表单设计器打破了通用抽象 UserControl

我有一个通用摘要UserControl class, SensorControl，我希望所有传感器控制面板都继承它。

问题

当尝试设计时EthernetSensorControl（我从 Visual Studio 继承的 UserControl 表单之一，表单设计器中显示以下错误：

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: DeviceSensorControl --- The base class 'Engine.Sensors.SensorControl' could not be loaded. Ensure the assembly has been referenced and that all projects have been built.

SensorControl class:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Windows.Forms;

namespace Engine.Sensors
{
    public abstract class SensorControl<SensorType>
        : UserControl where SensorType : class
    {
        protected SensorType _sensor;
        public SensorControl(SensorType sensor)
        {
            _sensor = sensor;
        }
    }
}

示例继承类，EthernetSensorControl:

namespace Engine.Sensors
{
    public partial class EthernetSensorControl
        : SensorControl<EthernetSensor>
    {
        public EthernetSensorControl(EthernetSensor sensor)
            : base(sensor)
        {
        }
    }
}

和调用堆栈：

at System.ComponentModel.Design.Serialization.CodeDomDesignerLoader.EnsureDocument(IDesignerSerializationManager manager) at System.ComponentModel.Design.Serialization.CodeDomDesignerLoader.PerformLoad(IDesignerSerializationManager manager) at Microsoft.VisualStudio.Design.Serialization.CodeDom.VSCodeDomDesignerLoader.PerformLoad(IDesignerSerializationManager serializationManager) at System.ComponentModel.Design.Serialization.BasicDesignerLoader.BeginLoad(IDesignerLoaderHost host)

一切都编译完毕，我可以看到显示的面板，但我无法设计它。我认为问题可能与partial类。有任何想法吗？

您不能设计继承某个控件或窗体abstract class.

（设计者需要实例化基类作为设计图面）

基类还需要有一个无参数构造函数供设计者调用。
这个构造函数可以是私有的。