I am building a Warehouse map in 2D with fabricjs where I am displaying their racking systems as a series of rectangles.
As a first "layer/group", i add all bays as groups containing a rectangle and text, both positioned at the same (x,y). They also have both an angle set to fit their orientation in the space.
As a second "layer/group", i add groups containing a circle and text, representing the bay's number of issues. The (x,y) also fits the bays. This way, all my issues are always on top of the bays and the center of the group fits the rotated corner of the bay.
On the first paint, all is well aligned. Once they're shown on the page, the user can create new issues, so I am trying to position the issue group fitting the original (x,y), but since it can all be panned and zoomed, I am having a hard time positioning it where it should be.
I've been looking at their explanations about transforms, but I can't figure who the boss should be and thinking that having nested groups may also be why I am all mixed up.
By doing:
const gMatrix = matchingBay.group.calcTransformMatrix(false);
const targetToCanvas = fabric.util.transformPoint(matchingBay.aCoords.tl, gMatrix);
I am on the bay, but on the "group" corner, which is not what I am looking for. (x,y) will be one of the corners of the rectangle in the group, that may have been rotated.
By specifying the original (x,y) in this code will get me way off the actual painting zone.
So, my question is, how do I get the transformed (x,y) so I can add my issue group at those coordinates?
[EDIT]
As the suggestion of spring, it made me realize I can use the rotated rectangle's transforms to find its coordinates, so I tried:
const rect = matchingBay.getObjects()[BAY_RECTANGLE_INX];
const gMatrix = rect.calcTransformMatrix();
const targetToCanvas = fabric.util.transformPoint(rect.aCoords.bl, gMatrix);
Bottom left corner is where I wish to add the new Circle. After rotation, the bottom left is now the bottom right. But I am still off. As shown, the red circle should be on the corner. It looks like the rotation has not been applied.
qrDecompose gives me something that seems right:
{angle: -90, scaleX: 1, scaleY: -1, skewX: 0, skewY: 0, translateX: 6099.626027314769, translateY: 4785.016008065199 }
I realized that I was not thinking it the right way. Since I have the rectangle already in hands, I just had to get its own transformation and resolve the corner by my own, the following fixed my issue:
{
[...]
const rect = matchingBay.getObjects()[BAY_RECTANGLE_INX];
const gMatrix = rect.calcTransformMatrix();
const decomposed = fabric.util.qrDecompose(gMatrix);
const trans = fabric.util.transformPoint(new fabric.Point((decomposed.scaleX * -rect.width) / 2, (decomposed.scaleY * rect.height) / 2), gMatrix);
const top = trans.y;
const left = trans.x;
[...]
}
Since the matrix is bringing the center point of the rectangle, I can get the corner by substracting its width and height and then transforming the coordinates to find out where the matrix puts it.
Related
I have a Text in fabricJs. I set top and left.
This sets the aCoords properly to those values.
However the oCoords dont match. And the Text is not displayed at the right position.
I suspect that I need to set to oCoords somehow. So that the Text is displayed at the right pixel coordinates (top & left) on the canvas.
aCoords and oCoords are two different things and should not be in sync.
In your comment you speak about scaled canvas.
Top and Left are 2 absolute values that represent the position of the object on the canvas. This position match with the canvas pixels when the canvas has a identity transform matrix.
If you apply a zoom, this coordinates diverge.
To get the position of pixel 300,100 of the scaled canvas on the unscaled canvas, you need to apply some basic math.
1) get the transform applied to the canvas
canvas.viewportTransform
2) invert it
var iM = fabric.util.invertTransform(canvas.viewportTransform)
3) multiply the wanted point by this matrix
var point = new fabric.Point(myX, myY);
var transformedPoint = fabric.util.transformPoint(point, iM)
4) set the object at that point.
I'm using a CIPerspectiveTransformWithExtent filter to apply homographies (perspective warp) to images on OS X. So far, so good, and I can get the desired warping applied to my images.
I'm still struggling however with the border behavior. I would like the output of the filter to be clipped outside the original image domain:
I managed to do it for the lower and left borders by shifting the origin of the inputExtent rectangle of the correct amount. For example, if the lower left corner is projected to x = -10 then using extent.origin.x = 10 will correctly clip the left border;
on the other hand, the upper and right borders are always shown in the output image. For example, if the rightmost corner is projected to x = width + 10, setting the extent via extent.origin.x = 0, extent.size.width = width; does not work an the rightmost corner remains visible.
Am I doing anything wrong here? or maybe I'm not trying the right way to achieve my goal?
I'm trying to separate the background of the graph grid in 3 areas using this code:
int[] data = {0xff000000, 0x80008000, 0xff000000};
bgBitmap = Bitmap.createBitmap(data, 1, 3, Bitmap.Config.ARGB_8888);
RectF rect = plot.getGraphWidget().getGridRect();
BitmapShader myShader = new BitmapShader(
Bitmap.createScaledBitmap(bgBitmap, 1, (int) rect.height(), false),
Shader.TileMode.REPEAT,
Shader.TileMode.REPEAT);
plot.getGraphWidget().getGridBackgroundPaint().setShader(myShader);
So scaling a 3 pixel bitmap to the graph height and repeating it over the whole domain area.
However the resulting graph show that the background seems to be shifted up a bit.
It looks like the shift size is about equal to the domain label height.
How can I fix this?
Hm cannot post picture because of 'reputation' sigh.
Link to the example graph: http://marcel.mesa.nl/androidplot.png
I think you're running into the issue mentioned near the end of this thread. Essentially, the origin of the shader is the top-left corner of the screen, not the top-left corner of component for which the background is being drawn using the shader. The solution is to translate to the top-left point of the graphWidget like this:
RectF rect = plot.getGraphWidget().getGridRect();
Matrix m = new Matrix();
m.setTranslate(rect.left, rect.top);
shader.setLocalMatrix(m); // where shader is your shader instance
I am drawing polygon by capturing mouse clicks on canvas and then passing these points to fabric.Polygon. So, in this manner I'm drawing multiple polygons.
What I need know is, I want to get the mouse co-ordinates (pixel points on canvas) for the polygon which is selected now?
I have tried with:
canvas.getActiveObject().get('points');
But this is giving some negative and some positive values.
So, can u please tell me a way to find out the polygon points?
Polygon points are relative to its center so you can get their "absolute" position like so:
var polygon = canvas.getActiveObject();
var polygonCenter = polygon.getCenterPoint();
var translatedPoints = polygon.get('points').map(function(p) {
return {
x: polygonCenter.x + p.x,
y: polygonCenter.y + p.y
};
});
Let's check how this looks:
translatedPoints.forEach(function(p) {
canvas.getContext().strokeRect(p.x-5, p.y-5, 10, 10);
});
I think this will only work if polygon's angle is at 0 (otherwise need to "rotate" points coordinates as well).
It looks like that from version 2.0 they changed the coordinates of the polygon. Before 2.0 points relative to the center of the polygon; after 2.0 they are absolute to the canvas;
Check out my response to the similar questions https://stackoverflow.com/a/53710375/4681279
Suppose we have been given the coordinates of the centre of a solid rectangular box, the box's length, breadth, height and a solid ball with a given centre and radius.
Is there a fast way to check if the box is a subset of the ball? The only simple method that comes to my mind is to check if each of the 8 corner vertices lies inside the sphere. If yes, the box is indeed a subset of the ball (by convexity property of the ball).
If you expect that most of the time the box will not be inside the sphere, you can do some quick tests:
if (sphere.center.x+sphere.radius<box.center.x-box.size.x) return false;
if (sphere.center.x-sphere.radius>box.center.x+box.size.x) return false;
etc.
If you expect that the box will usually be far inside the sphere, you can do other quick tests:
bsx = box.size.x;
bsy = box.size.y;
bsz = box.size.z;
box_radius = sqrt(bsx*bsx+bsy*bsy+bsz*bsz)/2;
brx = box.center.x-sphere.center.x;
bry = box.center.y-sphere.center.y;
brz = box.center.z-sphere.center.z;
box_dist = sqrt(brx*brx+bry*bry+brz*brz);
if (box_radius+box_dist<sphere.radius) return true;
You will still need more precise tests if the quick tests fail though.
There are only 4 corners to check, but in fact you only need to check 2 which are diagonal to each other. This follows from the property of the rectangle having straight sides, while the circle is convex, as you state.
Alternatively, consider that if the left upper corner and right lower corner are both within the circle, then the rectangle they form is a boudning box. The circle is obviously not, and thus the rectangle must be a sub-region of the circle.
EDIT: You might be talking about a box and a sphere, in which case the same idea applies, you just have to choose points which have different x, y, and z values