this.Visible is not working in Windows Forms

问题

I have a problem. I need to hide my window at window load. But

    private void Form1_Load(object sender, EventArgs e)
    {
        this.Visible = false;
    }

is not working. And property Visible remains true. Am I missing something?

回答1:

Yes, the Visible property is a big deal in Windows Forms, that's what actually gets the handle created and causes OnLoad() to run. In other words, the window doesn't exist until it gets visible. And it will ignore attempts to undo this.

It is pretty common to want to still create the handle but not make the window visible if you use a NotifyIcon. You can achieve this by overriding SetVisibleCore:

    protected override void SetVisibleCore(bool value) {
        if (!this.IsHandleCreated) {
            value = false;
            CreateHandle();
        }
        base.SetVisibleCore(value);
    }

Beware that OnLoad still won't run until the window actually gets visible so move code into the constructor if necessary. Just call Show() in the NotifyIcon's context menu event handler to make the window visible.

回答2:

It seems you can use the following:

    private void Form1_Load(object sender, EventArgs e)
    {
        this.Opacity = 0;
        this.ShowInTaskbar = false;
    }

I just tested it in a winforms app and it worked.

(Also just found this: Single Form Hide on Startup

回答3:

I think it is not good idea to set visibility from form's Load event. Instead, I would do it by add a public method:

public void LoadForm(...)
{
   // do the all the initializations
}

and call the method to load the form. The form should be not visible unless you explicitly show it:

MyForm instance = new MyForm();
instance.LoadForm(...);
// instance.Show(); or ShowDialog() for dialog form not sure exactly the syntax.

回答4:

Put your call in Windows event loop like this:

WindowsFormsSynchronizationContext.Current.Post((obj) => this.Hide(), null);

So Hide() will be invoked later, hence fix your problem.

回答5:

I believe this is because the window doesn't really exist until after this event. The best place to do this is outside the form:

if (x == 1)
{
    form1.ShowForm();
}
else
{
    // Don't show the form
}

If you really need to do it inside the form itself, then I think you need to use the Activated event.

EDIT:

You could also try something like:

    static void Main()
    {
        Application.EnableVisualStyles();
        Application.SetCompatibleTextRenderingDefault(false);

        Form1 myform = new Form1();
        if (myform.CheckStuff())
        {
            Application.Run(myform);
        }
    }

回答6:

Use this.Opacity = 0;

回答7:

this.Opacity = 0;
this.ShowInTaskbar = false;

回答8:

Use this.Hide() to hide your window. this.Close() to close

回答9:

Call Hide() in the Shown event, Hide() only sets Visible to false, and before the form is being shown the Visible property is already false...

来源：https://stackoverflow.com/questions/3742709/this-visible-is-not-working-in-windows-forms

标签

winforms

visible