Check opened FORM!

Question

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

Sujeet Suman · 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 { 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 Suman · Answer

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 Luong · Answer

Wow this really works thanks Sujeet! :) // Why didn't I think about this? :(

Sujeet Suman · Answer

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 .