Get position on canvas AFTER rotating/translating and restoring

问题

Okay, this is getting complicated...

Given situation:
I have a canvas with the dimensions of 800x600.
My mouse is at canvas position 100x200 (for example).

I save my canvas state.
Now I rotate and translate the canvas, I draw a square.
I restore my canvas state.

Is there any way to determine if my mouse is over the square?
I think I would have to translate/rotate my mouse position as well - in the opposite direction, but how would I do this?

回答1:

You can get hold of an objects world position/rotation by recursively applying this formula:

worldX = parentX + x * Math.cos( parentR ) - y * Math.sin( parentR );
worldY = parentY + x * Math.sin( parentR ) + y * Math.cos( parentR );
worldR = parentR + r;

A javascript implementation would be:

var deg2rad, rad2deg, getXYR;

deg2rad = function ( d ) { return d * Math.PI / 180 };
rad2deg = function ( r ) { return r / Math.PI * 180 };

getXYR = function ( node ) {
  var x, y, r,
      parentXYR, pX, pY, pR,
      nX, nY;

  x = y = r = 0;

  if ( node ) {
    parentXYR = getXYR( node.parent );
    pX = parentXYR.x;
    pY = parentXYR.y;
    pR = deg2rad( parentXYR.r );
    nX = node.x;
    nY = node.y;

    x = pX + nX * Math.cos( pR ) - nY * Math.sin( pR );
    y = pY + nX * Math.sin( pR ) + nY * Math.cos( pR );
    r = rad2deg( pR + deg2rad( node.r ) );
  }

  return { x:x, y:y, r:r };
};

Try it out with these objects:

el1 = {x:3,y:0,r:45};
el2 = {x:0,y:0,r:45};
el1.parent = el2;
getXYR(el1);

It won't be long before you want to calculate the shortest angle between two objects, if you get the deltaX (x2-x1) and deltaY (y2-y1) between the two objects you can get the angle with this function:

var getAngle = function ( dx, dy ) {
  var r = Math.atan2( dy, dx ) * 180 / Math.PI;
  return ( r > 180 )  ? r-360 :
         ( r < -180 ) ? r+360 :
         r;
}

In the long run it's better to learn using matrices instead. The equivalence of getting the world pos/rot is a world matrix. Here's some good info about matrices (in the SVG doc, but it's not relevant): http://www.w3.org/TR/SVG/coords.html#NestedTransformations

This is how you would do it with matrices (and the gl-matrix lib https://github.com/toji/gl-matrix):

var getWorldMatrix = function ( node ) {
    var parentMatrix;

    if ( !node )
        return mat4.identity();

    parentMatrix = getWorldMatrix( node.parent );
    return mat4.multiply( parentMatrix, node.matrix );
};

Oh, i forgot, now to finally register a click you just get the screen coordinates of the mouse and compare them to the objects position + the canvas viewport offset.

回答2:

Yes you have to either translate the mouse coordinates or retain a second set of coordinates for your shape. I recommend keeping a second set of coordinates as you will move the mouse more times than you transform the object. Try using an object like so

Box

function Box(x, y, w, h){
    this.x = x;
    this.y = y;
    this.tx = x;    //transformed x
    this.ty = y;    //transformed y
    this.w = w;
    this.h = h;

    this.mouseover = function(x, y){
        if (this.tx < x && this.tx + this.w > x && this.ty < y && this.ty + this.h > y){
            return true;
        }
        return false;
    }

    this.applyTransformation = function(transformation){
        switch(transformation){
        case 'rotation':
            //update tx/ty to match rotation
            break;
        case 'translation':
            //update tx/ty to match translation
            break;
       default:
            //do nothing or raise exception
    }
}

来源：https://stackoverflow.com/questions/8844525/get-position-on-canvas-after-rotating-translating-and-restoring