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I would like to create an isometric 3D cube with fillRect whose 3 faces have the same dimensions as the image below:
Edit: I want to do it with fillRect. The reason for this is that I will draw images on the 3 faces of the cube afterwards. This will be very easy to do since I will use exactly the same transformations as for drawing the faces.
Edit 2: I didn't specify that I want to avoid using an external library so that the code is as optimized as possible. I know that it is possible to calculate the 3 matrices beforehand to draw the 3 faces and make a perfect isometric cube.
Edit 3: As my example code showed, I want to be able to set the size of the side of the isometric cube on the fly (const faceSize = 150).
I have a beginning of code but I have several problems:
The faces are not all the same dimensions
I don't know how to draw the top face
const faceSize = 150;
const canvas = document.querySelector('canvas');
const ctx = canvas.getContext('2d');
const centerX = canvas.width / 2;
const centerY = canvas.height / 2;
// Top Face (not big enough)
ctx.save();
ctx.translate(centerX, centerY);
ctx.scale(1, .5);
ctx.rotate(-45 * Math.PI / 180);
ctx.fillStyle = 'yellow';
ctx.fillRect(0, -faceSize, faceSize, faceSize);
ctx.restore();
// Left Face (not high enough)
ctx.save();
ctx.translate(centerX, centerY);
ctx.transform(1, .5, 0, 1, 0, 0);
ctx.fillStyle = 'red';
ctx.fillRect(-faceSize, 0, faceSize, faceSize);
ctx.restore();
// Right Face (not high enough)
ctx.save();
ctx.translate(centerX, centerY);
ctx.transform(1, -.5, 0, 1, 0, 0);
ctx.fillStyle = 'blue';
ctx.fillRect(0, 0, faceSize, faceSize);
ctx.restore();
<canvas width="400" height="400"></canvas>
I used a large part of #enhzflep's code which I adapted so that the width of the cube is dynamically changeable.
All the code seems mathematically correct, I just have a doubt about the value 1.22 given as a parameter to scaleSelf. Why was this precise value chosen?
Here is the code:
window.addEventListener('load', onLoad, false);
const canvas = document.createElement('canvas');
function onLoad() {
//canvas.width = cubeWidth;
//canvas.height = faceSize * 2;
canvas.width = 400;
canvas.height = 400;
document.body.appendChild(canvas);
drawCube(canvas);
}
function drawCube() {
const scale = Math.abs(Math.sin(Date.now() / 1000) * canvas.width / 200); // scale effect
const faceSize = 100 * scale;
const radians = 30 * Math.PI / 180;
const cubeWidth = faceSize * Math.cos(radians) * 2;
const centerPosition = {
x: canvas.width / 2,
y: canvas.height / 2
};
const ctx = canvas.getContext('2d');
ctx.save();
ctx.fillStyle = '#000';
ctx.fillRect(0, 0, ctx.canvas.width, ctx.canvas.height);
const defaultMat = [1, 0, 0, 1, 0, 0];
// Left (red) side
const leftMat = new DOMMatrix(defaultMat);
leftMat.translateSelf(centerPosition.x - cubeWidth / 2, centerPosition.y - faceSize / 2);
leftMat.skewYSelf(30);
ctx.setTransform(leftMat);
ctx.fillStyle = '#F00';
ctx.fillRect(0, 0, cubeWidth / 2, faceSize);
// Right (blue) side
const rightMat = new DOMMatrix(defaultMat);
rightMat.translateSelf(centerPosition.x, centerPosition.y);
rightMat.skewYSelf(-30);
ctx.setTransform(rightMat);
ctx.fillStyle = '#00F';
ctx.fillRect(0, 0, cubeWidth / 2, faceSize);
// Top (yellow) side
const topMat = new DOMMatrix(defaultMat);
const toOriginMat = new DOMMatrix(defaultMat);
const fromOriginMat = new DOMMatrix(defaultMat);
const rotMat = new DOMMatrix(defaultMat);
const scaleMat = new DOMMatrix(defaultMat);
toOriginMat.translateSelf(-faceSize / 2, -faceSize / 2);
fromOriginMat.translateSelf(centerPosition.x, centerPosition.y - faceSize / 2);
rotMat.rotateSelf(0, 0, -45);
scaleMat.scaleSelf(1.22, (faceSize / cubeWidth) * 1.22);
topMat.preMultiplySelf(toOriginMat);
topMat.preMultiplySelf(rotMat);
topMat.preMultiplySelf(scaleMat);
topMat.preMultiplySelf(fromOriginMat);
ctx.setTransform(topMat);
ctx.fillStyle = '#FF0';
ctx.fillRect(0, 0, faceSize, faceSize);
ctx.restore();
requestAnimationFrame(drawCube);
}
Here's a quick n dirty approach to the problem. It's too hot here for me to really think very clearly about this question. (I struggle with matrix maths too)
There's 2 things I think worth mentioning, each of which has an effect on the scaling operation.
width and height of the finished figure (and your posted example image) are different.
I think it's the ratio of the distance between (opposite) corners of the untransformed rectangle which fills 1/4 of the canvas, and the finished yellow side which affect the scaling.
Also, note that I'm drawing a square of canvas.height/2 sidelength for the yellow side, whereas I was drawing a rectangle for the red and blue sides.
In the scaling section, width/4 and height/4 are both shorthand for (width/2)/2 and (height/2)/2. width/2 and height/2 give you a rectangle filling 1/2 of the canvas, with a centre (middle of the square) located at (width/2)/2, (height/2)/2 - height/4 means something different in the translation section (even though it's the same number)
With that said, here's the sort of thing I was talking about earlier.
"use strict";
window.addEventListener('load', onLoaded, false);
function onLoaded(evt)
{
let width = 147;
let height = 171;
let canvas = document.createElement('canvas');
canvas.width = width;
canvas.height = height;
document.body.appendChild(canvas);
drawIsoDemo(canvas);
}
function drawIsoDemo(destCanvas)
{
let ctx = destCanvas.getContext('2d');
let width = destCanvas.width;
let height = destCanvas.height;
ctx.fillStyle = '#000';
ctx.fillRect(0,0,width,height);
var idMatVars = [1,0, 0,1, 0,0];
// left (red) side
let leftMat = new DOMMatrix( idMatVars );
leftMat.translateSelf( 0, 0.25*height );
leftMat.skewYSelf(30);
ctx.save();
ctx.transform( leftMat.a, leftMat.b, leftMat.c, leftMat.d, leftMat.e, leftMat.f);
ctx.fillStyle = '#F00';
ctx.fillRect(0,0,width/2,height/2);
ctx.restore();
// right (blue) side
let rightMat = new DOMMatrix( idMatVars );
rightMat.translateSelf( 0.5*width, 0.5*height );
rightMat.skewYSelf(-30);
ctx.save();
ctx.transform( rightMat.a, rightMat.b, rightMat.c, rightMat.d, rightMat.e, rightMat.f);
ctx.fillStyle = '#00F';
ctx.fillRect(0,0,width/2,height/2);
ctx.restore();
// top (yellow) side
let topMat = new DOMMatrix( idMatVars );
let toOriginMat = new DOMMatrix( idMatVars );
let fromOriginMat = new DOMMatrix(idMatVars);
let rotMat = new DOMMatrix(idMatVars);
let scaleMat = new DOMMatrix(idMatVars);
toOriginMat.translateSelf(-height/4, -height/4);
fromOriginMat.translateSelf(width/2,height/4);
rotMat.rotateSelf(0,0,-45);
scaleMat.scaleSelf(1.22,((height/2)/width)*1.22);
topMat.preMultiplySelf(toOriginMat);
topMat.preMultiplySelf(rotMat);
topMat.preMultiplySelf(scaleMat);
topMat.preMultiplySelf(fromOriginMat);
ctx.save();
ctx.transform( topMat.a, topMat.b, topMat.c, topMat.d, topMat.e, topMat.f);
ctx.fillStyle = '#FF0';
ctx.fillRect(0,0,height/2,height/2);
ctx.restore();
}
If we overlay a circle on your isometric cube, we can see that the outer vertices are spaced equally apart. In fact it's always 60°, which is no wonder as it's a hexagon.
So all we have to do is obtaining the coordinates for the outer vertices. This is quite easy as we can make a further assumption: if you look at the shape again, you'll notice that the length of each of the cube's sides seems to be the radius of the circle.
With the help of a little trigonometry and a for-loop which increments by 60 degrees, we can put calculate and put all those vertices into an array and finally connect those vertices to draw the cube.
Here's an example:
let canvas = document.getElementById("canvas");
let ctx = canvas.getContext("2d");
function drawCube(x, y, sideLength) {
let vertices = [new Point(x, y)];
for (let a = 0; a < 6; a++) {
vertices.push(new Point(x + Math.cos(((a * 60) - 30) * Math.PI / 180) * sideLength, y + Math.sin(((a * 60) - 30) * Math.PI / 180) * sideLength));
}
ctx.fillStyle = "#ffffff";
ctx.beginPath();
ctx.moveTo(vertices[0].x, vertices[0].y);
ctx.lineTo(vertices[5].x, vertices[5].y);
ctx.lineTo(vertices[6].x, vertices[6].y);
ctx.lineTo(vertices[1].x, vertices[1].y);
ctx.lineTo(vertices[0].x, vertices[0].y);
ctx.fill();
ctx.fillStyle = "#a0a0a0";
ctx.beginPath();
ctx.moveTo(vertices[0].x, vertices[0].y);
ctx.lineTo(vertices[1].x, vertices[1].y);
ctx.lineTo(vertices[2].x, vertices[2].y);
ctx.lineTo(vertices[3].x, vertices[3].y);
ctx.lineTo(vertices[0].x, vertices[0].y);
ctx.fill();
ctx.fillStyle = "#efefef";
ctx.beginPath();
ctx.moveTo(vertices[0].x, vertices[0].y);
ctx.lineTo(vertices[3].x, vertices[3].y);
ctx.lineTo(vertices[4].x, vertices[4].y);
ctx.lineTo(vertices[5].x, vertices[5].y);
ctx.lineTo(vertices[0].x, vertices[0].y);
ctx.fill();
}
class Point {
constructor(x, y) {
this.x = x;
this.y = y;
}
}
drawCube(200, 150, 85);
canvas {
background: #401fc1;
}
<canvas id="canvas" width="400" height="300"></canvas>
EDIT
What you want to achieve is ain't that easily simply because the CanvasRenderingContext2D API actually does not offer a skewing/shearing transform.
Nevertheless with the help of a third-party library we're able to transform the three sides in an orthographic way. It's called perspective.js
Still we need to calculate the outer vertices but instead of using the moveTo/lineTo commands, we forward the coordinates to perspective.js to actually do the perspective distortion of some source images.
Here's another example:
let canvas = document.getElementById("canvas");
let ctx = canvas.getContext("2d");
class Point {
constructor(x, y) {
this.x = x;
this.y = y;
}
}
function drawCube(x, y, sideLength) {
let vertices = [new Point(x, y)];
for (let a = 0; a < 6; a++) {
vertices.push(new Point(x + Math.cos(((a * 60) - 30) * Math.PI / 180) * sideLength, y + Math.sin(((a * 60) - 30) * Math.PI / 180) * sideLength));
}
let p = new Perspective(ctx, images[0]);
p.draw([
[vertices[5].x, vertices[5].y],
[vertices[6].x, vertices[6].y],
[vertices[1].x, vertices[1].y],
[vertices[0].x, vertices[0].y]
]);
p = new Perspective(ctx, images[1]);
p.draw([
[vertices[0].x, vertices[0].y],
[vertices[1].x, vertices[1].y],
[vertices[2].x, vertices[2].y],
[vertices[3].x, vertices[3].y]
]);
p = new Perspective(ctx, images[2]);
p.draw([
[vertices[4].x, vertices[4].y],
[vertices[5].x, vertices[5].y],
[vertices[0].x, vertices[0].y],
[vertices[3].x, vertices[3].y]
]);
}
function loadImages(index) {
let image = new Image();
image.onload = function(e) {
images.push(e.target);
if (index + 1 < sources.length) {
loadImages(index + 1);
} else {
drawCube(200, 150, 125, e.target);
}
}
image.src = sources[index];
}
let sources = ["https://picsum.photos/id/1079/200/300", "https://picsum.photos/id/76/200/300", "https://picsum.photos/id/79/200/300"];
let images = [];
loadImages(0);
canvas {
background: #401fc1;
}
<script src="https://cdn.rawgit.com/wanadev/perspective.js/master/dist/perspective.min.js"></script>
<canvas id="canvas" width="400" height="300"></canvas>
I'm adding multiple rectangles in canvas which could collide with each other. The outer stroke should be displayed on the outer part of both rectangles or the rectangle shapes should be merged in to one producing the expected result.
See picture bellow
It has to be cut because it will display the content under the canvas. See live example with background image: https://jsfiddle.net/0qpgf5un/
In the code example bellow rectangles are being added on top of each other as you can see in the first example of the picture.
var canvas = document.getElementById('canvas');
var ctx = canvas.getContext('2d');
var offsetX = 150;
var offsetY = 150;
var w = 200;
var h = 100;
ctx.fillStyle = "red";
ctx.rect(0, 0, 600, 600);
ctx.fill();
ctx.clearRect(offsetX,offsetY, w, h);
ctx.strokeRect(offsetX, offsetY, w, h);
ctx.clearRect(offsetX-50,offsetY+50, w, h);
ctx.strokeRect(offsetX-50, offsetY+50, w, h);
Is there ways to achieve it without writing complex calculations of each path, since the collision of rectangles can be unintentional and diverse ?
Edit:
What I am trying to achieve is a similar functionality like in youtube's feedback form where when editing screenshot you can highlight items and the border then is merged.
Just add one more clearRect() (the first one)
var canvas = document.getElementById('canvas');
var ctx = canvas.getContext('2d');
var offsetX = 150;
var offsetY = 150;
var w = 200;
var h = 100;
ctx.fillStyle = "red";
ctx.rect(0, 0, 600, 600);
ctx.fill();
ctx.clearRect(offsetX,offsetY, w, h);
ctx.strokeRect(offsetX, offsetY, w, h);
ctx.clearRect(offsetX-50,offsetY+50, w, h);
ctx.strokeRect(offsetX-50, offsetY+50, w, h);
ctx.clearRect(offsetX,offsetY, w, h);
https://jsfiddle.net/kt3yjhpc/
You can skip clearing the first rectangle and then clear it after you stroke the second one.
The clearPrev function will clear the area inside the strokes of the initial rectangle.
let canvas = document.getElementById('canvas'),
ctx = canvas.getContext('2d'),
offsetX = 70,
offsetY = 20,
w = 200,
h = 100,
strokeWidth = 5;
ctx.fillStyle = '#F00'
ctx.rect(0, 0, 600, 600);
ctx.fill();
ctx.strokeStyle = '#0FF';
ctx.lineWidth = strokeWidth;
//ctx.clearRect(offsetX, offsetY, w, h); <-- Do not need to do this, if we clear below...
ctx.strokeRect(offsetX, offsetY, w, h);
ctx.clearRect(offsetX - 50, offsetY + 50, w, h);
ctx.strokeRect(offsetX - 50, offsetY + 50, w, h);
clearPrev(ctx, offsetX, offsetY, w, h); // Clear previous
function clearPrev(ctx, x, y, w, h) {
let startOffset = Math.round(ctx.lineWidth / 2) - 1,
endOffset = strokeWidth - 1;
ctx.clearRect(x + startOffset, y + startOffset, w - endOffset, h - endOffset);
}
<canvas id="canvas" width="290" height="190"></canvas>
When you want to clear the canvas with complex shapes, forget about clearRect, it's not the only one able to produce transparent pixels.
Instead, have a look at compositing.
So your shape is really border-line, but I think you'll benefit from using this already:
var canvas = document.getElementById('canvas');
var ctx = canvas.getContext('2d');
var offsetX = 150;
var offsetY = 150;
var w = 200;
var h = 100;
ctx.lineWidth = 2;
ctx.fillStyle = "red";
ctx.fillRect(0, 0, 600, 600);
// declare our complex shape as a single sub-path
ctx.beginPath()
ctx.rect(offsetX,offsetY, w, h);
ctx.rect(offsetX-50, offsetY+50, w, h);
// now we can paint it
// first the stroke, because we want to erase what's inside the fill-area
ctx.stroke();
// now to erase, we switch to destination-out compositing mode
ctx.globalCompositeOperation = 'destination-out';
// fill the inner path
ctx.fill();
// we're done
// If you wish to go back to normal mode later
ctx.globalCompositeOperation = 'source-over';
body { background: linear-gradient(blue,yellow); }
<canvas id="canvas" width="600" height="600"></canvas>
I have an arc on a canvas that moves around wherever the mouse is. it is stroked onto the canvas with the color black. if the mouse goes over something black, the circle disappears. i would like it if the circle could change color, depending on what it is being drawn over. could anyone help me?
here is some code:
ctx.beginPath()
ctx.clearRect(0, 0, brush.prePos.x + brush.size*2, brush.prePos.y + brush.size*2)
ctx.arc(pos.x, pos.y, brush.size / 4, 0, Math.PI*2)
ctx.stroke()
ctx.closePath()
You can try different compositing modes, particularly XOR. From the MDN example:
const canvas = document.getElementById('canvas');
const ctx = canvas.getContext('2d');
ctx.globalCompositeOperation = 'xor';
#SamiHult's answer is a good answer. Using globalCompositeOperation will do the trick. Here comes a demo:
const canvas = document.getElementById("canvas");
const ctx = canvas.getContext("2d");
let cw = canvas.width = 300,
cx = cw / 2;
let ch = canvas.height = 300,
cy = ch / 2;
// the mouse
let m = {}
// draw a black circle
ctx.beginPath();
ctx.arc(100,100,45,0,2*Math.PI);
ctx.fill();
canvas.addEventListener("mousemove",(evt)=>{
m = oMousePos(canvas, evt);
ctx.clearRect(0,0,cw,ch);
// draw a circle stroked black following the mouse
drawCircle(m);
// draw a black circle
ctx.beginPath();
ctx.arc(100,100,45,0,2*Math.PI);
ctx.fill();
// the important part:
ctx.globalCompositeOperation = "xor";
})
function drawCircle(p){
ctx.beginPath();
ctx.arc(p.x,p.y,10,0,2*Math.PI);
ctx.stroke();
}
function oMousePos(canvas, evt) {
var ClientRect = canvas.getBoundingClientRect();
return { //objeto
x: Math.round(evt.clientX - ClientRect.left),
y: Math.round(evt.clientY - ClientRect.top)
}
}
canvas {
border:1px solid;
}
<canvas id="canvas"></canvas>
I have canvas. I have a ctx. I want the ctx to draw in the center of the canvas, however
canvas.height/2
canvas.width/2
does not work
http://jsfiddle.net/3hnbarcg/3/
Also, I need proof It works before I can accept
Have a look at following code.
var centerPointWidth = 10;
var centerPointHeight = 10;
var canvas = document.getElementById("c");
var ctx = canvas.getContext("2d");
ctx.fillStyle = "#FF0000";
ctx.fillRect(0, 0, canvas.width, canvas.height);
ctx.fillStyle = "#000000";
ctx.fillRect((canvas.width / 2) - (centerPointWidth / 2), (canvas.height / 2) - (centerPointHeight / 2), centerPointWidth, centerPointHeight);
<canvas id="c"></canvas>
It works perfectly to draw a rectangle at the center of the canvas.
I made a cylinder gauge, very similar to this one:
It is drawn using about 7 or so functions... mine is a little different. It is very fleixble in that I can set the colors, transparency, height, width, whether there is % text shown and a host of other options. But now I have a need for the same thing, but all rotated 90 deg so that I can set the height long and the width low to generate something more like this:
I found ctx.rotate, but no mater where it goes all the shapes fall apart.. ctx.save/restore appears to do nothing, I tried putting that in each shape drawing function. I tried modifying, for example, the drawOval function so that it would first rotate the canvas if horizontal was set to one; but it appeared to rotate it every single iteration, even with save/restore... so the top cylinder would rotate and the bottom would rotate twice or something. Very tough to tell what is really happening. What am I doing wrong? I don't want to duplicate all this code and spend hours customizing it, just to produce something I already have but turned horizontal. Erg! Help.
Option 1
To rotate everything just apply a transform to the element itself:
canvas.style.transform = "rotate(90deg)"; // or -90 depending on need
canvas.style.webkitTransform = "rotate(90deg)";
Option 2
Rotate context before drawing anything and before using any save(). Unlike the CSS version you will first need to translate to center, then rotate, and finally translate back.
You will need to make sure width and height of canvas is swapped before this is performed.
ctx.translate(ctx.canvas.width * 0.5, ctx.canvas.height * 0.5); // center
ctx.rotate(Math.PI * 0.5); // 90°
ctx.translate(-ctx.canvas.width * 0.5, -ctx.canvas.height * 0.5);
And of course, as an option 3, you can recalculate all your values to go along the other axis.
Look at the rotate function in this example. You want to do a translation to the point you want to rotate around.
example1();
example2();
function rotate(ctx, degrees, x, y, fn) {
ctx.save();
ctx.translate(x, y);
ctx.rotate(degrees * (Math.PI / 180));
fn();
ctx.restore();
}
function rad(deg) {
return deg * (Math.PI / 180);
}
function example2() {
var can = document.getElementById("can2");
var ctx = can.getContext('2d');
var w = can.width;
var h = can.height;
function drawBattery() {
var percent = 60;
ctx.beginPath();
ctx.arc(35,50, 25,0,rad(360));
ctx.moveTo(35+percent+25,50);
ctx.arc(35+percent,50,25,0,rad(360));
ctx.stroke();
ctx.beginPath();
ctx.fillStyle = "rgba(0,255,0,.5)";
ctx.arc(35,50,25,0,rad(360));
ctx.arc(35+percent,50,25,0,rad(360));
ctx.rect(35,25,percent,50);
ctx.fill();
ctx.beginPath();
ctx.lineWidth = 2;
ctx.strokeStyle = "#666666";
ctx.moveTo(135,25);
ctx.arc(135,50,25, rad(270), rad(269.9999));
//ctx.moveTo(35,75);
ctx.arc(35,50,25,rad(270),rad(90), true);
ctx.lineTo(135,75);
ctx.stroke();
}
drawBattery();
can = document.getElementById("can3");
ctx = can.getContext('2d');
w = can.width;
h = can.height;
rotate(ctx, -90, 0, h, drawBattery);
}
function example1() {
var can = document.getElementById('can');
var ctx = can.getContext('2d');
var color1 = "#FFFFFF";
var color2 = "#FFFF00";
var color3 = "rgba(0,155,255,.5)"
var text = 0;
function fillBox() {
ctx.save();
ctx.fillStyle = color3;
ctx.fillRect(0, 0, can.width / 2, can.height);
ctx.restore();
}
function drawBox() {
ctx.save();
ctx.beginPath();
ctx.strokeStyle = ctx.fillStyle = color1;
ctx.rect(10, 10, 50, 180);
ctx.font = "30px Arial";
ctx.fillText(text, 25, 45);
ctx.stroke();
ctx.beginPath();
ctx.strokeStyle = color2;
ctx.lineWidth = 10;
ctx.moveTo(10, 10);
ctx.lineTo(60, 10);
ctx.stroke();
ctx.restore();
}
fillBox();
rotate(ctx, 90, can.width, 0, fillBox);
text = "A";
drawBox();
color1 = "#00FFFF";
color2 = "#FF00FF";
text = "B";
rotate(ctx, 90, can.width, 0, drawBox);
centerRotatedBox()
function centerRotatedBox() {
ctx.translate(can.width / 2, can.height / 2);
for (var i = 0; i <= 90; i += 10) {
var radians = i * (Math.PI / 180);
ctx.save();
ctx.rotate(radians);
ctx.beginPath();
ctx.strokeStyle = "#333333";
ctx.rect(0, 0, 50, 50)
ctx.stroke();
ctx.restore();
}
}
}
#can,
#can2,
#can3 {
border: 1px solid #333333
}
<canvas id="can" width="200" height="200"></canvas>
<canvas id="can2" width="200" height="100"></canvas>
<canvas id="can3" width="100" height="200"></canvas>