User (Legacy) 2 Posted February 4, 2005 Report Share Posted February 4, 2005 Hi, in the VS.NET GUI designer, I set my chart's Gallery to Curve. It generates the following 2 lines: this._chart.Gallery = SoftwareFX.ChartFX.Gallery.Curve; this._chart.GalleryObj.Native = true; The second line causes compile error because GalleryObject is of type object which doesn't have a property call Native. Is this a bug? Ming Quote Link to post Share on other sites
Software FX 0 Posted February 6, 2005 Report Share Posted February 6, 2005 No. This is not a bug. The type of GalleryObj is defined as object as it can not change when the gallery changes. You need to cast it to he appropriate type: ((SoftwareFX.ChartFX.GalleryObj.Surface) this._chart.GalleryObj).Native = true; FYI: Native = true is the default setting. -- FP Software FX Quote Link to post Share on other sites
User (Legacy) 2 Posted February 7, 2005 Author Report Share Posted February 7, 2005 As I mentioned before, the problem is that this code was generated by the VS.NET GUI Designer. Yes, I can manually make the cast but each time I make change the chart, it regenerate that code again. Can you reproduce that? "SoftwareFX Support" <noreply@softwarefx.com> wrote in message news:1CxD8NHDFHA.2488@webserver3.softwarefx.com... > No. This is not a bug. The type of GalleryObj is defined as object as it can > not change when the gallery changes. You need to cast it to he appropriate > type: > > ((SoftwareFX.ChartFX.GalleryObj.Surface) this._chart.GalleryObj).Native = > true; > > FYI: Native = true is the default setting. > > -- > FP > Software FX > > Quote Link to post Share on other sites
Software FX 0 Posted February 8, 2005 Report Share Posted February 8, 2005 No. What I get when I set it at design time is: this.chart1.Gallery = SoftwareFX.ChartFX.Gallery.Curve; ((SoftwareFX.ChartFX.GalleryObj.Curve)(this.chart1.GalleryObj)).Native = false; We have a TypeConverter that makes this possible. I don't know why this is not working on your side. What version of VS are you using ? What version (build) of ChartFX ? -- FP Software FX Quote Link to post Share on other sites
Software FX 0 Posted February 8, 2005 Report Share Posted February 8, 2005 No. What I get when I set it at design time is: this.chart1.Gallery = SoftwareFX.ChartFX.Gallery.Curve; ((SoftwareFX.ChartFX.GalleryObj.Curve)(this.chart1.GalleryObj)).Native = false; We have a TypeConverter that makes this possible. I don't know why this is not working on your side. What version of VS are you using ? What version (build) of ChartFX ? -- FP Software FX Quote Link to post Share on other sites
User (Legacy) 2 Posted February 8, 2005 Author Report Share Posted February 8, 2005 ChartFX 6.2.1342.0 VS 7.1.3088 .NET Framework 1.1.4322 "SoftwareFX Support" <noreply@softwarefx.com> wrote in message news:4lAuhJgDFHA.2536@webserver3.softwarefx.com... > No. What I get when I set it at design time is: > > this.chart1.Gallery = SoftwareFX.ChartFX.Gallery.Curve; > > ((SoftwareFX.ChartFX.GalleryObj.Curve)(this.chart1.GalleryObj)).Native = > false; > > > We have a TypeConverter that makes this possible. I don't know why this is > not working on your side. > > What version of VS are you using ? What version (build) of ChartFX ? > > -- > FP > Software FX > > Quote Link to post Share on other sites
User (Legacy) 2 Posted February 9, 2005 Author Report Share Posted February 9, 2005 Found the problem. We have a wrapper (inheritance) class around SoftwareFX.ChartFX.Chart class. I think that is what causes it not to cast probably. When I tested with direct SoftwareFX.ChartFX.Chart class, it works as you stated. Is there any way I can fix the wrapper class? It is a straight inheritance from SoftwareFX.ChartFX.Chart. Thanks again. "SoftwareFX Support" <noreply@softwarefx.com> wrote in message news:4lAuhJgDFHA.2536@webserver3.softwarefx.com... > No. What I get when I set it at design time is: > > this.chart1.Gallery = SoftwareFX.ChartFX.Gallery.Curve; > > ((SoftwareFX.ChartFX.GalleryObj.Curve)(this.chart1.GalleryObj)).Native = > false; > > > We have a TypeConverter that makes this possible. I don't know why this is > not working on your side. > > What version of VS are you using ? What version (build) of ChartFX ? > > -- > FP > Software FX > > Quote Link to post Share on other sites
Software FX 0 Posted February 10, 2005 Report Share Posted February 10, 2005 Yes. That is because the Designer attribute is probably not inherited. Try adding this to your control: Designer("SoftwareFX.ChartFX.Designer.ClientDesigner,ChartFX.Designer, Version=6.2.1342.0"), -- FP Software FX Quote Link to post Share on other sites
User (Legacy) 2 Posted February 10, 2005 Author Report Share Posted February 10, 2005 where in my contol?? "SoftwareFX Support" <noreply@softwarefx.com> wrote in message news:e%23XHZ34DFHA.2544@webserver3.softwarefx.com... > Yes. That is because the Designer attribute is probably not inherited. > > Try adding this to your control: > > Designer("SoftwareFX.ChartFX.Designer.ClientDesigner,ChartFX.Designer, > Version=6.2.1342.0"), > > > -- > FP > Software FX > > Quote Link to post Share on other sites
Software FX 0 Posted February 11, 2005 Report Share Posted February 11, 2005 In the attributes (at the top of the class). -- FP Software FX Quote Link to post Share on other sites
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