Opening and Working on a new form.

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Let's say I have a form and a button in it. When I click the button I want to open the new form and start working on it. I Tried this.

C#

private void ProceedButton_Click(object sender, EventArgs e)
{
    TestCreator TestCrt = new TestCreator();
    this.Hide();
    TestCrt.ShowDialog();
    this.Close();

}

But when TestCrt is opened , it is not functioning well.
When I run TestCreator from the main its working properly but when opening by showdialog its not. For example when I click a button in TestCrt the program is shutting.
Is there another better way?
Why am I having the problem?

Posted 21-Dec-12 2:25am

missak boyajian

Updated 21-Dec-12 2:48am

v4

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[no name] 21-Dec-12 8:28am

How it could be possible..?? Place your TestCrt code here..
And ya one more thing, first you hiding the form and after showing other form, you close previous one, is there something to do such ???

missak boyajian 21-Dec-12 8:44am

I already tried that same error. I updated the question.

[no name] 21-Dec-12 8:46am

See my solution below...

Zoltán Zörgő 21-Dec-12 8:34am

.Show is not modal! The this.Close(); statement is executed just after the other form is shown. If this is the main form, than you will exit the application also. Try .ShowDialog instead.

Zoltán Zörgő 21-Dec-12 8:46am

If the problem is the same with ShowDialog, than you probably posted far to few from your code. Are you using threads for example?

missak boyajian 21-Dec-12 8:55am

When fisrt running the form from the main its working properly but opening from another form its not.

missak boyajian 21-Dec-12 8:58am

And no i'm not using any threads. I just want to know if there is a problem with the code or running a new form needs another way.

2 solutions

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Thomas Daniels · Answer 1 · 2012-12-21T02:36:00

After this.Close();, you exit the application because this is the main form. Try this:

C#

private void ProceedButton_Click(object sender, EventArgs e)
{
    TestCreator TestCrt = new TestCreator();
    this.Hide();
    TestCrt.ShowDialog();
    this.Close();
 
}

If you call ShowDialog, your program wait until TestCrt is closed. If you call Show, this.Close(); is called immediately, also if TestCrt isn't closed.

Krunal Rohit · Answer 2 · 2012-12-21T02:40:00

Solution 2

C#

private void ProceedButton_Click(object sender, EventArgs e)
{
    TestCreator TestCrt = new TestCreator();
    this.Hide();
    TestCrt.Show();
    this.Close(); // remove this, because it closes the main form, that's your application exits..

}

Posted 21-Dec-12 2:40am

Krunal Rohit