Libgdx：设置另一个屏幕，但仍从旧的屏幕按键活跃 [英] Libgdx : setting another screen, but still buttons from old screen active

问题描述

在我libgdx 游戏，我有2个屏幕，菜单和列表。
当我点击菜单屏幕在标签上，我做了 setscreen（名单） 。
新的屏幕显示出来，并且在菜单画面非常久远的标签消失。

In my libgdx game, I have 2 screens, menu and list. When I click on a label in the menu screen, I do a setscreen(list). The new screen shows up, and the menu screen alongwith its labels disappears.

但是，当我在相同的位置单击（从菜单屏幕，标签分别为，但当然这些标签都没有显示，因为我已经改变屏幕）单击事件作出响应。为什么呢？

But when I click on the same positions (from menu screen where the labels were, but of course those labels are not showing as I have changed screens) the click event responds. Why?

请注意：我的列表屏幕目前没有任何控件事件处理程序。

note:My list screen currently has not event handlers for any widget.

当切换画面，我需要做任何事情，不仅仅是 setscreen（anotherscreen）停用oldscreen？

When switching screens do I need to do anything more than just setscreen(anotherscreen) to deactivate oldscreen?

推荐答案

我改变了这个：

我使用的屏幕级可变移动输入处理器到屏幕的show（）方法

I moved the input processor to the show() method of that screen using stage variable of that screen

    public void show() {
    ...
Gdx.input.setInputProcessor(stage);

    }

我只是在屏幕的构造器设置这个，所以我就算是改变屏幕，输入处理器仍然附着在最后创建屏幕的阶段之前

before I was setting this only in the constructor of the screen, so even If I was changing the screen, the input processor was still attached to the last created screen's stage

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