Set the KeyPreview attribute on your form to True, then use the KeyPress event at your form level to detect the Enter key. On detection call whatever code you would have for the "submit" button.
if (e.KeyCode.ToString() == "Return")
{
//do something
}
Simply use
this.Form.DefaultButton = MyButton.UniqueID;
**Put your button id in place of 'MyButton'.
You can designate a button as the "AcceptButton" in the Form's properties and that will catch any "Enter" keypresses on the form and route them to that control.
See How to: Designate a Windows Forms Button as the Accept Button Using the Designer and note the few exceptions it outlines (multi-line text-boxes, etc.)
private void textBox_KeyDown(object sender, KeyEventArgs e)
{
if (e.KeyCode == Keys.Enter)
button.PerformClick();
}
You can subscribe to the KeyUp
event of the TextBox
.
private void txtInput_KeyUp(object sender, KeyEventArgs e)
{
if (e.KeyCode == Keys.Enter)
DoSomething();
}
If you set your Form
's AcceptButton
property to one of the Button
s on the Form
, you'll get that behaviour by default.
Otherwise, set the KeyPreview
property to true
on the Form
and handle its KeyDown
event. You can check for the Enter
key and take the necessary action.
The Form has a KeyPreview property that you can use to intercept the keypress.
As previously stated, set your form's AcceptButton property to one of its buttons AND set the DialogResult property for that button to DialogResult.OK, in order for the caller to know if the dialog was accepted or dismissed.
Source: Stackoverflow.com