How to know more than one instance of a Form is running or not?

Veeramanikandan Dhanabalan

I have a Menustrip With 1 Menu Item Say Add Record.

When i click on Add Record Menu Item assume Form1 is opened.

My Problem is When i click on Add Record for 5 Times. it is opening Form1 5 Times.

How to open Form1 Once only even though i clicked for N Times.

Its the MDI form

Thanks in Advance..

6 Replies

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Sam Hobbs15y ago
I missed seeing the part that says it is a MDI form, but probably when I said "use an instance in the main form" in my first reply, then that is likely still valid. Do you know how to do that and to show/hide? Let me know if you have a specific question about doing that.
0
Vulpes15y ago
When you say that Form1 is an MDI form do you mean that it's an MDI parent or an MDI child?

If it's an MDI child, then it's not included in the Application.OpenForms collection and you'll have to do something like this instead:

bool openAlready = false;
foreach(Form f in this.MdiChildren)
{
if (f.Name == "Form1")
{
openAlready = true;
break;
}
}
if (!openAlready)
{
Form1 f1 = new Form1();
f1.Name = "Form1";
f1.MdiParent = this;
f1.Show();
}
0
Sam Hobbs15y ago
Make a static member someplace (in a class) for the form. Create the form once then use it many times. You could show it when you need to and hide it when you need to not show it. The "Program" class in the Program.cs file is a good place to put the form's instance, but you could put it in the main form. Actually, you don't need to use a static variable; you can just use an instance in the main form and then show/hide the form. Override the Form Close event and have it hide itself instead of closing it. The menu item could show the form.
0
Dorababu Meka15y ago
Try this

bool IsOpen = false;
foreach (Form f in Application.OpenForms)
{
if (f.Text == "Form2")
{
IsOpen = true;
f.Focus();
break;
}
}
if (IsOpen == false)
{
Form2 f2 = new Form2();
f2.MdiParent = this;
f2.Show();
}

Check this article http://www.c-sharpcorner.com/UploadFile/kirtan007/819/
0
Veeramanikandan Dhanabalan15y ago
@Vulpes

No.. not working....
0
Vulpes15y ago
Here's one way of doing it:

if (Application.OpenForms["Form1"] == null)
{
Form1 f1 = new Form1();
f1.Show();
}
0