我希望用户能够在 Canvas 周围拖动正方形的边缘。使用我当前的解决方案它可以工作但有故障,有时无法选择边缘。是否有一种干净的方法来判断一条线是否已被单击(例如,通过坐标)?这就是我目前正在测试的方式:
// check edge pressed, edge is the line between to
// coords e.g. (i) & (i = 1)
for (int i = 0; i < coords.size(); i++) {
p1 = coords.get(i);
if ((i + 1) > (coords.size() - 1)) p2 = coords.get(0);
else p2 = coords.get(i + 1);
// is this the line pressed
if (p1.x <= event.getX() + 5 && event.getX() - 5 <= p2.x && p1.y <= event.getY() + 5 && event.getY() - 5 <= p2.y) {
// points found, set to non temp
// variable for use in ACTION_MOVE
point1 = p1;
point2 = p2;
break;
} else if (p1.x >= event.getX() + 5 && event.getX() - 5 >= p2.x && p1.y >= event.getY() + 5 && event.getY() - 5 >= p2.y) {
// points found, set to non temp
// variable for use in ACTION_MOVE
point1 = p1;
point2 = p2;
break;
}
}
下面的代码//这是按下的行是最重要的,也是最有可能的问题。 +5 和 -5 用于为用户提供更大的点击区域。
这里是整个点击事件:
public void EditEdge() {
//TODO this works like shit
// Detect the two coordinates along the edge pressed and drag
// them
scene.setOnTouchListener(new View.OnTouchListener() {
private int startX;
private int startY;
private Point point1 = new Point(0, 0);
private Point point2 = new Point(0, 0);
@Override
public boolean onTouch(View v, MotionEvent event) {
switch (event.getAction()) {
case MotionEvent.ACTION_DOWN:
startX = (int) event.getX();
startY = (int) event.getY();
Point p1;
Point p2;
// check edge pressed, edge is the line between to
// coords e.g. (i) & (i = 1)
for (int i = 0; i < coords.size(); i++) {
p1 = coords.get(i);
if ((i + 1) > (coords.size() - 1)) p2 = coords.get(0);
else p2 = coords.get(i + 1);
// is this the line pressed
if (p1.x <= event.getX() + 5 && event.getX() - 5 <= p2.x && p1.y <= event.getY() + 5 && event.getY() - 5 <= p2.y) {
// points found, set to non temp
// variable for use in ACTION_MOVE
point1 = p1;
point2 = p2;
break;
} else if (p1.x >= event.getX() + 5 && event.getX() - 5 >= p2.x && p1.y >= event.getY() + 5 && event.getY() - 5 >= p2.y) {
// points found, set to non temp
// variable for use in ACTION_MOVE
point1 = p1;
point2 = p2;
break;
}
}
break;
case MotionEvent.ACTION_UP:
point1 = new Point(0, 0);
point2 = new Point(0, 0);
// scene.setOnTouchListener(scene.editModeOnTouchListener);
break;
case MotionEvent.ACTION_MOVE:
for (Point p: new Point[] {
point1, point2
}) {
int modX = (int)(p.x + (event.getX() - startX));
int modY = (int)(p.y + (event.getY() - startY));
p.set(modX, modY);
}
SetCoords(coords);
startX = (int) event.getX();
startY = (int) event.getY();
break;
default:
return false;
}
return true;
}
});
}
那么有没有更简单的方法来判断一条线是否被点击/穿过一个点或者这不是问题?
谢谢
最佳答案
使用线方程 y = mx + b
找出点是否在一条线上
float EPSILON = 0.001f;
public boolean isPointOnLine(Point linePointA, Point linePointB, Point point) {
float m = (linePointB.y - linePointA.y) / (linePointB.x - linePointA.x);
float b = linePointA.y - m * linePointA.x;
return Math.abs(point.y - (m*point.x+b)) < EPSILON);
}
关于android - 一条线是否包含一个点,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20887806/