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Create designer menu ???
Posted: Tue Jul 03, 2007 12:09 pm
by jayakumargr
Hai,
is there any possibilty to create designer menu in windows form ? if possible then how to create....???
(that is,i add menu names in my own format.and those menus are performing desinger menu functionalities)
Thanks in advance,
Jayakumar
Create designer menu ???
Posted: Wed Jul 04, 2007 8:07 am
by Guest
Use follow code for your button to access menu functionalities, for example, for create new report:
Code: Select all
StiMainMenuService mainmenu = StiMainMenuService.GetService(stiDesignerControl1);
mainmenu.InvokeReportNew();
where stiDesignerControl1 your StiDesignerControl placed to windows form
For create new page:
Code: Select all
StiMainMenuService mainmenu = StiMainMenuService.GetService(stiDesignerControl1);
mainmenu.InvokePageNew();
and etc.
Thank you.
Create designer menu ???
Posted: Wed Jul 04, 2007 9:12 am
by jayakumargr
Hai,
Thanks for your reply.it works fine.
i want to know,how to open dictionay window,report tree and properties window at run time ???
Please provide the solutions...
Thanks in advance,
Jayakumar
Create designer menu ???
Posted: Wed Jul 04, 2007 10:19 am
by Guest
For Property Grid use the following code:
Code: Select all
StiPropertiesPanelService properties = new StiPropertiesPanelService();
properties.Init(stiDesignerControl1);
For Report Tree:
Code: Select all
StiReportTreePanelService tree = new StiReportTreePanelService();
tree.Init(stiDesignerControl1);
For Dictionary:
Code: Select all
StiDictionaryPanelService dictionary = new StiDictionaryPanelService();
dictionary.Init(stiDesignerControl1);
Thank you.
Create designer menu ???
Posted: Wed Jul 04, 2007 11:32 am
by jayakumargr
Hai,
Thanks. Which namespace is used for StiPropertiesPanelService,StiReportTreePanelService and StiDictionaryPanelService ???
Thanks in Advance,
Jayakumar
Create designer menu ???
Posted: Wed Jul 04, 2007 1:19 pm
by Brendan
Create designer menu ???
Posted: Wed Jul 04, 2007 1:30 pm
by jayakumargr
Hai,
i wrote following,
StiReport rpt = new StiReport();
StiDesigner design = new StiDesigner(rpt);
StiReportTreePanelService tree = new StiReportTreePanelService();
tree.Init(design);
but it doesn't work...
How to show the Reporttree window ??
Thanks,
Jayakumar
Create designer menu ???
Posted: Thu Jul 05, 2007 3:40 am
by Guest
You must place StiDesignerControl on your windows form and write:
Code: Select all
StiReport rpt = new StiReport();
stiDesignerControl1.Report = rpt;
StiReportTreePanelService tree = new StiReportTreePanelService();
tree.Init(stiDesignerControl1);
Thank you.
Create designer menu ???
Posted: Thu Jul 05, 2007 12:06 pm
by jayakumargr
Hai,
i want to open report tree for existing(already opened designer) report designer. it is same as like designer view menu operation.
View --> Report Tree(ctrl+shift+l)
How to open report tree...???
Thanks in advance,
Jaykumar
Create designer menu ???
Posted: Mon Jul 09, 2007 8:57 am
by Guest
Use the following code:
Code: Select all
StiReportTreePanelService tree = StiReportTreePanelService.GetService(design);
StiPanelService service = (tree as StiPanelService);
if (service.DockingControl.IsVisible)
{
((StiDockingPanel)service.DockingControl.Parent).DoClose(service.DockingControl);
}
else
{
design.DockingManager.ClosedControls.Remove(service.DockingControl);
StiDockingPanel dockingPanel = new StiDockingPanel(design.DockingManager);
dockingPanel.Controls.Add(service.DockingControl);
dockingPanel.DockPanelToForm(design, service.DockStyle);
}
Of course, you may write
or
instead
Thank you.