在设计时无法在某些属性中“设置" [英] do not work “set” in some of property in design time

问题描述

为什么在设计时设置了"dataGrid"属性在我的代码中不起作用，但是属性字体正常工作.
我在表格上使用了这个组件. (我通过设计时调试对其进行了测试)

Why, in design time, does set "dataGrid" Property not work in my code, but property font is working correctly.
I used of this componet on the a Form. (I test it with design time debuging)

namespace Example
{
    public partial class Component1 : System.Windows.Forms.DataGridView
    {

        public Component1()
        {
        }

        private DataGridViewColumn _gridcolumn;

        [DesignerSerializationVisibility(DesignerSerializationVisibility.Content)]
        [TypeConverter(typeof ( ExpandableObjectConverter))]
        public DataGridViewColumn gridcolumn
        {
            get
            {
                if (_gridcolumn == null)
                    _gridcolumn = new DataGridViewColumn();
                return _gridcolumn;
            }

            set                                                 //Do not Work Set in designTime 
            {
                _gridcolumn = value;
            }
        }


        private System.Drawing.Font _MyFont;
        public System.Drawing.Font MyFont
        {
            get
            {
                if (_MyFont == null)
                    _MyFont = new System.Drawing.Font("tahoma", 8);
                return _MyFont;
            }
            set
            {
                _MyFont = value;                                            //Work correctly in design time
            }
        }

        int _with;
        public int withcustom
        {
            get
            {
                return _with;
            }
            set
            {
                _with = value;                                           //Work correctly in design time
            }
        }
 
    }
}

推荐答案

原因是设计人员希望设置自定义控件的所有可访问属性.这样做是由于设计器中Visual Studio中的属性"窗口.

尝试使用属性的instat函数；

Cause the designer wants to set all accesseble properties of your custom control. This is done because of the Properties-Window in Visual Studio in the Designer.

Try using functions instat of properties;

public DataGridViewColumn GetGridcolumn()
{
   // ...
}

public void SetGridcolumn(DataGridViewColumn column)
{
   // ...
}

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