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Home/ Questions/Q 6850909
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T01:10:57+00:00 2026-05-27T01:10:57+00:00

In my application , i m applying an animation on layout. This animation put

  • 0

In my application , i m applying an animation on layout. This animation put the layout off the screen and put it in from the other side :

private void slideAnimation(final int sens)
{

    Animation animOut = null;
    if(sens == -1) {
        animOut = AnimationUtils.loadAnimation(getApplicationContext(), R.anim.slide_out_left);
    } else {
        animOut = AnimationUtils.loadAnimation(getApplicationContext(), R.anim.slide_out_right);
    }

    animOut.setAnimationListener(new AnimationListener() {

        @Override
        public void onAnimationStart(Animation animation) {
            // TODO Auto-generated method stub

        }

        @Override
        public void onAnimationRepeat(Animation animation) {
            // TODO Auto-generated method stub

        }

        @Override
        public void onAnimationEnd(Animation animation) {
            Animation animIn = null;
            if(sens == -1) {
                animIn = AnimationUtils.loadAnimation(getApplicationContext(), R.anim.slide_in_right);
            } else {
                animIn = AnimationUtils.loadAnimation(getApplicationContext(), R.anim.slide_in_left);
            }
            camLayout.startAnimation(animIn);
        }
    });

    camLayout.startAnimation(animOut);
}

When needed i simply call slideAnimation().

It’s working fine , but sometime we can see animation in a runnable. Should i consider using an other solution to slide out and in my layout or my code is OK ?

Thanks

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T01:10:57+00:00Added an answer on May 27, 2026 at 1:10 am

    seems ok to me! I gather you are just trying to schedule one animation to play after the another. And this code will do the job. I am not sure but i suppose the animation end listener does already listen from a different thread. So i don’t think a new thread is required.

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