Login | Register   
RSS Feed
Download our iPhone app
Browse DevX
Sign up for e-mail newsletters from DevX

By submitting your information, you agree that devx.com may send you DevX offers via email, phone and text message, as well as email offers about other products and services that DevX believes may be of interest to you. DevX will process your information in accordance with the Quinstreet Privacy Policy.

Tip of the Day
Language: VB7
Expertise: Intermediate
Jun 8, 2002



Building the Right Environment to Support AI, Machine Learning and Deep Learning

Ensuring that a style is supported by a font family

' Not all fonts support the same styles. This function takes in 
' input a font family and a font style, and returns a font style
' which is safe for that particular font family, by removing
' the styles that are not supported
' Example:
'   GetSafeStyleForFontFamily(richTextBox1.SelectionFont.FontFamily,
'  '     richTextBox1.SelectionFont.Style)

Public Function GetSafeStyleForFontFamily(ByVal fontFam As FontFamily, _
    ByVal style As FontStyle) As FontStyle
    ' remove the styles not supported
    If (style And FontStyle.Regular) = FontStyle.Regular Then
        ' if the regular style is currently present, but not supported 
        ' by the new font family, remove it
        If Not fontFam.IsStyleAvailable(FontStyle.Regular) Then
            style = style And Not FontStyle.Regular
        End If
    End If
    ' do the same for bold, italic and underline
    If (style And FontStyle.Bold) = FontStyle.Bold Then
        If Not fontFam.IsStyleAvailable(FontStyle.Bold) Then
            style = style And Not FontStyle.Bold
        End If
    End If
    If (style And FontStyle.Italic) = FontStyle.Italic Then
        If Not fontFam.IsStyleAvailable(FontStyle.Italic) Then
            style = style And Not FontStyle.Italic
        End If
    End If
    If (style And FontStyle.Underline) = FontStyle.Underline Then
        If Not fontFam.IsStyleAvailable(FontStyle.Underline) Then
            style = style And Not FontStyle.Underline
        End If
    End If
    If (style And FontStyle.Strikeout) = FontStyle.Strikeout Then
        If Not fontFam.IsStyleAvailable(FontStyle.Strikeout) Then
            style = style And Not FontStyle.Strikeout
        End If
    End If
    Return style
End Function
Marco Bellinaso
Comment and Contribute






(Maximum characters: 1200). You have 1200 characters left.



Thanks for your registration, follow us on our social networks to keep up-to-date