How to draw a selection rectangle with drag and drop on a HTML canvas?

Question

I'd like to draw a selection rectangle (with mouse drag and drop, like with MS Paint) on a HTML canvas.

The following snippet works, but with two different canvas.

Question: How to do this on a single canvas, such that the image photo of the canvas can later be changed without removing the selection rectangle?

Should we keep 2 different canvas (if so, how to place the #2 exactly on top of the first one, without absolute positioning for both)?
Or is it possible in a simpler way with a single canvas? (and something like ctx.globalCompositeOperation = "source-over"; or a similar method)

var c1 = document.getElementById("c1"), c2 = document.getElementById("c2");
var ctx1 = c1.getContext("2d"), ctx2 = c2.getContext("2d");
ctx2.setLineDash([5, 5]);
var origin = null;
window.onload = () => { ctx1.drawImage(document.getElementById("img"), 0, 0); }
c1.onmousedown = e => { origin = {x: e.offsetX, y: e.offsetY}; };
window.onmouseup = e => { origin = null; };
c1.onmousemove = e => { 
    if (!!origin) { 
        ctx2.strokeStyle = "#ff0000";
        ctx2.clearRect(0, 0, c2.width, c2.height);
        ctx2.beginPath();
        ctx2.rect(origin.x, origin.y, e.offsetX - origin.x, e.offsetY - origin.y); 
        ctx2.stroke(); 
    } 
};

#img { display: none; }
#c1 { border: 1px solid green; }
#c2 { border: 1px solid blue; }

<img id="img" src="https://i.imgur.com/okSIKkW.jpg">
<canvas id="c1" height="200" width="200"></canvas>
<canvas id="c2" height="200" width="200"></canvas>

Cheeseparing

Answer 1

There might be a better solution with one single <canvas> and multiple layers (if so, feel free to post another answer), but I finally used a method similar to html5 - canvas element - Multiple layers with an overlay of 2 canvas:

var c1 = document.getElementById("c1"), c2 = document.getElementById("c2");
var ctx1 = c1.getContext("2d"), ctx2 = c2.getContext("2d");
ctx2.setLineDash([5, 5]);
var origin = null;
window.onload = () => { ctx1.drawImage(document.getElementById("img"), 0, 0); }
c2.onmousedown = e => { origin = {x: e.offsetX, y: e.offsetY}; };
window.onmouseup = e => { origin = null; };
c2.onmousemove = e => { 
    if (!!origin) { 
        ctx2.strokeStyle = "#ff0000";
        ctx2.clearRect(0, 0, c2.width, c2.height);
        ctx2.beginPath();
        ctx2.rect(origin.x, origin.y, e.offsetX - origin.x, e.offsetY - origin.y); 
        ctx2.stroke(); 
    } 
};

#img { display: none; }
#canvas-container { position: relative; }
canvas { position: absolute; left: 0; top: 0; }
#c1 { z-index: 0; }
#c2 { z-index: 1; }

<img id="img" src="https://i.imgur.com/okSIKkW.jpg">
<div id="canvas-container">
<canvas id="c1" height="200" width="200"></canvas>
<canvas id="c2" height="200" width="200"></canvas>
</div>

Cheeseparing

My solution does use single canvas. – Underproof

@Underproof Yes, I added a comment on the other answer. – Cheeseparing

Answer 2

The following solution uses a single canvas and rerenders on mouse movement.

https://jsfiddle.net/uhtdv0mj/3/

const c1 = document.getElementById("c1");
const ctx1 = c1.getContext("2d");
    
const drawImage = () => {
    ctx1.drawImage(document.getElementById("img"), 0, 0);
};
    
const drawSelection = (e) => {
    ctx1.strokeStyle = "#000";
    ctx1.beginPath();
    ctx1.rect(origin.x, origin.y, e.offsetX - origin.x, e.offsetY - origin.y); 
    ctx1.stroke();
};
    
const clear = () => {
    ctx1.clearRect(0, 0, c1.width, c1.height);
};
    
const render = (e) => {
    clear(); 
    drawImage();
        
    if (origin) drawSelection(e);
};
    
window.onload = drawImage;
    
let origin = null;
c1.onmousedown = e => { origin = {x: e.offsetX, y: e.offsetY}; };
c1.onmouseup = e => { origin = null; render(e); };
c1.onmousemove = render;

Underproof

Thank you for your answer. Interesting solution, but you re-draw both the full image + the selection on each mousemove, I would like to avoid repainting the whole image (which can be HD image with multiple megapixels in my case) on each mousemove. – Cheeseparing

@Cheeseparing I see, in that case 2 canvases on top of each-other is the solution, the back one containing the image, with a single canvas you'd have to re-draw. – Underproof