I have a windows form that pops up a dialog box if certian conditions are met when the form loads. The problem is the window does not stay on top and I can still click thing on the parent. However, there is a button on the form that when pressed opens the same window, when I do this it works as expected (like a dialog window).
Is there an issue with showing a dialog when a form is first loading?
Are you calling ShowDialog from the Form class? Because it will only set the parent window if called from another Form. Alternatively you can use the overload that has the IWin32Window parameter to specifically set the owner.
can you explain the issue further as this is my code which do not show the form it self until the dialog has been closed either you set the parent or not
private void Form1_Load(object sender, EventArgs e)
{
//your functionality goes here
AboutBox1 box = new AboutBox1();
box.ShowDialog();
}
}
on the other side you can also check with TopMost property
The ShowDialog method needs to be called from the form that you want to be it's parent/owner in order for it to be modal to that form. Alternatively I believe you can set the owner of a dialog directly but I have never needed to do that.
DaBomb,
To do what you want, you will have to call your modal dialog from the constructor of your main form, NOT from the Form_Load event.
Something like this:
public Form1()
{
InitializeComponent();
this.Show();
Form2 popupForm = new Form2();
popupForm.ShowDialog();
}
Related
I am showing some database records in a Dialog. When I click any particular record that record fills to Active Form. But I want to focus to a button when my dialog close. So I have written the following code on form closing evennt.
private void frmDG_RecordSelection_FormClosing(object sender, FormClosingEventArgs e)
{
RecordSelectionStatus.Text = "False";
Form TargetForm = Home.ActiveMdiChild;
Button SelectRefConsultant = (Button)TargetForm.Controls.Find("btnSelectRefConsultant_NI", true).SingleOrDefault();
SelectRefConsultant.Focus();
TargetForm.ActiveControl = SelectRefConsultant;
}
But it's not working. Focus still remain to it's previous place. What am I missing ?
I am assuming that the dialog is modal... Instead of doing this in FormClosing do it, after calling ShowDialog(). If not, try using the FormClosed event instead.
I think your code is not working because, while the Form is closing, it still has modal focus.
If frmDG_RecordSelection is also MDIChild, then Home.ActiveMDIChild is this form. That is being closed.
But if frmDG is just a Dialog, the problem is different.
This dialog is Closing. But it's still visible. You cannot set focus to control that is not visible.
So you will have to set focus after this frmDG is completely closed, and invisible... To be more specific, when your MDI form is visible.
It's far more easier to do this from your MDI Form. I don't know how you have programmmed it, but I suppose it's something like that:
//this is in your MDI form
void OnRecordSelected(...)
{
frmDG yourDialog = new frmDG();
frmDG.ShowModal();
frmDG.Dispose();
}
In this case, you will have to set focus after frmDG is disposed.
I would like to know the difference between form.show() and form.activate().
I have multiple forms that already opened and i would like to active my form that is behind another form which is the best way to call my desired form
form.show() or form.activate()?
The method Show() displays the form to the user.
The method Activate() brings the form to the front (it gives the form focus).
For example:
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
Form2 form2 = new Form2();
form2.Show();
this.Activate();
}
}
The above code will show form2 by calling form2.Show(); but form1 will be in front of form2 (in focus) because of the this.Activate(); call.
See MSDN documentation:
Show()
Activate()
From the documentation
Form.Activate Method ()
Activating a form brings it to the front if this is the active
application, or it flashes the window caption if this is not the
active application. The form must be visible for this method to have
any effect. To determine the active form in an application, use the
ActiveForm property or the ActiveMdiChild property if your forms are
in a Multiple-document interface (MDI) application.
Form.Show Method
Showing the control is equivalent to setting the Visible property to
true. After the Show method is called, the Visible property returns a
value of true until the Hide method is called.
Answer
I have multiple forms that already opened and i would like to active
my form that is behind another form which is the best way to call my
desired form form.show() or form.activate()?
If your form is already open Activate it probably the one you want
Tip : If you ever wonder what a .net method does, just go and type it into google, usually the help is the first thing that shows up, plus a myriad of other questions and answers
form.activate() activates the form, which means if you have input elements (such as text boxes), it will focus to that particular form regardless of any other form out there. Eg. If you have shown form 1,2 and 3. And if you activate form 2, the form 2 will get focused to the user.
If you use form.show() it will just display/show the form to the user. Thus the activate() is gets the highest priority in terms of user engagement.
As per msdn Form.Activate() &
Form.Show()
Activate() :-
Activating a form brings it to the front if this is the active application, or it flashes the window caption if this is not the active application. The form must be visible for this method to have any effect.
Show() :-
You can use this method to display a non-modal form. When you use this method, the Owner property of the form is set to owner. The non-modal form can use the Owner property to get information about the owning form. Calling this method is identical to setting the Owner property of the non-modal and then calling the Show() method.
Showing the form is equivalent to setting the Visible property to true. After the Show method is called, the Visible property returns a value of true until the Hide method is called.
you can visit above links for more information
Now if you make your purpose more clear we may help in you in "Specific Way"
After reading the edit "Activate" is best , and also my answer is exactly identical to #TheGeneral
Form.Show() creates a new example of a form, while Form.Activate() brings an already existing form in the foreground.
I have problem with Windows Forms.
When I show form2 from form1 like this (names of the variables changed):
form2.ShowDialog(form1);
then I have this exception:
System.InvalidOperationException: Form that is already visible cannot be displayed as a modal dialog
box. Set the form's visible property to false before calling showDialog.
Telling more - during debugging I see that after calling ShowDialog method debugger goes again to the same ShowDialog method - and this is why I have this exception. I suppose that form1 is loaded again and it is some kind of ShowDialog method bug? I have form2 Visible property set to false. I tried to use Hide method too - not working.
Edit:
More info - I use ShowDialog method after showing combo box selector from form1. When I click the last property in combo box by mouse - ShowDialog is working. If I go down by keyboard and click 'Enter' mentioned exception appears.
Try something like this. When you close Form2, control will go back to Form1 (or whoever called Form2):
private void button1_Click(object sender, EventArgs e)
{
Form2 form2 = new Form2();
form2.ShowDialog();
}
Or just
private void button1_Click(object sender, EventArgs e)
{
new form2().ShowDialog();
}
I have a windows form app. The main form has a textbox, and a button to launch another form. When the other form is launched, both forms are on screen (but the launched form is modal). The child form has a textbox and button, and when the button is pressed, I want the textbox on the main form (the parent) to be updated with the value in the textbox on the child form.
How is this functionality achieved?
Ideally you want to keep both forms from being dependent on each other, this could be achieved with interfaces:
public interface IMainView
{
public void UpdateValue(string val);
}
public interface IChildView
{
public void Show(IMainView parent);
}
have your main form implement IMainView and the child implement IChildView, the parent calls child.show(this) and the child calls parent.UpdateValue(blah);
Hope this helps.
If the child form is closed when the button is clicked, you could put a public property which wraps the value of the textbox on the child form. Then the main form can read this property after calling ShowDialog.
If you want this to happen without closing the child form, you can create a function on the main form to change the textbox. Then the child form would call that function.
The best ways to achive this situation are clockWize's and Hans Passants's advices.
But what about that?
Write a property for your textbox at parent form, like this.
public string TextBoxText
{
get { return txtTextBox.Text;}
set { txtTextBox.Text = value;}
}
When you are opening the child form set the owner.
ChildForm f = new ChildForm();
f.Owner = this;
f.Show();
Create an event handler to child forms button click event.
public Button1_Click(object sender; EventArgs e)
{
ParentForm f = (ParentForm)this.Owner;
f.TextBoxText = txtChildTextBox.Text;
}
i didn't compile code; so may have errors :)
}
When a button is pressed to close the launched form, returning you to the main form- the launched form's text box is still in scope.
Closing a form is merely changing the object's state, not disposing of it. So in the button eventhandler that launches the form from the main form, the next line after launching your modal window, it can access the text from the object it launched as the textbox is a child of that form's object. Unless you're launching your modal window in another thread, which I wouldn't figure you are since it's modal, when it is closed, it should go to the next line in the buttons eventhandler that launched it.
your main form may have code something like this right now (haven't done winforms in a while so bear with me if I miss something):
public void Button1_Click(object sender, ClickEventArgs e)
{
SomeFormIWantToLaunch launchForm = new SomeFormIWantToLaunch();
launchForm.ShowDialog(this);
}
You need to just add after launchForm.ShowDialog(this); something like:
this.SomeTextBox.Text = launchForm.ATextBox.Text;
I am using Visual Studio 2010, C# .NET 4.0. I have 3 forms: Form1, Form2, Form3.
In Form1 I have a button to open Form2:
private void button1_Click(object sender, EventArgs e)
{
Form2 f = new Form2();
f.Show();
}
In Form2 I have a private Form3 variable always pointing to the same Form3:
private Form3 f = new Form3();
And a button to open it as a dialog:
private void button1_Click(object sender, EventArgs e)
{
f.ShowDialog();
}
In Form3 I just have a button to hide the form:
private void button1_Click(object sender, EventArgs e)
{
this.Hide();
}
The problem is that having the situation that Form2 is in front of Form1, and Form3 in front of Form2, when I click the button of Form3 to hide it, it not only hides itself but sends Form1 to the back of all of the other Windows.
This only happens when there is a window of another program (such as Windows Explorer) in the background of Form1. It seems like a bug. What do you think?
Yes, this cannot work properly by design. A dialog disables all of the windows that your program displays. So that it is modal. When you hide the dialog, there are no windows left that can get the focus. Windows is forced to find another window to give the focus to. That will be a window owned by another application. Your own windows will now hide behind it.
There are more side effects, the dialog will also close. Necessary because otherwise the user can never get back to your program anymore since all windows are disabled. This is all unsurprising behavior. Bug would be a strong word, but it would of course work better if it first re-enabled all windows before closing the dialog. But closing the dialog is already undesirable behavior.
Don't call Hide() for a dialog. Just set the DialogResult property to DialogResult.Cancel to achieve the exact same effect, minus the focus problem. You do have to reset it back to None if you want to display the dialog again. That's a real bug.
By the documentation. Form.Close method doesn't dispose forms shown by Form.ShowDialog method. Quote:
The two conditions when a form is not disposed on Close is when (1) it is part of a multiple-document interface (MDI) application, and the form is not visible; and (2) you have displayed the form using ShowDialog. In these cases, you will need to call Dispose manually to mark all of the form's controls for garbage collection.
So, maybe there are ways to return focus to your application (e.g. via Windows API). But it is much more convenient to call Form.Close manually on dialog windows.