Check opened FORM!

Quang Dinh Luong

I have a new problem! :(

My app have 2 form, I add a button to Form1 so as to open Form2. And now the problem appears, the more I click that button, the more Form2 are opened.

Ex: Click 10 times , 10 Form2 on the screen!

I'v tried this:

Form2 formTwo = new Form2();

if (formTwo.Visible == false) {formTwo.OpenDIalog();}

else {formTwo.BringToFront();}

But it doesn't work! :(

I've also tried to set Enabled of that button properties to false after Form2 open. But it will make that button's text disapper. So please ignore this!

Is there a way to check if the form2 is opened or not to do different action

Expect model:

if(Form2 is not opend) {//Do not open again}

else{open form2} //I know this, just need the red one

4 Replies

Know the answer? Post it — somebody with the same question will find it here.

Sujeet Suman11y ago
Accepted answer
Quang if anyway you want to use "show()" only then you can follow below process:

Step-(01.) On Form1

public static int count = 0;
private void button1_Click(object sender, EventArgs e)
{
   Form2 frm02 = new Form2();
   if (count < 1)
   {
      frm02.Show();
      count++;
    }
}

Step-(02.) On Form2
private void Form2_FormClosing(object sender, FormClosingEventArgs e)
{
   Form1.count = 0;
}

If this is helpful for you then please accept my Answer
Sujeet Suman11y ago
Quang we are human. It may be at some point I am thing in right direction where as at some point you. And this credit also goes to you because you made me think. Whatever you are really very practical, innovative & enthusiastic guy.
Quang Dinh Luong11y ago
Wow, this really works, thanks Sujeet! :)
// Why didn't I think about this? :(
Sujeet Suman11y ago
Quang on Form1 button click event open Form2 as "ShowDialog()". Then automatic Form2 will be on front. This is standard way to open a dialog/second form on a button click event of first form.

So write below code on Form1 button click event:

private void button1_Click(object sender, EventArgs e)
{
   Form2 frm2 = new Form2();
frm2.ShowDialog();
}

If this is helpful for you then please accept my Answer.