ToolTip.ShowAlways Property
Definition
Important
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Gets or sets a value indicating whether a ToolTip window is displayed, even when its parent control is not active.
public:
property bool ShowAlways { bool get(); void set(bool value); };public bool ShowAlways { get; set; }member this.ShowAlways : bool with get, setPublic Property ShowAlways As BooleanProperty Value
true if the ToolTip is always displayed; otherwise, false. The default is false.
Examples
The following code example creates an instance of the ToolTip class and associates the instance with the Form that the instance is created within. The code then initializes the delay properties AutoPopDelay, InitialDelay, and ReshowDelay. In addition the instance of the ToolTip class sets the ShowAlways property to true to enable ToolTip text to be displayed regardless of whether the form is active. Finally, the example associates ToolTip text with two controls on a form, a Button and a CheckBox. The code example requires that the method defined in the example is located within a Form that contains a Button control named button1 and a CheckBox control named checkBox1, and that the method is called from the constructor of the Form.
// This example assumes that the Form_Load event handling method
// is connected to the Load event of the form.
void Form1_Load( Object^ sender, System::EventArgs^ e )
{
// Create the ToolTip and associate with the Form container.
ToolTip^ toolTip1 = gcnew ToolTip;
// Set up the delays for the ToolTip.
toolTip1->AutoPopDelay = 5000;
toolTip1->InitialDelay = 1000;
toolTip1->ReshowDelay = 500;
// Force the ToolTip text to be displayed whether or not the form is active.
toolTip1->ShowAlways = true;
// Set up the ToolTip text for the Button and Checkbox.
toolTip1->SetToolTip( this->button1, "My button1" );
toolTip1->SetToolTip( this->checkBox1, "My checkBox1" );
}
// This example assumes that the Form_Load event handling method
// is connected to the Load event of the form.
private void Form1_Load(object sender, System.EventArgs e)
{
// Create the ToolTip and associate with the Form container.
ToolTip toolTip1 = new ToolTip();
// Set up the delays for the ToolTip.
toolTip1.AutoPopDelay = 5000;
toolTip1.InitialDelay = 1000;
toolTip1.ReshowDelay = 500;
// Force the ToolTip text to be displayed whether or not the form is active.
toolTip1.ShowAlways = true;
// Set up the ToolTip text for the Button and Checkbox.
toolTip1.SetToolTip(this.button1, "My button1");
toolTip1.SetToolTip(this.checkBox1, "My checkBox1");
}
' This example assumes that the Form_Load event handling method
' is connected to the Load event of the form.
Private Sub Form1_Load(sender As Object, e As System.EventArgs) Handles MyBase.Load
' Create the ToolTip and associate with the Form container.
Dim toolTip1 As New ToolTip()
' Set up the delays for the ToolTip.
toolTip1.AutoPopDelay = 5000
toolTip1.InitialDelay = 1000
toolTip1.ReshowDelay = 500
' Force the ToolTip text to be displayed whether or not the form is active.
toolTip1.ShowAlways = True
' Set up the ToolTip text for the Button and Checkbox.
toolTip1.SetToolTip(Me.button1, "My button1")
toolTip1.SetToolTip(Me.checkBox1, "My checkBox1")
End Sub
Remarks
With the ShowAlways property, you can display a ToolTip window even when the container of the ToolTip is not active. You can use this feature in a modeless window application to enable ToolTip windows to be displayed regardless of which modeless window is active. This feature is also useful when you want to create a control by using the UserControl, which contains a number of controls within it that display ToolTip windows. Because the UserControl is often not the active window on a form, setting this property to true enables the controls within the UserControl to display ToolTip windows at any time.
