I've included a zoom functionality similar to the one explained at this website:
http://www.flashandmath.com/howtos/zoom/
This works perfectly on my background image(a map, that is), but I want to keep the symbols on my map the same size while zooming in.
I probably could work this out by changing all the children's size when calling the zoom-function, but I am hoping there is some kind of easy code adapt in my children class to make the size of the instances unchangable. Is there?
Thanks!
One crude way, so you don't have to calculate the symbols scale, would be to remove the symbols from the mapDisplayObject so they're no longer a child and instead put symbol placeholders. Then match each symbol's x and y to each place holder, using localToGlobal...
If your children are not scaled or skewed or rotated you can iterate all of them and set transformation matrix to 1/parentScale. Something like:
for each (var child:DisplayObject in parent) {
var matrix:Matrix = child.transform.matrix;
matrix.a = 1/parentScale;
matrix.d = 1/parentScale;
child.transform.matrix = marix;
}
Related
I'm brand new to AS3, trying to take a bitmap from the library and display it in the timeline with actionscript. Here's the code I'm using so far as per Adobe's site:
addChild(new Bitmap(new myBitmap(100, 100)));
and this works, but changing the dimensions in those brackets doesn't. How can I change the dimensions?
Also, how can I change the position of the bitmap? Right now it's at 0,0.
You first get a handler of that bitmap:
var theBitmap:Bitmap=new myBitmap();
Then, you can change its coordinates:
theBitmap.x=100;
theBitmap.y=100;
addChild(theBitmap);
In order to do scaling, you can also alter its scaleX and scaleY properties as with any other displayed object. In order to make a Bitmap object to be of certain size on screen, you have to scale it so that scaling parameter equals (new dimensions)/(old dimensions). But, if you want something else, you have to explain it further.
I am trying to get the derived position and scaling of a 3D Sprite and set them to a 2D Sprite.
I have managed to do the first part like this:
var p:Point = sprite3d.local3DToGlobal(new Vector3D(0,0,0));
billboard.x = p.x;
billboard.y = p.y;
But I can't get the scaling part correctly. I am trying this:
var mat:Matrix3D = sprite3d.transform.getRelativeMatrix3D(stage); // get derived matrix(?)
var scaleV:Vector3D = mat.decompose()[2]; // get scaling vector from derived matrix
var scale:Number = scaleV.length;
billboard.scaleX = scale;
billboard.scaleY = scale;
...but the result is apparently wrong.
PS. One might ask what I am trying to achieve. I am trying to create "billboard" 3D sprites, i.e. sprites which are affected by all 3D transformations except rotations, thus they always face the "camera".
The documentation says that you get the vector correctly, but its coefficients don't seem to be added together to form a single length value. Try first an unscaled sprite, and check if you're receiving a sqrt(3) value as its length. If yes, then you should use 0th element of the vector as X scale, and 1th as Y scale. I'm not sure what to do with 2nd element (in this case it'll be a Z scale, either divide both scales by it, or multiply by it). Hope that helped.
I'm trying to position an image on top of another image based upon the make-up of the smaller image. The smaller image is a cut-out of a larger image and I need it to be positioned exactly on the larger image to make it look like a single image, but allow for separate filters and alphas to be applied. As the images are not simple rectangles or circles, but complex satellite images, I cannot simply redraw them in code. I have quite a few images and therefore do not feel like manually finding the position of each image every and hard setting them manually in actionscript. Is there any way for me to sample a small 5-10 sq. pixel area against the larger image and set the x and y values of the smaller image if a perfect match is found? All the images are in an array and iterating through them has already been set, I just need a way to sample and match pixels. My first guess was to loop the images pixel by pixel right and down, covering the whole bitmap and moving to the next child in the array once a match was found, leaving the matched child where it was when the perfect match was found.
I hope I understood your question correctly.
There may be an option that uses copypixels to achieve what you want. You can use the bitmapdata.rect value to determine the size of the sample you want, and loop through the bigger bitmap using thet rectangle and a moving point. Let's see if I can code this out...
function findBitmapInBitmap(tinyimg:BitmapData, largeimg:BitmapData):Point {
var rect:Rectangle = tinyimg.rect;
var xbound:uint = largeimg.rect.width;
var ybound:uint = largeimg.rect.height;
var imgtest:BitmapData = new BitmapData(tinyimg.rect.width, tinyimg.rect.height);
for (var ypos:uint = 0, y <= ybound, y++) {
for (var xpos:uint = 0, x <= xbound, x++) {
imgtest.copyPixels(largeimg, rect, new Point(xpos, ypos);
if (imgtest.compare(tinyimg) == 0) return new Point(xpos, ypos);
}
}
return new Point(-1,-1); // Dummy value, indicating no match.
}
Something along those lines should work - I'm sure there's room for code elegance and possible optimization. However, it seems like something like this method would be very slow, since you'd have to check each pixel for a match.
There is a better way. Split your big image into layers, and use the blitting technique to composite them at runtime. In your case, you could create a ground texture without satellites, and then create the satellites separately, and use the copyPixels method to place them whereever you want. Google "blitting in as3" to find some good tutorials. I'm currently working on a game project that uses this technique and it's a very good method.
Good luck!
Edit: Forgot to code in a default return statement. Using this method, you'd have to return an invalid point (like (-1,-1)) and check for it outside the function. Alternatively, you could just copy your small bitmap to the big one within the function, which would be much more logical, but I don't know your requirements.
You need to find pixel sequence in the big image. BitmapData.getPixel gives you pixel value. So get first pixel from small image, find it in big image, then continue comparing until you find full match. If you have trouble to code that, feel free to ask.
For the actual comparison, there's BitmapData.compare which returns the number 0 if the BitmapData objects are equivalent.
I want to move a display object from one container to another, but have it appear in the same place on screen.
I thought I'd understood this years ago, but the following does not work:
function moveToNewContainer(obj:DisplayObject, newParent:DisplayObjectContainer):void {
var pos:Point = new Point(obj.x, obj.y);
var currentParent:DisplayObjectContainer = obj.parent;
pos = currentParent.localToGlobal(pos);
currentParent.removeChild(obj);
newParent.addChild(obj);
pos = newParent.globalToLocal(pos);
obj.x = pos.x;
obj.y = pos.y;
}
This doesn't position the object in the same place as I would have expected.
Does anyone know what I am doing wrong, please?
Thanks,
James
Using localToGlobal/globalToLocal and setting the x and y properties like you showed calculates the correct position for the object in its new parent, but does not adjust for other aspects of the transformation such as scaling or rotation. In other words, the object's registration point will indeed remain in the same place, but the object may be rotated, scaled, or sheared differently.
The solution to your problem will need to take into account the transform.concatenatedMatrix properties of the old and new parents--you'll need to multiply the object's transformation matrix by one and then by the inverse of the other, or something along those lines. Leave a comment if you need help working out the math.
There is nothing wrong with your code, provided that both containers have no transformations applied. If your clips are scaled, rotated, etc.. you need to handle that in addition to the coordinate space transformations that localToGlobal and globalToLocal do.
You have to check if your containers are actually placed on stage. If your new container isn't added as a child to stage, function globalToLocal fails, just because it doesnt know how to correctly calculate that data.
I have a task:
I need to place about 100 sprites on one canvas (with prepared grid on it). I need to place them as invisible (circles) stones, on the board, and make visible only on mouseover.
The problem I come across is following, I can't place those objects accurately into the nodes on the grid.
E.g.
if I define stones (it's just a sprite, as I said earlier) this way:
var stone:StoneSprite = new StoneSprite();
stone.x = this.x + 2*cellWidth;
stone.graphics.beginFill( 0x000000 );
stone.graphics.drawCircle(stone.x , this.y + cellWidth, cellWidth/3 );
stone.graphics.endFill();
rawChildren.addChild(stone);
They don't sit on the node...
See image:
http://img.skitch.com/20091014-kuhfyjeg1g5qmrbyxbcerp4aya.png
And if I do it this way:
var stone:StoneSprite = new StoneSprite();
stone.graphics.beginFill( 0x000000 );
stone.graphics.drawCircle(this.x + 2*cellWidth , this.y + cellWidth, cellWidth/3 );
stone.graphics.endFill();
rawChildren.addChild(stone);
The stone is displayed correctly in the grid node... See image 2:
http://img.skitch.com/20091014-f595tksjxramt98s7yfye591bh.png
So I wonder what is the difference between these 2 approaches.
Also, I think I need to pass correct coordinates to the stone class... In case I would like to change some properties of the stone object. E.g. visibility, or radius.
Could you please suggest, what's wrong in defining coordinates as stone.x, stone.y
How to fix the problem with incorrect positioning.
Would really appreciate ideas about the problem, I am trying to solve for so long :(
Assume x & y are 30 and cellWidth is 30.
First Example:
stone.x = 30 + 60; //90
drawCircle(90, 60, 10);
This means if you were to draw a rectangle around your circle, it would be at [170,50]. (x,y).
Second Example:
stone.x = 0;
drawCircle(90, 60, 10)
This means the rectangle around your circle is at [80,50];
In the first example, you are moving the sprite to position x==90. Then drawing a circle whose center is at x==90 inside the sprite. So relative to this, you're at x==180. But because a circle's x,y coords are the center, subtract 10 for the radius to get the boundary x position.
In the second example, the sprite defaults to position x==0 relative to this and you're drawing the circle inside the sprite at position x==90. (therefore it begins at x==80).
I am not sure what's causing the issue - might be some padding induced by the container - can't say without testing. But I believe that adding a Sprite (say board) to canvas.rawChildren and using it as the parent for the grid and stones would fix the issue.