I am learning HTML5 and JavaScript and am attempting to draw an animated image. I thought the easiest way to do this would be to create an image with the frames in a row, as below.
Image http://html5stuff.x10.mx/HTML5%20Test/alien_green_strip8.png
Then a only part of the image would be draw at a time. I followed this tutorial.
This is a link to what I have made:
html5stuff.x10.mx/HTML5%20Test/page.html
The problem is, the image isn't being drawn. It's something within the drawSprite function, because when I change it to a simple "ctx.drawImage(sprite.source, x, y)", it does draw the image (just as a whole without the animation, obviously). Please note that though there is an option for rotating the image, I have not yet added support for that. Also, keys.js is not being used yet though it is included.
The reason is because sprite.imagenum is not defined when drawSprite is called.
This is because in some places you use imagenum and others imgnum, so correct that typo and you're good to go!
TOTALLY OPTIONAL:
But now that thats answered lets take a look at your js to get a better idea of how to structure this. You have:
function Sprite(){
var imagenum = null; //The number of images
var width = null; //The width of each image
var height = null; //The height on each image
var xoffset = null; //The origin on each image
var yoffset = null;
var source = null; //The location of each image
}
function drawSprite(sprite, subimg, x, y, w, h, angle){
ctx.drawImage(sprite.source, Math.floor(subimg) * sprite.width, 0, w * sprite.imagenum, h, x - sprite.xoffset * (w/sprite.width), y - sprite.yoffset * (h/sprite.height), w, h);
}
All those var statements are actually doing nothing. It should be:
function Sprite(){
this.imagenum = null; //The number of images
this.width = null; //The width of each image
this.height = null; //The height on each image
this.xoffset = null; //The origin on each image
this.yoffset = null;
this.source = null; //The location of each image
}
in order to correctly set them as you were envisioning. Also, you can rewrite drawSprite so that the sprites are drawing themselves, so that you don't need to pass them as an argument:
// now we can use "this" instead of "sprite"
Spite.prototype.draw = function(subimg, x, y, w, h, angle){
// put this on several lines just so we can see it easier
ctx.drawImage(this.source,
Math.floor(subimg) * this.width,
0,
w * this.imagenum, h,
x - this.xoffset * (w/this.width),
y - this.yoffset * (h/this.height),
w, h);
}
Then instead of:
drawSprite(img, index, 128, 128, 32, 32, 0); // img is a sprite
We can write:
img.draw(index, 128, 128, 32, 32, 0); // img is a sprite
Related
I'm using slightly modified resize code based on example I found. However, on resize the everything is flipped. I would like to either flip it back or prevent it from flipping in the first place.
Here is my resize code:
private static void ResizePath(SKPath buildingPath, IEnumerable<Room> rooms)
{
var info = new SKImageInfo(512, 600, SKImageInfo.PlatformColorType, SKAlphaType.Premul);
var drawSpaceRect = SKRect.Create(info.Size);
//I need to find the size of the path
var buildingPathRect = buildingPath.TightBounds;
//I want to find the largest rectangle that can fit on my canvas maintaining the path's aspect ratio
var sketchRect = drawSpaceRect.AspectFit(buildingPathRect.Size);
//Now I need to transform the path to draw within the sketchRect
//First translate original path to its own origin
var firstTranslateM = SKMatrix.MakeTranslation(-buildingPathRect.Left, -buildingPathRect.Top);
//Next handle scaling. Since I maintained aspect ratio, I should be able to use either
//width or height to figure out scaling factor
var scalingFactor = sketchRect.Width/buildingPathRect.Width;
var scaleM = SKMatrix.MakeScale(scalingFactor, scalingFactor);
//Next I need to handle translation so path is centered on canvas
var secondTranslateM = SKMatrix.MakeTranslation(sketchRect.Left, sketchRect.Top);
//Finally I need to handle transforming the path to rotate 180 degrees
var rotationMatrix = SKMatrix.MakeRotationDegrees(180, sketchRect.MidX, sketchRect.MidY);
//Now combine the translation, scaling, and rotation into a single matrix by matrix multiplication/concatentation
var transformM = SKMatrix.MakeIdentity();
SKMatrix.PostConcat(ref transformM, firstTranslateM);
SKMatrix.PostConcat(ref transformM, scaleM);
SKMatrix.PostConcat(ref transformM, secondTranslateM);
SKMatrix.PostConcat(ref transformM, rotationMatrix);
//Now apply the transform to the path
foreach (var r in rooms)
{
r.Path.Transform(transformM);
}
}
Here is an example of what I want (ignore the line numbers):
Flipped to:
Any help would be appreciated.
This transformation should do what you are looking for. The terminology would be flip horizontal or reflect horizontal.
var Ma = new SKMatrix {Values = new float[] {-1, 0, 0, 1, 0, 0, 0, 0, 0}};
pathToFlip.Transform(Ma);
I'm reading "create terrain from heightmap" example from ThreeJs Cookbook
This example load GrandCanyon: http://lh5.ggpht.com/_-B0hFoGrn-w/SvHiYk39yAI/AAAAAAAABOQ/6IGZwifUYGA/GrandCanyon.png
And create a 3D terrain: http://www.smartjava.org/tjscb/02-geometries-meshes/02.06-create-terrain-from-heightmap.html
There are some code pieces I can not understand:
// draw on canvas
ctx.drawImage(img, 0, 0);
var pixel = ctx.getImageData(0, 0, width, depth);
var geom = new THREE.Geometry;
var output = [];
for (var x = 0; x < depth; x++) {
for (var z = 0; z < width; z++) {
// get pixel
// since we're grayscale, we only need one element
var yValue = pixel.data[z * 4 + (depth * x * 4)] / heightOffset;
var vertex = new THREE.Vector3(x * spacingX, yValue, z * spacingZ);
geom.vertices.push(vertex);
}
}
why is yValue calculated with that value ? why don't we use var yValue = pixel.data[z * 4 + (depth * x )] or something like that ?
And do we really need spacingX and spacingZ ?
Source code is here: https://github.com/josdirksen/threejs-cookbook/blob/master/02-geometries-meshes/02.06-create-terrain-from-heightmap.html
Could you please help me ?
Thank you very much!
You don't NEED spacingX and spacingZ, no. You could adjust scale in other ways, like applying a scale matrix to the entire THREE.Geometry after you've populated the vertices. Up to you, really.
As fort the yValue, the indexing is to adjust for the way the data for the texture is laid out. There are four channels, usually RGBA, but in this case we only need one of them as a height.
I have a situation where i am generating several graphs in web page and showing them in canvas and my requirement is that on click of download button,i should be able to export all canvas images to pdf.
I have successfully done this for single canvas element using html2canvas and Jspdf but cannot figure out how to do the same for all.
I followed this JSFiddle code for generating pdf from Html2canvas and jspdf.
jsfiddle
$(document).ready(function() {
var d_canvas = document.getElementById('canvas');
var context = d_canvas.getContext('2d');
context.moveTo(20, 20);
context.lineTo(100, 20);
context.fillStyle = "#999";
context.beginPath();
context.arc(100, 100, 75, 0, 2 * Math.PI);
context.fill();
context.fillStyle = "orange";
context.fillRect(20, 20, 50, 50);
context.font = "24px Helvetica";
context.fillStyle = "#000";
context.fillText("Canvas", 50, 130);
$('#ballon').draggable();
$('#download').click(function() {
html2canvas($("#canvas"), {
onrendered: function(canvas) {
var imgData = canvas.toDataURL(
'image/png');
var doc = new jsPDF('p', 'mm');
doc.addImage(imgData, 'PNG', 10, 10);
doc.save('sample-file.pdf');
}
});
});
});
Kindly help,thanks in advance.
It was very simple ,I just changed the argument to this line
html2canvas($("#canvas"), {
Instead of passing seperate canvases and then trying to export them to single pdf rather i kept different canvases in on Div and passed the Div id to the above mentioned line and both canvases were exported to single pdf file
There should be no need to use html2canvas for this. It will only deliver you another canvas element at a cost. You can use the original canvas element and toDataURL() directly with jsPdf.
Example (partly pseudo)
This will collect all canvases in the page and put them in a PDF. The pseudo part is the missing variables for width, deltas, factor etc. But you should get the gist of it.
Note: The size for images must be given in the same unit you're using for the document, so you need to convert pixel positions and sizes into millimeter representation using a pre-calculated factor based on document DPI (not shown here, but this may help).
var x = someX,
y = someY,
dx = somDeltaForX,
dy = somDeltaForY,
i,
canvases = document.querySelectorAll("canvas"),
pdf = new jsPDF('p', 'mm'),
f = convertionFactorFromPixelstoMM;
for(i = 0; i < canvases.length; i++) {
var url = canvases[i].toDataURL("image/jpeg", 0.75);
doc.addImage(url, "JPEG", x * f, y * f, canvases[i].width * f, canvases[i].height * f);
x += dx; // tip: dx could also be based on previous canvas width in non-uniform sizes
if (x > widthOfPage) {
x = 0;
y += dy;
}
}
I'm looking for few days a solution to draw rectangle on image frame. Basically I'm using CvInvoke.cvRectangle method to draw rectangle on image because I need antialiased rect.
But problem is when I need to rotate a given shape for given angle. I can't find any good solution.
I have tryed to draw rectangle on separate frame then rotate hole frame and apply this new image on top of my base frame. But in this solution there is a problem with antialiasing. It's not working.
I'm working on simple application that should allow draw few kinds of shape, resize them and rotation for given angle.
Any idea how to achive this?
The best way I found to draw a minimum enclosing rectangle on the contour is using the Polylines() function which uses vertices that are returned from MinAreaRect() function. There are surely other ways to do it as well. Here is the code walk down:
// Find contours
var contours = new Emgu.CV.Util.VectorOfVectorOfPoint();
Mat hierarchy = new Mat();
CvInvoke.FindContours(image, contours, hierarchy, RetrType.Tree, ChainApproxMethod.ChainApproxSimple);
// According to your metric, get an index of the contour you want to find the min enclosing rectangle for
int index = 2; // Say, 2nd index works for you.
var rectangle = CvInvoke.MinAreaRect(contours[index]);
Point[] vertices = Array.ConvertAll(rectangle.GetVertices(), Point.Round);
CvInvoke.Polylines(image, vertices, true, new MCvScalar(0, 0, 255), 5);
The result can be visualized in the image below, in red is the minimum enclosing rectangle.
I use C# and EMGU.CV(4.1), and I think this code will not be difficult to transfer to any platform.
Add function in the in your helper:
public static Mat DrawRect(Mat input, RotatedRect rect, MCvScalar color = default(MCvScalar),
int thickness = 1, LineType lineType = LineType.EightConnected, int shift = 0)
{
var v = rect.GetVertices();
var prevPoint = v[0];
var firstPoint = prevPoint;
var nextPoint = prevPoint;
var lastPoint = nextPoint;
for (var i = 1; i < v.Length; i++)
{
nextPoint = v[i];
CvInvoke.Line(input, Point.Round(prevPoint), Point.Round(nextPoint), color, thickness, lineType, shift);
prevPoint = nextPoint;
lastPoint = prevPoint;
}
CvInvoke.Line(input, Point.Round(lastPoint), Point.Round(firstPoint), color, thickness, lineType, shift);
return input;
}
This draws roteted rectangle by points. Here used rounding points by method Point.Round becose RotatedRect has points in float coordinates and CvInvoke.Line takes points as integer.
Use:
var mat = Mat.Zeros(200, 200, DepthType.Cv8U, 3);
mat.GetValueRange();
var rRect = new RotatedRect(new PointF(100, 100), new SizeF(100, 50), 30);
DrawRect(mat, rRect,new MCvScalar(255,0,0));
var brect = CvInvoke.BoundingRectangle(new VectorOfPointF(rRect.GetVertices()));
CvInvoke.Rectangle(mat, brect, new MCvScalar(0,255,0), 1, LineType.EightConnected, 0);
Result:
You should read the OpenCV documentation.
There is a RotatedRectangle class that you can use for your task. You can specify the angle by which the rectangle will be rotated.
Here is a sample code (taken from the docs) for drawing a rotated rectangle:
Mat image(200, 200, CV_8UC3, Scalar(0));
RotatedRect rRect = RotatedRect(Point2f(100,100), Size2f(100,50), 30);
Point2f vertices[4];
rRect.points(vertices);
for (int i = 0; i < 4; i++)
line(image, vertices[i], vertices[(i+1)%4], Scalar(0,255,0));
Rect brect = rRect.boundingRect();
rectangle(image, brect, Scalar(255,0,0));
imshow("rectangles", image);
waitKey(0);
Here is the result:
Is it possible to move an image (or image object) without clearing the whole background?
I wish to create an app that allows the user to "paint", using a device that is not the mouse. I would like to have a cursor to follow the users movement with the input device, without having to clear the already painted picture.
Is this possible? And how?
It depends how you handle drawing.
I would suggest using PImage as a canvas to draw into and another PImage to store the pixels of your brush. The 'brush' can be a loaded image, or at the start of your sketch you could make the brush using drawing commands, then store those as a PImage using get().
You will need to clear everything because you want to draw your cursor, but you will also draw your canvas, and you'll store 'brush strokes' only when the mouse is pressed (or some device specific method) by using the copy() or the blend() function (depending on your brush PNG - with or without transparency, etc.)
Here's a quick sketch to illustrate this:
PImage canvas;
PImage brush;
void setup(){
size(800,800);
stroke(128);
smooth();
canvas = createImage(width,height,ARGB);
brush = loadImage("brush.png");
}
void draw(){
background(255);
image(canvas,0,0);
//draw cursor
line(mouseX-5,mouseY-5,mouseX+5,mouseY+5);
line(mouseX+5,mouseY-5,mouseX-5,mouseY+5);
//blend brush pixels into canvas if mouse is pressed
if(mousePressed) canvas.blend(brush, 0, 0, brush.width, brush.width, (int)(mouseX-brush.width*.5), (int)(mouseY-brush.height*.5), brush.width, brush.width,MULTIPLY);
}
Note that you need an image into your sketch's data folder.
You can try it here:
You can run a javascript version bellow:
var canvas;
var brush;
function setup(){
createCanvas(800,800);
stroke(128);strokeWeight(3);
smooth();
canvas = createImage(width,height);
brush = getGradientImg(64,64,random(360),random(100),85);
}
function draw(){
background(255);
image(canvas,0,0);
//draw cursor
line(mouseX-5,mouseY-5,mouseX+5,mouseY+5);
line(mouseX+5,mouseY-5,mouseX-5,mouseY+5);
//blend brush pixels into canvas if mouse is pressed
if(isMousePressed) canvas.blend(brush, 0, 0, brush.width, brush.width, (int)(mouseX-brush.width*.5), (int)(mouseY-brush.height*.5), brush.width, brush.width,MULTIPLY);
//image(brush,mouseX,mouseY);
}
//*
function getGradientImg(w,h,hue,satMax,brightness){
push();//isolate drawing styles such as color Mode
colorMode(HSB,360,100,100);
var gradient = createImage(w,h);//create an image with an alpha channel
var np = w * h;//total number of pixels
var np4 = np*4;
var cx = floor(gradient.width * 0.5);//center on x
var cy = floor(gradient.height * 0.5);//center on y
gradient.loadPixels();
for(var i = 0 ; i < np4; i+=4){//for each pixel
var id4 = floor(i * .25);
var x = id4%gradient.width;//compute x from pixel index
var y = floor(id4/gradient.width);//compute y from pixel index
var d = dist(x,y,cx,cy);//compute distance from centre to current pixel
//map the saturation and transparency based on the distance to centre
gradient.pixels[i] = hue;
gradient.pixels[i+1] = map(d,0,cx,satMax,0);
gradient.pixels[i+2] = brightness;
gradient.pixels[i+3] = map(d,0,cx,255,0);
}
gradient.updatePixels();//finally update all the pixels
pop();
console.log(gradient);
return gradient;
}
//*/
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