I have a windows form application where I show information about products and product categories in a datagridview. I would like to create a popup window so when I right click on a product and choose add to category a popup window appears and in that I show all the categories in a dropdownbox and when I click a button the product add to the category.
I want to create a popup window with a dropdownbox and a button. How do I do that in a window form application?
You can create a regular form, and call the myPopupForm.ShowDialog() method. The ShowDialog method blocks the main form, so the user can select a category, upon which you Close() the popup window; execution will continue on the main form.
More information can be found on http://msdn.microsoft.com/en-us/library/c7ykbedk.aspx#Y800.
Form2 form = new Form2();
This method will be like the Message.Show method, but you can add Buttons, TextBoxes, etc. in the Designer tools.
form.ShowDialog();
This method will just bring up another form.
form.Show();
AddHandler Me.Click,AddressOf Me_Click
Private Sub Me_Click(ByVal sender As Object, ByVal e As EventArgs)
Dim popupForm As PopupForm = New PopupForm()
popupForm.Show(Me)
End Sub
You may get an exception window.Click continue-popup window will come.For further clicks get as many popups as you click
Regards
kvinvisibleguy
Related
I have shown a new form in button click on parent form like below,
Here the background parent form gets disabled when a new child form is activated.
Is there any available options to show the child form without disabling the parent form?
Regards,
The Show function shows the form in a non modal form. This means that you can click on the parent form.
ShowDialog shows the form modally, meaning you cannot go to the parent form
Application.Run() runs the main parent form, and makes that form the main form. Application.Run() is usually found in main.
If it is fully disabled (no interaction possible), you are using .ShowDialog() on the child form instead of .Show(). If you use .Show() you will be able to use both forms
Example:
ChildForm childForm = new ChildForm();
childForm.Show();
after you open the new form and call this.Activate(); it will refocus on the parent window, but this will cause it to loose focus for a fraction of a second
I have a Splash Screen to my program.
And when the splash end , exec the code
this.Hide();
frmLogin o = new frmLogin();
o.show();
And it works but the splash screen go invisible and when i close the program by my custom exit button it's only closing the current form.
But my splash screen is still hidden and appears the app name to task manager.
How i can close that currently opened form and the invisible ones with my custom button ?
You are launching your whole application from the splash screen form.
It would be better avoid this behaviour separating splash screen form from main form, and opening up from this last form the splash screen, as well as you have done with frmLogin.
However you can workaround this problem using this.Hide() and at the exit of the program using Application.Exit() within your "custom exit button" event.
While in design view, select the frmLogin button. On the right, the Properties window should update. Select the Events tab. Look for FormClosing in the list of events. Select the right column and type a method name of your choice. If you typed xBtnPress, the designer will generate a stub that will look like this:
private void xBtnPress(object sender, FormClosingEventArgs e) {
}
If you wish to close your application upon this event, call Application.Exit() from within the event handler.
I want the form which will open when the user click a button, I don't want the user to click anywhere else outside this form(preventing him from clicking any other buttons on the parent form), I want to restrict his actions inside this form until he close it (For sure this form is top most), and I want the form to make alert(flicker) to any clicks outside it so the user will understand that he have to close it first.
When opening the form instead of
Form form = new Form();
form.Show();
use
Form form = new Form();
form.ShowDialog()
Super easy fix!
I assume your code to show the form right now is
form.Show();
To create/show an instance of the form that is modal (restricts clicking to that form) all you need to change is that line to:
form.ShowDialog();
Liam
In my form i have button to show other form.
But i want the previously form cannot be clicked before the new form is closed , How to create that ?
Because if the previously form is clicked, and i click the button again, the form is show multiple.
this my code in button click :
MDACS_AOP_CFSTL_InputActivity addProblem = new MDACS_AOP_CFSTL_InputActivity(ParameterSesi, ParameterNamaKaryawan, ParameterTanggal);
//addProblem.Close();
addProblem.Show();
You should use
addProblem.ShowDialog(this);
This will open the dialog as child of the parent (this) dialog. You cant click the parent dialog but you can still see it.
You need to use
addProblem.ShowDialog();
instead of
addProblem.Show();
so that it will open a modal dialog.
You could use
MDACS_AOP_CFSTL_InputActivity addProblem = new MDACS_AOP_CFSTL_InputActivity(ParameterSesi, ParameterNamaKaryawan, ParameterTanggal);
this.Hide;
addProblem.ShowDialog();
this.Show();
The show(); command will not be executed until the dialog is close. So it will stay hidden
I have two open winforms, say that winform1 gets a modal dialog, this means that all forms(winform1 and winform2) will be "disabled". If we minimize all forms and then bring up winform1, then the modal dialog will be shown above it. If we again minimize all the forms but this time brings up winform2, it will look like the finform2 is ready to be used while its really disabled like winform1.
What I need is to clearly show that the modal dialog needs to be handled before using winform2 again.
Is there anything built in to handle this or am I on my own here?
In your winform2.Activated event handler, call this:
static void FocusModalForm()
{
foreach (Form form in Application.OpenForms)
if (form.Modal)
{
form.WindowState = FormWindowState.Normal;
form.BringToFront();
}
}
e.g.
Form f2 = new Form();
f2.Activated += (_, __) => FocusModalForm();
f2.Show();
You may need to do the same thing for winform1's Activated event. It depends how winform2 gets created. Just try it and if you find that winform1 (or any other nonmodal form) is still able to get in front of the modal form, just call FocusModalForm() from its Activated event.
I tried this in Windows 7. I tried hiding all windows (click the Show Desktop button on the taskbar) and then selecting form2 directly from the taskbar, and I also tried just selecting form2 from the taskbar without hiding all windows. Form3 always stayed on top.
I have a similiar app (vb.net) where win1 calls win2 & win2 displays win3 and it works as you would like but wins 2 & 3 are both modal. I don't know if that is why it works or not. Perhaps that is an option for you?
If you do this:
var winform2 = new Winform2();
winform2.Show(winform1);
Then winform2 will always be shown above winform1, but it won't be modal. May be this can help you.