java 如何检查直线和矩形之间的交点？
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How to check intersection between a line and a rectangle?
提问by Code Doggo
The title says it all, Ive been searching around and couldnt find anything that was straight and to the point. How would I take a line with points (x1,y1) & (x2, y2) and check its intersection between a rectangle (xR,yR)? I saw in the Line2D package that there were some intersection methods but not sure how to set it all up. Can someone show me a correct way of setting it up to check for an intersection (collision)?
标题说明了一切，我一直在四处寻找，找不到任何直截了当的内容。我如何取一条带点 (x1,y1) & (x2, y2) 的线并检查它在矩形 (xR,yR) 之间的交点？我在 Line2D 包中看到有一些交集方法，但不确定如何设置。有人可以告诉我设置它以检查交叉路口（碰撞）的正确方法吗？
采纳答案by MadProgrammer
Using the available classes from the 2D Graphics API.
使用 2D Graphics API 中的可用类。
Rectangle r1 = new Rectangle(100, 100, 100, 100);
Line2D l1 = new Line2D.Float(0, 200, 200, 0);
System.out.println("l1.intsects(r1) = " + l1.intersects(r1));
What this doesn't tell you, is where...
这并没有告诉你，是哪里...
回答by loopkin
A rectangle is 4 lines. You could compute the intersect between your line and the 4 lines of the rectangle.
一个矩形是 4 条线。您可以计算您的线与矩形的 4 条线之间的交点。
given the equations of two lines, they would intersect when x and y are equal.
给定两条线的方程，当 x 和 y 相等时，它们会相交。
y = m1x + b1 y = m2x + b2
y = m1x + b1 y = m2x + b2
solving the equation you should get:
解方程你应该得到：
x = b2 - b1 / (m1 - m2);
x = b2 - b1 / (m1 - m2);
Note that if m1 == m2, the lines are parallel and will never intersect, watch out for the divided by 0 in this case.
请注意，如果 m1 == m2，则这些线是平行的并且永远不会相交，在这种情况下要注意除以 0。
Then, since you are dealing with segments ratter than infinite lines, check if the intersect falls off within your segments (check if both X and Y are within each segment's boundaries).
然后，由于您处理的线段比无限线更粗，请检查线段内的相交是否脱落（检查 X 和 Y 是否都在每个线段的边界内）。
回答by user2875504
Returns null if lines do not intersect. Modified some c code from another response to similar question to make it Java. Haven't bothered to look into how/why it works, but does the job I needed it to.
如果线不相交，则返回 null。从另一个对类似问题的回答中修改了一些 c 代码，使其成为 Java。没有费心去研究它是如何/为什么工作的，但是我需要它做的工作。
static Point get_line_intersection(Line2D.Double pLine1, Line2D.Double pLine2)
{
Point
result = null;
double
s1_x = pLine1.x2 - pLine1.x1,
s1_y = pLine1.y2 - pLine1.y1,
s2_x = pLine2.x2 - pLine2.x1,
s2_y = pLine2.y2 - pLine2.y1,
s = (-s1_y * (pLine1.x1 - pLine2.x1) + s1_x * (pLine1.y1 - pLine2.y1)) / (-s2_x * s1_y + s1_x * s2_y),
t = ( s2_x * (pLine1.y1 - pLine2.y1) - s2_y * (pLine1.x1 - pLine2.x1)) / (-s2_x * s1_y + s1_x * s2_y);
if (s >= 0 && s <= 1 && t >= 0 && t <= 1)
{
// Collision detected
result = new Point(
(int) (pLine1.x1 + (t * s1_x)),
(int) (pLine1.y1 + (t * s1_y)));
} // end if
return result;
}
