Open view in external window..

Question

Hi,
try this:
define a new window
and at this to the link action:
//Create a new Window and open it
IWDWindowInfo windowInfo = (IWDWindowInfo)wdComponentAPI.getComponentInfo().findInWindows("MyNewWindow");
IWDWindow window = wdComponentAPI.getWindowManager().createWindow( windowInfo, true);
window.setWindowPosition(x, y);
window.open();
// Save WindowInstance in Context
wdContext.nodePopup().bind(wdContext.nodePopup().createPopupElement());
wdContext.currentPopupElement().setWindowInstance(window);
Regards
Sebastian

Answer 1

>
Husalban RM wrote:
> Hi all,
>
> This is the scenario:
>
> In my WD i have a initial view with a selection criteria for search documents (like POWL) when you select one and push the button 'Display' I start a navigation to another view to display the details...
>
> Now the client wants to change that. When the user push the button 'Display' a new external window must be opened.
>
> With the method CREATE_EXTERNAL_WINDOW of the class IF_WD_WINDOW_MANAGER i need an URL that i haven't for call this view.
>
> Then, is necessary to create a new Webdynpro Application in my component in order to do it? Or is there any other way to do it?
>
> Thanks in advance.
Hi,
Should  it be absolutely a external window ?
You can use a create_window method of the IF_WD_WINDOW_MANAGER to show it on a popup window.