I have developed a C# Windows Form.
At first, the Windows Form works fine.
However, one day the Windows Form starts up always minimized and I have no idea.
I checked the WindowState is Normal not Minimized.
How Can I fix it, Thanks!
Edit:
I comment each code block to narrow the scope to locate the problem point.
And I found that I used a Drive Detector in my MainForm.
When that instance was created, the call Window form must be passed as a parameter to the constructor.
Otherwise, the Drive Detector will create a hidden form. However, the MainForm will be minimized.
The below code will NOT create a hidden form.
driveDetector = new DriveDetector(this);
The below code will create a hidden form, it will interfere the call Windows Form.
driveDetector = new DriveDetector();
try to add this code in form load event and test
this.WindowState = FormWindowState.Normal;
you should use WindowState = FormWindowState.Maximized if you wish to open your windows in full screen by default. You can do this programmatically in Form load event.
There are other various options available too from which you can control on How to open your windows form.
1.Check whether you have set the size of form to smaller one.
2.Try re-build your solution.
3.Add Form Load Event from your Events Properties of Form and add following code to it
this.WindowState = FormWindowState.Normal;
Try to do it in form activated event
bool bIsLoaded = false;
private void Form1_Activated(object sender, EventArgs e)
{
if (!bIsLoaded)
{
this.WindowState = FormWindowState.Maximized;
bIsLoaded = true;
}
}
Just try to add it from the code level to say the windows state as follows.
private void Form1_Load(object sender, EventArgs e)
{
this.WindowState = FormWindowState.Normal;
}
I comment each code block to narrow the scope to locate the problem point.
And I found that I used a Drive Detector in my MainForm.
When that instance was created, the call Window form must be passed as a parameter to the constructor.
Otherwise, the Drive Detector will create a hidden form. However, the MainForm will be minimized.
The below code will NOT create a hidden form.
driveDetector = new DriveDetector(this);
The below code will create a hidden form, it will interfere the call Windows Form.
driveDetector = new DriveDetector();
Try this:
Topmost = true;
In your Form_Load event
Related
I have a form and through this I press a button and a new form opens. If I click minimize this new form, the form I call this form from is minimized as well. How could I make sure the form from which I call the new form is not minimized?
Maybe I have to enable some property in the form or something.
Of course, I tried with the following code and with the form propety singleFixed but the two forms are minimized:
private void bminimize_Click(object sender, EventArgs e)
{
WindowState = FormWindowState.Minimized;
}
Maybe have I to create this new form as subform or something like this?
EDIT: How I call this new form:
private void Button_Click(object sender, EventArgs e)
{
DateTime rnow = DateTime.Now;
Chronometer chrono = new chronometer();
var resultchrono = chrono.ShowDialog();
if (resultchrono == DialogResult.OK)
{
...
}
You're using ShowDialog(), which is documented thus, emphasis mine:
This version of the ShowDialog method does not specify a form or control as its owner. When this version is called, the currently active window is made the owner of the dialog box. If you want to specify a specific owner, use the other version of this method.
If you need the dialog result, you'll need to use the ShowDialog(owner) form of the call with say, the desktop window handle.
Its because of: chrono.ShowDialog();. Show the dialog with chrono.Show(); . But you will have to handle returned values differently.
If you need the DialogResult in the same method in which you open the window, then use AKX´s answer.
I've had a look around but none of the answers make any sense to me. I have a menu form which has buttons on; when users come to use the menu form, you can open other forms from the menu. Currently, I can get the form to open, but the menu form stays open too.
private void BtnAddNewCar_Click(object sender, EventArgs e)
{
AddCompanyCar carForm = new AddCompanyCar();
carForm.ShowDialog();
}
The code above opens the form AddCompanyCar from the menu. How do I add to this code so that the form 'Menu' closes when AddCompanyCar opens?
Are you sure want to do this as it impacts usability. If you're using WinForms, then just create a container window, and replace the panels instead. Might be easy and best way
If not and you wanna go-ahead, can take a look on this example
Why not just hide it, then show it again when ShowDialog() returns?
private void BtnAddNewCar_Click(object sender, EventArgs e)
{
this.Visible = false;
AddCompanyCar carForm = new AddCompanyCar();
carForm.ShowDialog(); // execution stops here until "carForm" is dismissed
this.Visible = true;
}
by closing the main window, you destroy the context in which you were previously working. As others suggest, simply hide the main window so you can return to it.
Is there any way of stopping the default opening animation of a form from Windows? I mean stopping it just for one form, not all of them. I need the showing of a form to be instantly, without that growing effect.
I tried to document about WinApi, but I couldn't find anything. Thank you!
You have to take a couple of steps to achieve this:
In the constructor, set your BorderStyle to None. This will prevent the animation:
public Form1()
{
InitializeComponent();
this.FormBorderStyle = FormBorderStyle.None;
}
Of course, now you want to show the border again, so you have to set it back as soon as your form is shown, so the user never sees this. Therefore create a method:
private void Form_Loaded(object sender, EventArgs e)
{
this.FormBorderStyle = FormBorderStyle.Sizable;
}
Now you have to set the Shown event of your form to your Form_Loaded method. You can do this from the form designer event properties window.
If you want to do it for the entire application you can change it using the SystemParametersInfo Method.
I had to set the location property after showing the form. It works!
this.Hide();
this.Size = this.Size;
this.Location = this.Location;
this.Show();
I have 2 forms ...when i start the application..and use the close "X" from the title bar the entire application closes...now when i select an option from the 1st form in my case it is a button "ADD" as its a phonebook application..it goes to the 2nd form as i have used 1stform.hide() and 2ndform.show()...now when i do "X" from the title bar it doesnt shutdown completely as the 1stform is not closed....how to program it in such a way tht any stage the entire application should close
Your first form is set as the startup form. That means whenever it gets closed, your entire application is closed. And conversely, your application does not close until it gets closed. So when you hide the startup form and show the second form, the user closing the second form does not trigger your application closing because they have only closed a secondary, non-modal dialog.
I recommend changing your design so that the startup form is also the main form of your application. No sense trying to work around built-in functionality that can actually be useful. You want the application to quit when the main form is closed, no matter what other child forms are opened.
But the quick-and-dirty solution in your case is to make a call to Application.Exit. That will close all of the currently open forms and quit your application immediately. As I said just above, I don't so much recommend this approach because having to call Application.Exit from every form's FormClosed event handler is a sign that something has gone seriously wrong in your design.
If the single startup form paradigm doesn't work out for you, you should look into taking matters into your own hands and customizing the Main method in your Program.cs source file. See the answers given to this related question for some ideas on how that might work for you.
What you can do is to use the Form's FormClosing event, and add the following code:
Application.Exit();
This will stop the entire application, and close all windows. However, if a background thread is running, the process itself will survive. In this case you can use:
Environment.Exit();
Add a Application.Exit on every forms's Closing event
like this:
Create an closing event handler first
private void Form_ClosingEventhandler()(object sender, CancelEventArgs e)
{
//Perform any processing if required like saving user settings or cleaning resources
Application.Exit();
}
then bind this event to any form you create.
//Where you create new form and show it.
Form1 frm= new Form1();
//set other properties
frm.Closing += new EventHandler(Form_ClosingEventhandler);
Form2 frm2= new Form2();
//set other properties
frm2.Closing += new EventHandler(Form_ClosingEventhandler);
Surely you don't want to shut down the entire application after the user adds a phone number? You just need to make sure that your main window becomes visible again. Write that like this:
private void AddButton_Click(object sender, EventArgs e) {
var frm = new AddPhoneNumber();
frm.StartPosition = FormStartPosition.Manual;
frm.Location = this.Location;
frm.Size = this.Size; // optional
frm.FormClosing += delegate { this.Show(); };
frm.Show();
this.Hide();
}
I am changing the Visibility of a Form to false during the load event AND the form still shows itself. What is the right event to tie this.Visible = false; to? I'd like to instantiate the Form1 without showing it.
using System;
using System.Windows.Forms;
namespace TestClient
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private void Form1_Load(object sender, EventArgs e)
{
this.Visible = false;
}
}
}
Regardless of how much you try to set the Visible property before the form has been shown, it will pop up. As I understand it, this is because it is the MainForm of the current ApplicationContext. One way to have the form automatically load, but not show at application startup is to alter the Main method. By default, it looks something like this (.NET 2.0 VS2005):
[STAThread]
static void Main()
{
Application.EnableVisualStyles();
Application.SetCompatibleTextRenderingDefault(false);
Application.Run(new Form1());
}
If you instead do something like this, the application will start, load your form and run, but the form will not show:
[STAThread]
static void Main()
{
Application.EnableVisualStyles();
Application.SetCompatibleTextRenderingDefault(false);
Form1 f = new Form1();
Application.Run();
}
I am not entirely sure how this is useful, but I hope you know that ;o)
Update: it seems that you do not need to set the Visible property to false, or supply an ApplicationContext instance (that will be automatically created for you "under the hood"). Shortened the code accordingly.
I know this is an old question, but I just stumbled upon it and am pretty surprised no one has mentioned SetVisibleCore:
bool isVisibleCore = false;
protected override void SetVisibleCore(bool value)
{
base.SetVisibleCore(isVisibleCore);
}
In that code snippet, as long as isVisibleCore remains false, the form will remain invisible. If it's set to false when the form is instantiated, you won't get that brief flash of visibility that you'd get if you set Visible = false in the Shown event.
It took me some time to find a properly working Solution.
Set the properties named WindowState to Minimized and ShowInTaskbar to False under properties window. Once your form is completly loaded, Call below lines of code.
this.ShowInTaskbar = true;
this.WindowState = System.Windows.Forms.FormWindowState.Maximized;
//this.WindowState = System.Windows.Forms.FormWindowState.Normal;
PS: This solution is Tested on Visual C# 2008 Express Edition
How about setting the Opacity property to 0 on design and back to 100 when you want to show the form?
a solution i can live with
so the form is created and on_load is called on createtion.
set WindowState to minimize then on load set visible to false and windowstate to normal
private void Form1_Load(object sender, EventArgs e)
{
this.Visible = false;
this.WindowState = FormWindowState.Normal;
}
what did not worked:
the SetVisibleCore override solution did not created the form
as also the
Application {
Form1 f = new Form1();
Application.Run();
):
For a flicker-free Shown solution, also set the form's location off-screen during load:
private Point startuplocation;
private void Form1_Load(object sender, EventArgs e)
{
this.startuplocation = this.Location;
this.Location = new Point(-1000, -1000);
}
private void Form1_Shown(object sender, EventArgs e) //fires first time shown
{
this.Visible = false;
this.Location = this.startuplocation;
}
Just create an instance of Form1 and do not call methods to show/display it. But I bet you're doing something wrong.
Try on the VisibleChanged event.
The shown event may give you want you want. Although the form will "flash" for a second before hiding.
private void Form1_Shown(object sender, EventArgs e)
{
this.Visible = false;
}
What I would suggest would be to instantiate the form in an event the precedes the _Show event, such as the constructor, after the IntializeComponent() call.
If this is your main form, there may not be a better place then the Shown event. But in that case you will get flicker.
I couldn't find a good place to stop a running main form from showing at least quickly. Even a timer activated in the load event won't do it.
If it is a secondary form just create it but don't show it.
Have you tried
this.Hide();
in the form_load or form_activated events
Set the visibilty on the constructor, after init and then this.Show() later
InitializeComponent() is setting this.Visible = true, since you specified that the form should be visible in the designer (or it defaulted to that). You should set Visible to false in the designer, and it won't be called by InitializeComponent(). You can then make it visible any time you like.
Having .Visible = false or Hide() in the Load event will cause your form to show briefly, as there is time between when it becomes physically visible and when the Load event gets fired, in spite of the fact that the documentation says the contrary.
Are you calling Show() or ShowDialog() somewhere? I'm not sure if this behavior is still present, but at least in past versions of the framework a call to ShowDialog() did not trigger the Load event, so perhaps that is your issue (though I think calling ShowDialog() then hiding a modal form would be a bad practice!)
If you have to have the handle created (and the handles for controls) for whatever it is you're trying to do, a better idea would be to set the StartLocation to Manual, then set the Position property to an offscreen location. This will create and show the form, while making it invisible to the user.
Yeah the really one elegant way in perspective to your code than your applications visual is to flicker the form by hiding in the constructor/load event.
I set these three property settings for the form:
ShowInTaskbar = false
ShowIcon = false
WindowState = Minimized