ShowDialog problem

[BbEsy]

Hello Im using show dialog.. this code

settings f = new settings();
 
f.ShowDialog();
this.Enabled = false;
if (f.ShowDialog() == DialogResult.OK)
{
load_reg_key();
this.Enabled = true;
}
f.Close();

and when i click on button..

private void button1_Click(object sender, EventArgs e)
{
this.DialogResult = DialogResult.OK;  
this.Close();
}

then form close and imediatly show again.. After second click on button . form closed and program resume... Where i making problem ?

[GaryTexmo]

f.ShowDialog();
this.Enabled = false;
if (f.ShowDialog() == DialogResult.OK)

That's your problem right there... you're calling ShowDialog() twice. When that method is called it actually shows the dialog, and returns the result. Remove the first ShowDialog() call and it should work just fine for you.

[tlhintoq]

That's because you are telling it to show the dialog twice: Once in line 3 and again in line 5.

Try something more like this.

settings f = new settings();// I presume this is a form you made
DialogResult MyResult = f.ShowDialog();// Code stops here until the dialog closes
// Now you have recovered the result from the dialog into a variable you can work with.
if (MyResult == DialogResult.OK)
{
     LoadRegKey();
     this.Enabled = true;
// Hmmm... Here "this" would be the main form. 
// if 'this' was previous disabled then how did you call the settings dialog?
}
 

[BbEsy]

many thanks for your fast reply.. It works :]]]