show(ex.Message); }
        }
    }