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How do I load a Window from a XAML file dynamically?

Platform: WPF| Category: Window

You can load a xaml file dynamically as follows:

[C#]
// This file should then be next to the exe. If not specify a relative path.
FileStream xamlFile = new FileStream('ChartWindow.xaml', FileMode.Open, FileAccess.Read);

Window win = XamlReader.Load(xamlFile) as Window;

win.Show();

Note that any code-behind associated with the above xaml will not be loaded/processed.

In case, you have an assembly reference to a custom dll in the above xaml file, you should reference that assembly with a fully qualified namespace as follows:

[XAML]
xmlns:syncfusion='clr-namespace:Syncfusion.Windows.Chart;assembly=Syncfusion.Chart.WPF,Version=6.303.0.6,Culture=neutral,PublicKeyToken=3d67ed1f87d44c89'

The above assembly should then be present in the GAC for the runtime to find it.

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