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如何获得in的WindowStartupLocation属性?WindowUserControl

public partial class ActivityDetailsView : UserControl
{
    public ActivityDetailsView(ActivityDetailsViewModel viewModel)
    {
        InitializeComponent();
        DataContext = viewModel;
        Loaded += ActivityDetailsView_Loaded;
        JobOrderLineItem.SelectionChanged += JobOrder_SelectionChanged;

        //this.WindowStartupLocation doesnt exist
    }
}
4

1 回答 1

0

这个怎么样?

public partial class ActivityDetailsView : Window
{
    // ...
}

如果你UserControl应该表现得像Window为什么不从中派生?

于 2013-04-17T08:16:46.100 回答