How to bring a childform over a parentform?

Question

0.00/5 (No votes)

See more:

I am having two Forms 'ParentForm' and 'ChildForm'
On 'ParentForm'=) buttonClick 'ChildForm' needs to display like a Popup.
So i wrote like

C#

private void button_Click(object sender, EventArgs e)
       {
           ChildForm popup = new ChildForm();
           DialogResult dialogresult = popup.ShowDialog();
           if (dialogresult == DialogResult.OK)
           {
               popup.Dispose();
           }
       }

But this won't coming in front of ParentForm.
How can i make it display like wise?

What I have tried:

C#

popup.TopMost = true;
popup.Focus();
popup.BringToFront();
popup.Activate();

Posted 8-Jan-21 5:30am

Member 14978771

Updated 8-Jan-21 15:34pm

v3

Add a Solution

3 solutions

Add a Solution

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Answer 1 · 2021-01-08T05:43:00

ShowDialog makes it a modal form - so any code you write after this line:

C#

DialogResult dialogresult = popup.ShowDialog();

Will not be executed until the user closes the Child form. So if we fix your code so it will compile (you need a space between the type "ChildForm" and the variable name "popup") you should find it appears somewhere and you won't be able to put the focus back to the parent.

And since it's a modal form, it will be above the parent provided you set a location for it to appear in. So start like this:

C#

ChildForm popup = new ChildForm();
popup.StartPosition = FormStartPosition.CenterParent;
if (popup.ShowDialog(this) == DialogResult.OK)
   {
   ...

Luc Pattyn · Answer 2 · 2021-01-08T15:34:00

you will get improved behavior by using

popup.ShowDialog(this);

which gives ownership of the child form to the parent form, creating a stronger relationship between both forms.

BTW: whether to call Dispose() or not should not depend on the return value!

And then, a better way to call Dispose() automatically is by using a using statement, as in:

C#

private void button_Click(object sender, EventArgs e) {
   using (ChildForm popup = new ChildForm()) {
       DialogResult dialogresult = popup.ShowDialog(this);
       if (dialogresult == DialogResult.OK) {
           ... // do something with popup result properties if and as required
       }
   }
}

This will always dispose of the ChildForm when you leave the using block, even when an Exception gets thrown.

:)

BillWoodruff · Answer 3 · 2021-01-08T13:20:00

private void button_Click(object sender, EventArgs e)
{
   ChildForm popup = new ChildForm();
   DialogResult dialogresult = popup.ShowDialog();
   if (dialogresult == DialogResult.OK)
   {
       popup.Dispose();
   }
}

You create a new ChildForm instance inside a Button Click EventHandler. Outside the scope of the Button Click EventHandler that instance doesn't exist, and you have no access to what the DialogResult was.

Since this is ASP.NET, you need to clarify if you are using web Forms ... "ASP.Net WebForms is designed to only have one form per page" ...or new Windows. Define clearly what you mean by "Parent Child" here.

Do you need to show the "other" Form modally ?

How to bring a childform over a parentform?

3 solutions

Solution 1

Solution 3

Solution 2

Add your solution here

Preview 0

How to bring a childform over a parentform?

3 solutions

Solution 1

Solution 3

Solution 2

Add your solution here

Preview 0

Existing Members

...or Join us