Is it possible home widget for the custom menu?
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Added a new menu item to the home menu.
Normally, the action in the settings is onclick, but I want to show the content of the new menu item as a widget when the home menu is focused, like the Kodi default skin.
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Here is my code
xml:
<control type="group" id="4000">
<visible>String.IsEqual(Container(9000).ListItem.Property(kenProperty),ken)</visible>
<include content="WidgetListPoster" condition="Library.HasContent(movies)">
<param name="content_path" value="

?"/> <- I don't know!!!
<param name="widget_header" value="KEN-TEST1"/>
<param name="widget_target" value="videos"/>
<param name="list_id" value="4100"/>
</include>
When I add new main menu, it generates automatically @script-skinshortcuts-includes.xml
xml:
<item id="4">
<property name="id">$NUMBER[4]</property>
<label>KEN-1</label>
<label2>VIDEO ADDON</label2>
<icon>special://skin/extras/icons/addtile.png</icon>
<thumb>D:\KODI\user_movie_icon.png</thumb>
<property name="labelID">ken-1</property>
<property name="defaultID" />
<property name="kenProperty">ken</property>
<property name="translatedPath">ActivateWindow(Videos,"library://video/Plex-4/4_all.xml",return)</property>
<onclick>ActivateWindow(Videos,"library://video/Plex-4/4_all.xml",return)</onclick>
<property name="path">ActivateWindow(Videos,"library://video/Plex-4/4_all.xml",return)</property>
<property name="list">library://video/Plex-4/4_all.xml</property>
<visible>String.IsEqual(System.ProfileName,Master user)</visible>
<property name="submenuVisibility">ken-1</property>
<property name="group">mainmenu</property>
<property name="hasSubmenu">False</property>
</item>
I don't know how to get contents path "<property name="list">library://video/Plex-4/4_all.xml</property> from <item id="4"> and put <param name="content_path" value="?"/> into
Can someone let me know?