C# Get control by name

Question

I have a ToolStripMenuItem called "myMenu"

How can i access this like so:

/* normally i would do: */
this.myMenu... etc

/* but how do i access it like this: */
String name = myMenu;
this.name...

This is because i am dynamically generating ToolStripMenuItems from an xml file and need to reference menuitems by their dynamically generated names

Answer 1

Try this

this.Controls.Find()

Answer 2

Control GetControlByName(string Name)
{
    foreach(Control c in this.Controls)
        if(c.Name == Name)
            return c;

    return null;
}

Disregard this, I reinvent wheels.

Answer 3

Since you're generating them dynamically, keep a map between a string and the menu item, that will allow fast retrieval.

// in class scope
private readonly Dictionary<string, ToolStripMenuItem> _menuItemsByName = new Dictionary<string, ToolStripMenuItem>();

// in your method creating items
ToolStripMenuItem createdItem = ...
_menuItemsByName.Add("<name here>", createdItem);

// to access it
ToolStripMenuItem menuItem = _menuItemsByName["<name here>"];

Answer 4

Have a look at the ToolStrip.Items collection. It even has a find method available.

Answer 5

=]] LOOOOOOOOOOOOOOOOOL

my answer:

string name = "the_name_you_know";

Control ctn = this.Controls[name];

ctn.Text = "Example...";

Answer 6

this.Controls["name"];

This is the actual code that is ran:

public virtual Control this[string key]
{
    get
    {
        if (!string.IsNullOrEmpty(key))
        {
            int index = this.IndexOfKey(key);
            if (this.IsValidIndex(index))
            {
                return this[index];
            }
        }
        return null;
    }
}

vs:

public Control[] Find(string key, bool searchAllChildren)
{
    if (string.IsNullOrEmpty(key))
    {
        throw new ArgumentNullException("key", SR.GetString("FindKeyMayNotBeEmptyOrNull"));
    }
    ArrayList list = this.FindInternal(key, searchAllChildren, this, new ArrayList());
    Control[] array = new Control[list.Count];
    list.CopyTo(array, 0);
    return array;
}

private ArrayList FindInternal(string key, bool searchAllChildren, Control.ControlCollection controlsToLookIn, ArrayList foundControls)
{
    if ((controlsToLookIn == null) || (foundControls == null))
    {
        return null;
    }
    try
    {
        for (int i = 0; i < controlsToLookIn.Count; i++)
        {
            if ((controlsToLookIn[i] != null) && WindowsFormsUtils.SafeCompareStrings(controlsToLookIn[i].Name, key, true))
            {
                foundControls.Add(controlsToLookIn[i]);
            }
        }
        if (!searchAllChildren)
        {
            return foundControls;
        }
        for (int j = 0; j < controlsToLookIn.Count; j++)
        {
            if (((controlsToLookIn[j] != null) && (controlsToLookIn[j].Controls != null)) && (controlsToLookIn[j].Controls.Count > 0))
            {
                foundControls = this.FindInternal(key, searchAllChildren, controlsToLookIn[j].Controls, foundControls);
            }
        }
    }
    catch (Exception exception)
    {
        if (ClientUtils.IsSecurityOrCriticalException(exception))
        {
            throw;
        }
    }
    return foundControls;
}

Answer 7

this.Controls.Find(name, searchAllChildren) doesn't find ToolStripItem because ToolStripItem is not a Control

  using SWF = System.Windows.Forms;
  using NUF = NUnit.Framework;
  namespace workshop.findControlTest {
     [NUF.TestFixture]
     public class FormTest {
        [NUF.Test]public void Find_menu() {
           // == prepare ==
           var form = new SWF.Form();
           var menuStrip = new SWF.MenuStrip();
           var fileTool = new SWF.ToolStripMenuItem();
           fileTool.Name = "fileTool";
           fileTool.Text = "File";
           menuStrip.Items.Add(fileTool);
           form.Controls.Add(menuStrip);

           // == execute ==
           var ctrl = form.Controls.Find("fileTool", true);

           // == not found! ==
           NUF.Assert.That(ctrl.Length, NUF.Is.EqualTo(0)); 
        }
     }
  }