Property als Hyperlink darstellen in einem GridList (WIN) 1


Um ein Property als Hyperlink anzuzeigen erstellen Sie folgenden Window Controller im Module.Win Projekt unter Controllers. In Verbindung mit dem HyperLinkPropertyEditor ergibt das eine perfekte Kombination.

using System;
using System.Drawing;
using DevExpress.ExpressApp;
using DevExpress.XtraEditors;
using DevExpress.XtraGrid.Views.Grid;
using DevExpress.XtraEditors.Repository;
using DevExpress.ExpressApp.Win.Editors;
using DevExpress.XtraGrid.Views.Grid.ViewInfo;

namespace HyperLinkPropertyEditor.Win
{
public class HyperLinkGridListViewController : ViewController
{
private GridListEditor gridListEditor;
public HyperLinkGridListViewController()
{
TargetViewType = ViewType.ListView;
}
protected override void OnViewControlsCreated()
{
base.OnViewControlsCreated();
gridListEditor = ((ListView)View).Editor as GridListEditor;
if (gridListEditor != null)
gridListEditor.GridView.MouseDown += GridView_MouseDown;
}
protected override void OnDeactivated()
{
if (gridListEditor != null)
gridListEditor.GridView.MouseDown -= GridView_MouseDown;
base.OnDeactivated();
}
private void GridView_MouseDown(object sender, System.Windows.Forms.MouseEventArgs e)
{
GridView gv = (GridView)sender;
GridHitInfo hi = gv.CalcHitInfo(new Point(e.X, e.Y));
if (hi.InRowCell)
{
RepositoryItemHyperLinkEdit repositoryItemHyperLinkEdit = hi.Column.ColumnEdit as RepositoryItemHyperLinkEdit;
if (repositoryItemHyperLinkEdit != null)
{
HyperLinkEdit editor = (HyperLinkEdit)repositoryItemHyperLinkEdit.CreateEditor();
editor.ShowBrowser(HyperLinkPropertyEditor.GetResolvedUrl(gv.GetRowCellValue(hi.RowHandle, hi.Column)));
}
}
}
}
}

Schreibe einen Kommentar

Ein Gedanke zu “Property als Hyperlink darstellen in einem GridList (WIN)