XtraInputBox.Show<T>(XtraInputBoxArgs) Method
Displays an input box with the specified settings.
Namespace: DevExpress.XtraEditors
Assembly: DevExpress.XtraEditors.v24.1.dll
NuGet Package: DevExpress.Win.Navigation
Declaration
Parameters
Name | Type | Description |
---|---|---|
args | XtraInputBoxArgs | An object that specifies the input box settings. |
Type Parameters
Name | Description |
---|---|
T | The returned value’s type. |
Returns
Type | Description |
---|---|
Nullable<T> | The value entered by a user, or |
Example
The code snippet below shows how to display a dialog box with custom settings.
using DevExpress.XtraEditors;
// Display an iput box with the specifiedx prompt, title, and default response.
XtraInputBox.Show("Enter a new value", "Change Settings", "Default");
XtraInputBox.Show(this, "Enter a new value", "Change Settings", "Default");
// Initialize a new XtraInputBoxArgs instance.
XtraInputBoxArgs args = new XtraInputBoxArgs();
// Specify settings.
args.Caption = "Shipping options";
args.Prompt = "Delivery date";
args.DefaultButtonIndex = 0;
// Initialize a new DateEdit instance.
DateEdit editor = new DateEdit();
// Specify settings.
editor.Properties.CalendarView = DevExpress.XtraEditors.Repository.CalendarView.TouchUI;
editor.Properties.Mask.EditMask = "MMMM d, yyyy";
// Assign the editor to the input box settings.
args.Editor = editor;
// Specify the editor's default value.
args.DefaultResponse = DateTime.Now.Date.AddDays(3);
// Display an input box and assign the result to a variable.
var result = XtraInputBox.Show(args);
// You can also use the generic method and specify the return value's type.
DateTime resultDateTime = XtraInputBox.Show<DateTime>(args);
See Also