MDI Child form problem....

Question

hi forum i have created mdi parent & child form. when i click menu option to open child form it opens. now i kept it open and again i click menu option & i got new instance of form. ( another same form i got ) what to do to stop this problem ?? i m really confuse. using flag it is done. but other simple way ????? please guide me thank u bhavin gajjar

soman · Answer

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 frmChildif found show that one else create new instance of child... If u find difficulty in understanding the above.. just mess me again. Cheers soman

onlybhavin · Answer

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

shimtanny · Answer

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