Libgdx : setting another screen, but still buttons from old screen active

问题

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.

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?

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

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

回答1:

I changed this :

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);

}

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

回答2:

The answer above adiddnt work for me because I dont work with a InputProcessor in the 2nd screen.

My Solution was to set the Inputprocessor to null after setScreen.

Gdx.app.getApplicationListener().setScreen(new Screen()); Gdx.input.setInputProcessor(null);

回答3:

Just broke my head over this. I changed screens in the show method but had Gdx.input.setInputProcessor(stage) called after that. The method still completed after the show method of the new screen ran and thus reverting back to the previous stage.

So make sure you call Gdx.input.setInputProcessor(stage) before you change screens or return after you change screens.

来源：https://stackoverflow.com/questions/18170274/libgdx-setting-another-screen-but-still-buttons-from-old-screen-active