在 C# 中按名称获取 Windows 窗体控件

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

Use the Control.ControlCollection.Find method.

Try this:

this.Controls.Find()

Answer 2

string name = "the_name_you_know";

Control ctn = this.Controls[name];

ctn.Text = "Example...";

Answer 3

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

    return null;
}

Disregard this, I reinvent wheels.

Answer 4

Assuming you have the menuStrip object and the menu is only one level deep, use:

ToolStripMenuItem item = menuStrip.Items
    .OfType<ToolStripMenuItem>()
    .SelectMany(it => it.DropDownItems.OfType<ToolStripMenuItem>())
    .SingleOrDefault(n => n.Name == "MyMenu");

For deeper menu levels add more SelectMany operators in the statement.

if you want to search all menu items in the strip then use

ToolStripMenuItem item = menuStrip.Items
    .Find("MyMenu",true)
    .OfType<ToolStripMenuItem>()
    .Single();

However, make sure each menu has a different name to avoid exception thrown by key duplicates.

To avoid exceptions you could use FirstOrDefault instead of SingleOrDefault / Single , or just return a sequence if you might have Name duplicates.

Answer 5

Using the same approach of Philip Wallace , we can do like this:

    public Control GetControlByName(Control ParentCntl, string NameToSearch)
    {
        if (ParentCntl.Name == NameToSearch)
            return ParentCntl;

        foreach (Control ChildCntl in ParentCntl.Controls)
        {
            Control ResultCntl = GetControlByName(ChildCntl, NameToSearch);
            if (ResultCntl != null)
                return ResultCntl;
        }
        return null;
    }

Example:

    public void doSomething() 
    {
            TextBox myTextBox = (TextBox) this.GetControlByName(this, "mytextboxname");
            myTextBox.Text = "Hello!";
    }

I hope it help! :)

Answer 6

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 fileTool = new SWF.ToolStripMenuItem();
           fileTool.Name = "fileTool";
           fileTool.Text = "File";

           var menuStrip = new SWF.MenuStrip();
           menuStrip.Items.Add(fileTool);

           var form = new SWF.Form();
           form.Controls.Add(menuStrip);

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

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

Answer 7

One of the best way is a single row of code like this:

In this example we search all PictureBox by name in a form

PictureBox[] picSample = 
                    (PictureBox)this.Controls.Find(PIC_SAMPLE_NAME, true);

Most important is the second paramenter of find .

if you are certain that the control name exists you can directly use it:

  PictureBox picSample = 
                        (PictureBox)this.Controls.Find(PIC_SAMPLE_NAME, true)[0];

Answer 8

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 9

You can use find function in your Form class. If you want to cast (Label) ,(TextView) ... etc, in this way you can use special features of objects. It will be return Label object.

(Label)this.Controls.Find(name,true)[0];

name : item name of searched item in the form

true : Search all Children boolean value

Answer 10

Assuming you have Windows.Form Form1 as the parent form which owns the menu you've created. One of the form's attributes is named .Menu . If the menu was created programmatically, it should be the same, and it would be recognized as a menu and placed in the Menu attribute of the Form.

In this case, I had a main menu called File . A sub menu, called a MenuItem under File contained the tag Open and was named menu_File_Open . The following worked. Assuming you

// So you don't have to fully reference the objects.
using System.Windows.Forms;

// More stuff before the real code line, but irrelevant to this discussion.

MenuItem my_menuItem = (MenuItem)Form1.Menu.MenuItems["menu_File_Open"];

// Now you can do what you like with my_menuItem;

Answer 11

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 12

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

Answer 13

You can do the following:

private ToolStripMenuItem getToolStripMenuItemByName(string nameParam)
   {
      foreach (Control ctn in this.Controls)
         {
            if (ctn is ToolStripMenuItem)
               {
                   if (ctn.Name = nameParam)
                      {
                         return ctn;
                      }
                }
         }
         return null;
    }

Answer 14

A simple solution would be to iterate through the Controls list in a foreach loop. Something like this:

foreach (Control child in Controls)
{
    // Code that executes for each control.
}

So now you have your iterator, child , which is of type Control . Now do what you will with that, personally I found this in a project I did a while ago in which it added an event for this control, like this:

child.MouseDown += new MouseEventHandler(dragDown);