如何从 NotesXSPViewEntry 获取父 NotesViewEntry? [英] How do I get the parent NotesViewEntry from the NotesXSPViewEntry?

问题描述

使用 SSJS 我有一个 NotesXSPViewEntry 的句柄，它是视图中的一个类别，现在我需要从该条目中获取父类别作为 NotesViewEntry，所以我尝试使用 NotesXSPViewEntry.getParent() 但它返回一个 notesViewNavigator一个 NotesViewEntry.

Using SSJS I have a handle to a NotesXSPViewEntry which is a category in view,now I need to get the parent category from that entry as a NotesViewEntry, so I tried to use NotesXSPViewEntry.getParent() but that returns a notesViewNavigator instead of a NotesViewEntry.

如何从 notesXSPViewEntry 获取父 NotesViewEntry?或者如何将 notesXSPViewEntry 转换为 NotesViewEntry?

How do I get the parent NotesViewEntry from the notesXSPViewEntry? or how do I convert a notesXSPViewEntry to a NotesViewEntry?

推荐答案

一个解决方案:使用 getNavigatorPosition() 获取位置.使用 getParent() 获取 NotesViewNavigator，使用该位置作为参数调用 gotoPos()，然后 getPrev() 应返回上一个条目和 getParent() 父条目.

One solution: get the position with getNavigatorPosition(). The get the NotesViewNavigator with getParent(), call gotoPos() with that position as parameter, then getPrev() should return the previous entry and getParent() the parent entry.

您也许可以不使用 gotoPos() 调用:getParent().getCurrent().getParent() 应该返回 XSPViewEntry 的导航器的当前条目的父条目.

You might be able to leave the gotoPos() call out: getParent().getCurrent().getParent() should return XSPViewEntry's navigator's current entry's parent entry.

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