how to Access to Other form when it's still in "Show()"?
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hi my project have 2 form and i call show() method to showing Form2 i want to pass a parameter (or property) to form2 when it showing or find that form2 is close or show in form1:confused: Does anyone know how i can do this action? Thanks;
I think I understand your question, try using
ShowDialog()
instead ofShow()
. If I don't understand your question, reply with more details.I get all the news I need from the weather report - Paul Simon (from "The Only Living Boy in New York")
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hi my project have 2 form and i call show() method to showing Form2 i want to pass a parameter (or property) to form2 when it showing or find that form2 is close or show in form1:confused: Does anyone know how i can do this action? Thanks;
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hi my project have 2 form and i call show() method to showing Form2 i want to pass a parameter (or property) to form2 when it showing or find that form2 is close or show in form1:confused: Does anyone know how i can do this action? Thanks;
hi my project have 2 forms ...in form 1 i have a button and button1_Click is private void button1_Click(object sender, EventArgs e) { this.DesktopLocation = new Point(DesktopLocation.X - 200, DesktopLocation.Y); new Form2(this.DesktopLocation.X + this.Width, this.DesktopLocation.Y).Show(); } now form2 is Show() and i want to know IN FORM1 that form2 is close or still running .... and when form2 is close event or property fire in form1 to Form1.DesktopLocation return to orginl location... here is source code of my project[] Does anyone know how i can do this action? Thanks;
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hi my project have 2 forms ...in form 1 i have a button and button1_Click is private void button1_Click(object sender, EventArgs e) { this.DesktopLocation = new Point(DesktopLocation.X - 200, DesktopLocation.Y); new Form2(this.DesktopLocation.X + this.Width, this.DesktopLocation.Y).Show(); } now form2 is Show() and i want to know IN FORM1 that form2 is close or still running .... and when form2 is close event or property fire in form1 to Form1.DesktopLocation return to orginl location... here is source code of my project[] Does anyone know how i can do this action? Thanks;