How to create custom PropertyGrid editor item which opens a form?

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I have a List<> (my custom class). I want to display a specific item in this list in a box on the PropertyGrid control. At the end of the box I would like the [...] button. When clicked, it would open up a form which, among other things, would allow them to pick one of the items from the List. When closed, the PropertyGrid would be updated to reflect the selected value.

Any help appreciated.

propertygrid winforms c#
1 Answers

You need to implement a modal UITypeEditor, using the IWindowsFormsEditorService service to display it:

using System.ComponentModel;
using System.Drawing.Design;
using System.Windows.Forms;
using System.Windows.Forms.Design;
using System;

class MyType
{
    private Foo foo = new Foo();
    public Foo Foo { get { return foo; } }
}

[Editor(typeof(FooEditor), typeof(UITypeEditor))]
[TypeConverter(typeof(ExpandableObjectConverter))]
class Foo
{
    private string bar;
    public string Bar
    {
        get { return bar; }
        set { bar = value; }
    }
}
class FooEditor : UITypeEditor
{
    public override UITypeEditorEditStyle GetEditStyle(ITypeDescriptorContext context)
    {
        return UITypeEditorEditStyle.Modal;
    }
    public override object EditValue(ITypeDescriptorContext context, System.IServiceProvider provider, object value)
    {
        IWindowsFormsEditorService svc = provider.GetService(typeof(IWindowsFormsEditorService)) as IWindowsFormsEditorService;
        Foo foo = value as Foo;
        if (svc != null && foo != null)
        {            
            using (FooForm form = new FooForm())
            {
                form.Value = foo.Bar;
                if (svc.ShowDialog(form) == DialogResult.OK)
                {
                    foo.Bar = form.Value; // update object
                }
            }
        }
        return value; // can also replace the wrapper object here
    }
}
class FooForm : Form
{
    private TextBox textbox;
    private Button okButton;
    public FooForm() {
        textbox = new TextBox();
        Controls.Add(textbox);
        okButton = new Button();
        okButton.Text = "OK";
        okButton.Dock = DockStyle.Bottom;
        okButton.DialogResult = DialogResult.OK;
        Controls.Add(okButton);
    }
    public string Value
    {
        get { return textbox.Text; }
        set { textbox.Text = value; }
    }
}
static class Program
{
    [STAThread]
    static void Main()
    {
        Application.EnableVisualStyles();
        Form form = new Form();
        PropertyGrid grid = new PropertyGrid();
        grid.Dock = DockStyle.Fill;
        form.Controls.Add(grid);
        grid.SelectedObject = new MyType();
        Application.Run(form);
    }
}

Note: if you need to access something about the context of the property (the parent object etc), that is what the ITypeDescriptorContext (in EditValue) provides; it tells you the PropertyDescriptor and Instance (the MyType) that is involved.

Marc_Gravell_233540
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