MDI Child form problem....

Hi just go through code, It will make sure that, u r having only one instance.. frmChild childForm = null; foreach(Form form in this.MdiChildren) { if(form is frmChild ) { childForm = (frmChild)form; break; } } if( childForm != null) { childForm.Show(); childForm.Focus(); } else { childForm = new frmChild(); childForm .MdiParent=this; childForm .Show(); childForm .Focus(); } Here frmChild is the child form, in foreach loop just checks for the instance of that frmChild,if found show that one, else create new instance of child... If u find difficulty in understanding the above.. just mess me again. Cheers soman

hi, thank u simon. but problem is not u said to me. see i m telling u my steps. in mdi form i have menu file. in that sub menu new. now when i click on new option child form opens. now keep in mind i m not closeing child form. i again click new option & the result is. i got another instance of child form. so on my screen 2 child forms appers. i want if 1 instance is open then another can't be open. please guide me. thank u, bhavin

Hi Create child form once (in constructor or pageload). Then in menu only make show, and in child form do not close, but hide it. Simon