c# - 如何继承通用窗体并在 Visual Studio 设计器中打开它？

Question

在我的应用程序中，我有一个 BaseForm其中有一个通用成员:

public partial class BaseForm<T> : Form where T : Presenter
{
    protected T Presenter;

    public BaseForm()
    {
        InitializeComponent();
    }
}

现在我需要的是有一个从我的 BaseForm 继承的表单

public partial class SampleForm : BaseForm<SamplePresenter>
{
    public SampleForm()
    {
        InitializeComponent();
        Presenter = new SamplePresenter();
    }
}

问题是 Visual Studio 设计器没有显示我的 SampleForm源自 BaseForm<T> .

它给出了这个警告:

Warning 1 The designer could not be shown for this file because none of the classes within it can be designed. The designer inspected the following classes in the file:

SampleForm --- The base class 'Invoice.BaseForm' could not be loaded. Ensure the assembly has been referenced and that all projects have been built. 0 0

我该如何克服这个问题？

附言我看了this post但并没有真正了解如何解决这个问题。

Answer 1

如该帖子所述，设计者不支持这一点。

你需要这个基类:

public partial class SampleFormIntermediate : BaseForm<SamplePresenter>
{
    public SampleFormIntermediate()
    {
        InitializeComponent();
        Presenter = new SamplePresenter();
    }
}

并且您需要为 Visual Studio 设计器使用此类:

public partial class SampleForm : SampleFormIntermediate
{
}

通过这种方式，Visual Studio“了解”在设计器中打开什么以及如何打开它。