How can I get a view display into random position in the screen (adaptable to different screen sizes)

Question

I already have a method to display the button in random position of the screen but it is NOT adaptable to other screen sizes. How can I do it (inside: r.nextInt()).

int buttonHeight;
int buttonWidth;
buttonHeight = button.getHeight();
buttonWidth = button.getWidth();
int xLeft = r.nextInt(480 - buttonHeight);
int yUp = r.nextInt(500 - buttonWidth);
int xRight = r.nextInt(670 + buttonHeight);
int yDown = r.nextInt(1400 + buttonWidth);

button.setX(xLeft);
button.setY(yUp);
button.setX(xRight);
button.setY(yDown);

I just want the random number to be adaptable to every screen size.

Possible solution: https://stackoverflow.com/questions/22144690/how-to-display-a-button-in-random-screen-position — WasabiTea, Jan 08 '19 at 23:28
actually almost worked, the only problem is that it gets off the screen sometimes. I don´t why. — Roberto Chávez, Jan 09 '19 at 00:16

score 0 · Answer 1 · answered Jan 09 '19 at 01:17

Android coordinates are based on the view's top left corner. So the x and y values you feed to the view are actually x and y of the view's top left corner. Try limit the nextInt()'s argument to something like, random.nextInt(maxScreenWidth - buttonWidth) and random.nextInt(maxScreenHeight-buttonHeight) and assign these coordinates to your button's x and y.

How can I get a view display into random position in the screen (adaptable to different screen sizes)

1 Answers1