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.
Related
This question already has answers here:
How to prevent clicks outside the current form?
(2 answers)
Closed 1 year ago.
I am creating an application where I have a Form 1 (Main Form). Inside this main form there is a button that opens up a different and smaller form (Form 2). I don't want to hide Form 1 when showing Form 2 but because of the size of Form 2, the user is able to click Form 1 hiding Form 2 with its size (not hidden like Form2.hide();) however this is something I don't want to do. I want to prohibit the user to click Form 1 if Form 2 is currently open. Is there any event or function I can use to do this? I have seen in it in other applications but I don't even know how to look for it.
In addition to using the ShowDialog method as suggested in question's comments, you can disable the form itself if you want for example the user be able to switch between forms and view or copy some text.
Form.ShowDialog Method: Shows the form as a modal dialog box.
Here is a definition:
Wikipedia: A modal window creates a mode that disables the main window but keeps it visible, with the modal window as a child window in front of it. Users must interact with the modal window before they can return to the parent application. This avoids interrupting the workflow on the main window. Modal windows are sometimes called heavy windows or modal dialogs because they often display a dialog box.
If you don't want to use a modal form, you can initialize the form2 instance like that:
form2.FormClosed += (_s, _e) => this.Enabled = true;
Thus now you can call:
this.Enabled = false;
form2.Show();
You can also check the ShowInTaskBar property of the form2.
Be careful not to add the lambda event handler several times to the same instance: if form2 is just hidden on close, only one FormClosed += is needed, but is required for every instance of a form you want to manage this way.
If you need to disable only certain controls, use them instead of this:
private void SetControlsEnabled(bool state)
{
myControl1.Enabled = state;
myControl2.Enabled = state;
myPanelHavingSomeControls = state; // this changes all inners too
...
}
form2.FormClosed += (_s, _e) => SetControlsEnabled(true);
SetControlsEnabled(false);
form2.Show();
This is the exact use case for a modal form, which disables the parent form and provides visual clues from the OS to tell the user of that relationship. Something along the lines of this will do:
//Code in the event handler that opens the child form
using (Form2 childForm = new Form2())
{
//Put any initialization code here
childForm.ShowDialog(this);
//Any cleanup or using some return value from the form after it closed
}
I have a Main Form and a Other Form.
The Main Form is always open, and at some time it will launch Other Form.
I tried:
form.TopMost = true;
But this only places the form on top. The Form behind (Main Form) still can be accessible.
How can I get the same behavior like when I do OpenFileDialog, and disable the main form behind it?
(Thanks in advance)
You need to make your form modal. To do this, use ShowDialog instead of Show to display it (the same way you do with a dialog).
Also, note that forms shown with ShowDialog are not actually closed and disposed when you click their Close button, so you should dispose them manually. The usual way to handle their lifetime is to use a using construct:
using (var form = new SomeForm())
{
form.ShowDialog();
// do stuff after the dialog is closed
}
form.showdialog(); where form is the the top form to be launched.so while lauching the top form just add form.showdialog()
Is there a way to show a modal form without ShowDialog metod calling? By showing a modal form I mean that the Form overlays the current Form and prevent user input to the bacground Form. The Form that is to be shown is a MessageBox style form that is not fullscreen.
Thanks
Dominik
I would assume that you could set the "dialog" form to stay on top (TopMost property) and then disable the main form in order to prevent input.
This is only a partial solution as the main form will still be able to be focused, closed etc. You will need to handle all these cases yourself in order to prevent them.
Is there any particular reason you don't want to use ShowDialog?
You could try to do this:
MyForm frm = new MyForm(); // this would be your modeless dialog
frm->show(this); // "this" being the instance that invokes it.
I have a problem with modality of the forms under C#.NET. Let's say I have main form #0 (see the image below). This form represents main application form, where user can perform various operations. However, from time to time, there is a need to open additional non-modal form to perform additional main application functionality supporting tasks. Let's say this is form #1 in the image. On this #1 form there might be opened few additional modal forms on top of each other (#2 form in the image), and at the end, there is a progress dialog showing a long operation progress and status, which might take from few minutes up to few hours. The problem is that the main form #0 is not responsive until you close all modal forms (#2 in the image). I need that the main form #0 would be operational in this situation. However, if you open a non-modal form in form #2, you can operate with both modal #2 form and newly created non modal form. I need the same behavior between the main form #0 and form #1 with all its child forms. Is it possible? Or am I doing something wrong? Maybe there is some kind of workaround, I really would not like to change all ShowDialog calls to Show...
Image http://img225.imageshack.us/img225/1075/modalnonmodalproblem.png
Modal forms do exactly what "modal" means, they disable all other windows in the app. That's rather important, your program is in a somewhat perilous state. You've got a chunk of code that is waiting for the dialog to close. Really Bad Things could happen if those other windows were not disabled. Like the user could start the modal dialog again, now your code is nested twice. Or she could close the owner window of the dialog, now it suddenly disappears.
These are the exact kind of problems you'd run into if you call Application.DoEvents() inside a loop. Which is one way to get a form to behave modal without disabling other windows. For example:
Form2 mDialog;
private void button1_Click(object sender, EventArgs e) {
mDialog = new Form2();
mDialog.FormClosed += (o, ea) => mDialog = null;
mDialog.Show(this);
while (mDialog != null) Application.DoEvents();
}
This is dangerous.
It is certainly best to use modal forms the way they were designed to stay out of trouble. If you don't want a modal form then simply don't make it modal, use the Show() method. Subscribe to its FormClosing event to know that it is about to close:
private void button1_Click(object sender, EventArgs e) {
var frm = new Form2();
frm.FormClosing += new FormClosingEventHandler(frm_FormClosing);
frm.Show();
}
void frm_FormClosing(object sender, FormClosingEventArgs e) {
var frm = sender as Form2;
// Do something with <frm>
//...
}
The first thing that comes to mind would be something like this. You could disable form 1 when you launch form 2 and then have form 1 handle the closed event of the second form to re-enable itself. You would NOT open modal 2 using show dialog.
Now keep in mind, from a user perspective this is going to be quite cumbersome, you might look at doing a MDI application to get all windows inside of a single container.
Your main form will not be responsive until any modal dialogs that are in the same process space are closed. There is not work around for that.
It looks to me like you could use an MDI application setting the Form #0 IsMdiContainer property to true.
Then, you could do something alike:
public partial class Form0 {
public Form0 {
InitializeComponent();
this.IsMdiContainer = true; // This will allow the Form #0 to be responsive while other forms are opened.
}
private void button1_Click(object sender, EventArgs e) {
Form1 newForm1 = new Form1();
newForm1.Parent = this;
newForm1.Show();
}
}
Using the ShowDialog() as you stated in your question will make all of the forms Modal = true.
By definition, a modal form is:
When a form is displayed modally, no input (keyboard or mouse click) can occur except to objects on the modal form. The program must hide or close a modal form (usually in response to some user action) before input to another form can occur. Forms that are displayed modally are typically used as dialog boxes in an application.
You can use this property [(Modal)] to determine whether a form that you have obtained from a method or property has been displayed modally.
So, a modal form shall be used only when you require immediate assistance/interaction from the user. Using modal forms otherwise makes believe that you're perhaps running into a wrong direction.
If you do not want your main form to be an MDI container, then perhaps using multithreading is one solution through a simple BackgroundWorker class is the key to what you want to achieve. Thus, it looks to me like a design smell...
What is it you want to do, apart of making your main form responsive, etc.
What is it you have to do?
Explaining what you have to do, we might be able to guide you altogether into the right, or at least perhaps better, direction.
Actually the answer is very simple. Try
newForm.showDialog();
This will open a new form, while the parent one is inaccessible.
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();
}