touchDragged 如何在 libgdx 中工作?

How does touchDragged work in libgdx?(touchDragged 如何在 libgdx 中工作?)

问题描述

我目前正在学习 libgdx 游戏编程，现在我已经学会了如何使用 touchDown，但我不知道如何使用 touchDragged.计算机如何知道手指被拖动的方向(用户是否向左拖动或对)

I am currently learning libgdx game programming,now i have learnt how to use touchDown but iam not getting idea how to use touchDragged.How will the computer knows in which direction the finger is dragged(whether the user has dragged to left or right)

推荐答案

计算机不知道.或者至少界面不会告诉你这些信息.它看起来像这样:

The computer doesn't know that. Or at least the interface won't tell you this information. It looks like this:

public boolean touchDragged(int screenX, int screenY, int pointer);

和touchDown差不多:

It is nearly the same like touchDown:

public boolean touchDown(int screenX, int screenY, int pointer, int button);

touchDown 事件发生后，只有 touchDragged 事件会发生(对于同一个指针)，直到 touchUp 事件被触发.如果您想知道指针移动的方向，您必须通过计算最后一个接触点和当前接触点之间的增量(差值)来自己计算.可能看起来像这样:

After a touchDown event happened, only touchDragged events will occur (for the same pointer) until a touchUp event gets fired. If you want to know the direction in which the pointer moved, you have to calculate it yourself by computing the delta (difference) between the last touchpoint and the current one. That might look like this:

private Vector2 lastTouch = new Vector2();

public boolean touchDown(int screenX, int screenY, int pointer, int button) {
    lastTouch.set(screenX, screenY);
}

public boolean touchDragged(int screenX, int screenY, int pointer) {
    Vector2 newTouch = new Vector2(screenX, screenY);
    // delta will now hold the difference between the last and the current touch positions
    // delta.x > 0 means the touch moved to the right, delta.x < 0 means a move to the left
    Vector2 delta = newTouch.cpy().sub(lastTouch);
    lastTouch = newTouch;
}

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