Ticket Q331938
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How to initialize the Series and DataFilters at runtime

created 14 years ago

Please, i need to create n instance of series but i'm not able to add all settings:
my code is
diagramma.Series.Add(New DevExpress.Xpf.Charts.AreaSeries2D With {.DisplayName = "Fatture" _
                                                                             , .Transparency = 0.5 _
                                                                             , .ArgumentDataMember = "DataEmissione" _
                                                                             , .ValueDataMember = "Importo" _
                                                                             , .ArgumentScaleType = DevExpress.Xpf.Charts.ScaleType.DateTime _
                                                                             , .AnimationAutoStartMode = DevExpress.Xpf.Charts.AnimationAutoStartMode.PlayOnce _
                                                                             })
but i'm not able to add settings for tags
<dxc:AreaSeries2D.SeriesAnimation>
   <dxc:Area2DUnwindAnimation Duration="00:00:01" />
</dxc:AreaSeries2D.SeriesAnimation>
Coul'd you write a full example?
Thanks

Comments (1)

    please, i need <dxc:DataFilter ColumnName="Year" Condition="Equal" Value="2002"/> too

    Answers approved by DevExpress Support

    created 14 years ago (modified 11 years ago)

    Hi Davide,
    Thank you for contacting us. Please refer to the sample code snippet illustrating how to initialize the SeriesAnimation and DataFilters properties at runtime:

    Visual Basic
    Dim series As New DevExpress.Xpf.Charts.AreaSeries2D() Dim animation As New DevExpress.Xpf.Charts.Area2DUnwindAnimation() animation.Duration = TimeSpan.FromSeconds(1) series.SeriesAnimation = animation series.DataFilters.Add(New DataFilter()) series.DataFilters(0).ColumnName = "Year" series.DataFilters(0).Condition = DataFilterCondition.Equal series.DataFilters(0).Value = "2002"

    Should you need any further clarification, feel free to ask. I will be glad to help you.
    Thanks, Susan

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