我不是你的問題完全肯定,但希望這些建議將有助於:
- 您可以在顯示對象上使用localToGlobal()函數將該顯示容器中的旋轉,平移和縮放點返回到舞臺。例如,$ p:Point = myMovieClip.localToGlobal(new Point(10,10));
- A Matrix也是一個很好的和簡單的方法來旋轉一個點。例如,var $ mtx:Matrix = new Matrix(); $ mtx.tx = 10; $ mtx.ty = 10; $ mtx.rotate();現在$ mtx.tx和$ mtx.ty有移位的結果
這些大概不會回答你的問題,但我想我會提到它們,以防萬一,我進入更復雜的東西了。就像wvxvw說的,你不能真正解決你想要做的方程而沒有其他變量。我寫了一些代碼,顯示瞭如何在一個線段比較X到點時,發現Y:
import flash.display.Shape;
import flash.geom.Point;
import flash.display.Graphics;
import flash.events.MouseEvent;
var $s:Shape = new Shape();
addChild($s);
var borderStart:Point = new Point(stage.stageWidth/2, stage.stageHeight/2);
var borderRotation:Number = 45;
var borderLength:Number = 800;
var borderRad:Number = borderRotation * (Math.PI/180);
var borderEnd:Point = new Point(borderStart.x + Math.cos(borderRad) * borderLength, borderStart.y + Math.sin(borderRad) * borderLength);
stage.addEventListener(MouseEvent.MOUSE_MOVE, update);
function update(e:MouseEvent):void{
var $g:Graphics = $s.graphics;
$g.clear();
//Drawing the rotated border
$g.lineStyle(3, 0xff0000, .5);
$g.moveTo(borderStart.x, borderStart.y);
$g.lineTo(borderEnd.x, borderEnd.y);
//Finding if and where mouseX collides with our border
if (stage.mouseX >= Math.min(borderStart.x, borderEnd.x) && stage.mouseX <= Math.max(borderStart.x, borderEnd.x)){
var $x:Number = stage.mouseX;
//SOLVING HERE : Solve collision with X
var $percent:Number = ($x - borderStart.x)/(borderLength * Math.cos(borderRad));
var $y:Number = borderStart.y + Math.sin(borderRad) * borderLength * $percent;
//Drawing to our collision
$g.lineStyle(1, 0xffff00, .6);
$g.moveTo($x, 0);
$g.lineTo($x, $y);
$g.lineStyle(2, 0xffff00, 1);
$g.drawCircle($x, $y, 3);
trace("----\nCollision @\t" + "x: " + $x + "\ty:" + Math.round($y));
}
}
希望這會就如何解決您的特定問題的一些見解。