FontDialog.ShowColor Property
Important
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Gets or sets a value indicating whether the dialog box displays the color choice.
public:
property bool ShowColor { bool get(); void set(bool value); };
public bool ShowColor { get; set; }
member this.ShowColor : bool with get, set
Public Property ShowColor As Boolean
true
if the dialog box displays the color choice; otherwise, false
. The default value is false
.
The following code example uses ShowDialog to display a FontDialog. This code requires that a Form has already been created with a TextBox and button placed on it. It also requires that the fontDialog1
has been created. The Font contains the size information but not the color information.
private:
void button1_Click( Object^ /*sender*/, System::EventArgs^ /*e*/ )
{
fontDialog1->ShowColor = true;
fontDialog1->Font = textBox1->Font;
fontDialog1->Color = textBox1->ForeColor;
if ( fontDialog1->ShowDialog() != ::DialogResult::Cancel )
{
textBox1->Font = fontDialog1->Font;
textBox1->ForeColor = fontDialog1->Color;
}
}
private void button1_Click(object sender, System.EventArgs e)
{
fontDialog1.ShowColor = true;
fontDialog1.Font = textBox1.Font;
fontDialog1.Color = textBox1.ForeColor;
if(fontDialog1.ShowDialog() != DialogResult.Cancel )
{
textBox1.Font = fontDialog1.Font ;
textBox1.ForeColor = fontDialog1.Color;
}
}
Private Sub button1_Click(sender As Object, e As System.EventArgs)
fontDialog1.ShowColor = True
fontDialog1.Font = textBox1.Font
fontDialog1.Color = textBox1.ForeColor
If fontDialog1.ShowDialog() <> DialogResult.Cancel Then
textBox1.Font = fontDialog1.Font
textBox1.ForeColor = fontDialog1.Color
End If
End Sub
Produit | Versions |
---|---|
.NET Framework | 1.1, 2.0, 3.0, 3.5, 4.0, 4.5, 4.5.1, 4.5.2, 4.6, 4.6.1, 4.6.2, 4.7, 4.7.1, 4.7.2, 4.8, 4.8.1 |
Windows Desktop | 3.0, 3.1, 5, 6, 7, 8, 9 |
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