You can use: f. Controls[name]; Where f is your form variable.
Control c = TheForm("Form1"); Once we have this, we can gain access to ALL the child controls including the children in other container controls on the form. The next method to declare is the one that locates the control to access on the form returned by the TheForm method.
Windows Forms controls are reusable components that encapsulate user interface functionality and are used in client-side, Windows-based applications. Not only does Windows Forms provide many ready-to-use controls, it also provides the infrastructure for developing your own controls.
The Name property can be used at run time to evaluate the object by name rather than type and programmatic name. Because the Name property returns a String type, it can be evaluated in case-style logic statements ( Select statement in Visual Basic, switch statement in Visual C# and Visual C++).
Use the Control.ControlCollection.Find method.
Try this:
this.Controls.Find()
string name = "the_name_you_know";
Control ctn = this.Controls[name];
ctn.Text = "Example...";
Control GetControlByName(string Name)
{
foreach(Control c in this.Controls)
if(c.Name == Name)
return c;
return null;
}
Disregard this, I reinvent wheels.
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.
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));
}
}
}
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