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Image magnifier in Angular

I am trying to implement a image maginfier on hover.I tried to replicate the code as in w3schools which is purely of Javascript.I am trying to implement the following code in Angular
https://www.w3schools.com/howto/tryit.asp?filename=tryhow_js_image_magnifier_glass
I used the above method in typescript and called it from ngOnInit in Angular but i am not able to get any result from the method.I have ensured the id is passed correctly and validated the method is being called .But still not able to get any result .I wish not to use any npm packages for magnifier since most of them had bugs.
component.ts
ngOnInit(){
this.magnify(imgID, zoom)
}
magnify(imgID, zoom) {
var img, glass, w, h, bw;
img = document.getElementById(imgID);
/*create magnifier glass:*/
glass = document.createElement("DIV");
glass.setAttribute("class", "img-magnifier-glass");
/*insert magnifier glass:*/
img.parentElement.insertBefore(glass, img);
/*set background properties for the magnifier glass:*/
glass.style.backgroundImage = "url('" + img.src + "')";
glass.style.backgroundRepeat = "no-repeat";
glass.style.backgroundSize = (img.width * zoom) + "px " + (img.height * zoom) + "px";
bw = 3;
w = glass.offsetWidth / 2;
h = glass.offsetHeight / 2;
/*execute a function when someone moves the magnifier glass over the image:*/
glass.addEventListener("mousemove", moveMagnifier);
img.addEventListener("mousemove", moveMagnifier);
/*and also for touch screens:*/
glass.addEventListener("touchmove", moveMagnifier);
img.addEventListener("touchmove", moveMagnifier);
function moveMagnifier(e) {
var pos, x, y;
/*prevent any other actions that may occur when moving over the image*/
e.preventDefault();
/*get the cursor's x and y positions:*/
pos = getCursorPos(e);
x = pos.x;
y = pos.y;
/*prevent the magnifier glass from being positioned outside the image:*/
if (x > img.width - (w / zoom)) {x = img.width - (w / zoom);}
if (x < w / zoom) {x = w / zoom;}
if (y > img.height - (h / zoom)) {y = img.height - (h / zoom);}
if (y < h / zoom) {y = h / zoom;}
/*set the position of the magnifier glass:*/
glass.style.left = (x - w) + "px";
glass.style.top = (y - h) + "px";
/*display what the magnifier glass "sees":*/
glass.style.backgroundPosition = "-" + ((x * zoom) - w + bw) + "px -" + ((y * zoom) - h + bw) + "px";
}
function getCursorPos(e) {
var a, x = 0, y = 0;
e = e || window.event;
/*get the x and y positions of the image:*/
a = img.getBoundingClientRect();
/*calculate the cursor's x and y coordinates, relative to the image:*/
x = e.pageX - a.left;
y = e.pageY - a.top;
/*consider any page scrolling:*/
x = x - window.pageXOffset;
y = y - window.pageYOffset;
return {x : x, y : y};
}
}

Here is a working stackblitz as per your requirements. It shows the implementation of image zoom functionality mentioned on w3school.
https://stackblitz.com/edit/angular-w3school-image-magnification
You have not shown your html and css files. So, I am not totally sure but the following might be the reason why zoom is not working for you.
Problem is that img-magnifier-glass div element is created using classical DOM method 'document.createElement'. And then, class 'img-magnifier-glass' is applied to it, again using a classical DOM method (setAttribute). But, in angular, styles are encapsulated. So, if you have added a class definition of '.img-magnifier-glass' in app.component.css then that class won't be applied to glass div since it is not mentioned in the template (app.component.html). See this for more info - https://github.com/angular/angular/issues/7845
To fix this, you can either move definition of class '.img-magnifier-glass' to styles.css (Where global styles are defined)
or you can keep the class in app.component.css but use pseudo-selector ::ng-deep with it. Applying the ::ng-deep pseudo-class to any CSS rule completely disables view-encapsulation for that rule. Any style with ::ng-deep applied becomes a global style.
::ng-deep .img-magnifier-glass {
}
or you can stop style encapsulation for component by specifying
#Component({
// ...
encapsulation: ViewEncapsulation.None, //<<<<< this one!
styles: [
// ...
]
})
It will be better will be if you use Renderer2 (https://angular.io/api/core/Renderer2) instead for creating dynamic elements like glass element here. Renderer2 will take care of correctly encapsulating class applied on elements created using it.

I ran into the same issue but for image zoomer which has the same implementation of magnifier. I got it working by tweaking the css and ts code (which was taken from the magnifier and combined with the zoomer). the answer above is not what I am looking for coz I need the zoomed img to be in a different box and not on top of the img itself. Here is the stackblitz code which I modified to suit what I need.

For implementing image magnification feature like W3School and Amazon in Angular, you can use npm package ng-img-magnifier.
Here is the working DEMO.
<ng-img-magnifier
[thumbImage]='img' [fullImage]='img2'
[top]='top' [right]='right'
[lensWidth]='lensewidth' [lensHeight]='lensheight'
[imgWidth]='imgWidth' [imgHeight]='imgheight'
[resultWidth]='resultWidth' [resultHeight]='resultheight'
>
</ng-img-magnifier>
This package comes with full customization options.
Hopefully this will resolve your issue.

Find new left and top corresponding to original left and top on CSS rotate

Suppose my div has left:200px and top:400px, after I apply a rotate transform of suppose 90 deg the above top and left positions no more point to the old positions. Now how can we calculate the new top and left for the transformed div which are equivalent to the left and top positions of the non-transformed div after rotation.

Edited answer
Besides the starting position of the corner point (top-left in your example), and the rotation angle, we also need to know the position of the reference point of the rotation. This is the point around which we rotate the div (CSS calls it transform-origin). If you don't specify it, then normally, the centre of mass of the element is used.
I don't know of any JavaScript method that simply calculates it for you, but I can show you its Math, and a simple JS implementation.
Math
P: original position of the corner point, with (Px, Py) coordinates
O: reference point of the rotation, with (Ox, Oy) coordinates
Calculate the original position of P, relative to O.
x = Px - Ox
y = Py - Oy
Calculate the rotated position of P, relative to O.
x' = x * cos(angle) - y * sin(angle)
y' = x * sin(angle) + y * cos(angle)
Convert this position back to the original coordinate system.
Px' = x' + Ox
Py' = y' + Oy
If you're not aware of the formulas in step #2, you can find an explanation here.
JavaScript implementation
function rotatedPosition(pLeft, pTop, oLeft, oTop, angle){
// 1
var x = pLeft - oLeft;
var y = pTop - oTop;
// 2
var xRot = x * Math.cos(angle) - y * Math.sin(angle);
var yRot = x * Math.sin(angle) + y * Math.cos(angle);
// 3
var pLeftRot = xRot + oLeft;
var pTopRot = yRot + oTop
return {left: pLeftRot, top: pTopRot};
}
rotatedPosition requires you to define the original position of the point and the reference point, plus the angle.
In case you need a method which takes only a single argument, the div element itself, and computes the rest for you, then you can do something like:
function divTopLeftRotatedPosition(div){
var pLeft = // ...
var pTop = // ...
var width = // ...
var height = // ...
var angle = // ...
return rotatedPosition(pLeft, pTop, pLeft + width / 2, pTop + height / 2, angle);
}

How do I find the position of an element in raphael?

I am using raphael to animate a ball and need to find out the position after stopping the animation, here is the code I have:
ball.stop();
console.log(ball.attrs.transform);
// t399.6625490203161,180r180
how do I convert this string into the x,y position of the ball?

You can use Element.getBBox for this:
var x = ball.getBBox().x;
var y = ball.getBBox().y;
var r = ball.getBBox().height / 2;
console.log(x, y, r);
But beware:
The coordinates (element.getBBox().x, element.getBBox().y) will refer to the upper left corner of the bounding box.
If the ball is a circle, that point will not even be contained in the circle and it might be more sensible to work with the coordinates of the center.
Given the bounding box bbox, to those would be
var x = bbox.x + bbox.width / 2,
y = bbox.y + bbox.height / 2;

Calculate the bounding box's X, Y, Height and Width of a rotated element via JavaScript

Basically I'm asking this question for JavaScript: Calculate Bounding box coordinates from a rotated rectangle
In this case:
iX = Width of rotated (blue) HTML element
iY = Height of rotated (blue) HTML element
bx = Width of Bounding Box (red)
by = Height of Bounding Box (red)
x = X coord of Bounding Box (red)
y = Y coord of Bounding Box (red)
iAngle/t = Angle of rotation of HTML element (blue; not shown but
used in code below), FYI: It's 37 degrees in this example (not that it matters for the example)
How does one calculate the X, Y, Height and Width of a bounding box (all the red numbers) surrounding a rotated HTML element (given its width, height, and Angle of rotation) via JavaScript? A sticky bit to this will be getting the rotated HTML element (blue box)'s original X/Y coords to use as an offset somehow (this is not represented in the code below). This may well have to look at CSS3's transform-origin to determine the center point.
I've got a partial solution, but the calculation of the X/Y coords is not functioning properly...
var boundingBox = function (iX, iY, iAngle) {
var x, y, bx, by, t;
//# Allow for negetive iAngle's that rotate counter clockwise while always ensuring iAngle's < 360
t = ((iAngle < 0 ? 360 - iAngle : iAngle) % 360);
//# Calculate the width (bx) and height (by) of the .boundingBox
//# NOTE: See https://stackoverflow.com/questions/3231176/how-to-get-size-of-a-rotated-rectangle
bx = (iX * Math.sin(iAngle) + iY * Math.cos(iAngle));
by = (iX * Math.cos(iAngle) + iY * Math.sin(iAngle));
//# This part is wrong, as it's re-calculating the iX/iY of the rotated element (blue)
//# we want the x/y of the bounding box (red)
//# NOTE: See https://stackoverflow.com/questions/9971230/calculate-rotated-rectangle-size-from-known-bounding-box-coordinates
x = (1 / (Math.pow(Math.cos(t), 2) - Math.pow(Math.sin(t), 2))) * (bx * Math.cos(t) - by * Math.sin(t));
y = (1 / (Math.pow(Math.cos(t), 2) - Math.pow(Math.sin(t), 2))) * (-bx * Math.sin(t) + by * Math.cos(t));
//# Return an object to the caller representing the x/y and width/height of the calculated .boundingBox
return {
x: parseInt(x), width: parseInt(bx),
y: parseInt(y), height: parseInt(by)
}
};
I feel like I am so close, and yet so far...
Many thanks for any help you can provide!
TO HELP THE NON-JAVASCRIPTERS...
Once the HTML element is rotated, the browser returns a "matrix transform" or "rotation matrix" which seems to be this: rotate(Xdeg) = matrix(cos(X), sin(X), -sin(X), cos(X), 0, 0); See this page for more info.
I have a feeling this will enlighten us on how to get the X,Y of the bounding box (red) based solely on the Width, Height and Angle of the rotated element (blue).
New Info
Humm... interesting...
Each browser seems to handle the rotation differently from an X/Y perspective! FF ignores it completely, IE & Opera draw the bounding box (but its properties are not exposed, ie: bx & by) and Chrome & Safari rotate the rectangle! All are properly reporting the X/Y except FF. So... the X/Y issue seems to exist for FF only! How very odd!
Also of note, it seems that $(document).ready(function () {...}); fires too early for the rotated X/Y to be recognized (which was part of my original problem!). I am rotating the elements directly before the X/Y interrogation calls in $(document).ready(function () {...}); but they don't seem to update until some time after(!?).
When I get a little more time, I will toss up a jFiddle with the example, but I'm using a modified form of "jquery-css-transform.js" so I have a tiny bit of tinkering before the jFiddle...
So... what's up, FireFox? That ain't cool, man!
The Plot Thickens...
Well, FF12 seems to fix the issue with FF11, and now acts like IE and Opera. But now I am back to square one with the X/Y, but at least I think I know why now...
It seems that even though the X/Y is being reported correctly by the browsers for the rotated object, a "ghost" X/Y still exists on the un-rotated version. It seems as though this is the order of operations:
Starting with an un-rotated element at an X,Y of 20,20
Rotate said element, resulting in the reporting of X,Y as 15,35
Move said element via JavaScript/CSS to X,Y 10,10
Browser logically un-rotates element back to 20,20, moves to 10,10 then re-rotates, resulting in an X,Y of 5,25
So... I want the element to end up at 10,10 post rotation, but thanks to the fact that the element is (seemingly) re-rotated post move, the resulting X,Y differs from the set X,Y.
This is my problem! So what I really need is a function to take the desired destination coords (10,10), and work backwards from there to get the starting X,Y coords that will result in the element being rotated into 10,10. At least I know what my problem is now, as thanks to the inner workings of the browsers, it seems with a rotated element 10=5!

I know this is a bit late, but I've written a fiddle for exactly this problem, on an HTML5 canvas:
http://jsfiddle.net/oscarpalacious/ZdQKg/
I hope somebody finds it useful!
I'm actually not calculating your x,y for the upper left corner of the container. It's calculated as a result of the offset (code from the fiddle example):
this.w = Math.sin(this.angulo) * rotador.h + Math.cos(this.angulo) * rotador.w;
this.h = Math.sin(this.angulo) * rotador.w + Math.cos(this.angulo) * rotador.h;
// The offset on a canvas for the upper left corner (x, y) is
// given by the first two parameters for the rect() method:
contexto.rect(-(this.w/2), -(this.h/2), this.w, this.h);
Cheers

Have you tried using getBoundingClientRect() ?
This method returns an object with current values of "bottom, height, left, right, top, width" considering rotations

Turn the four corners into vectors from the center, rotate them, and get the new min/max width/height from them.
EDIT:
I see where you're having problems now. You're doing the calculations using the entire side when you need to be doing them with the offsets from the center of rotation. Yes, this results in four rotated points (which, strangely enough, is exactly as many points as you started with). Between them there will be one minimum X, one maximum X, one minimum Y, and one maximum Y. Those are your bounds.

My gist can help you
Bounding box of a polygon (rectangle, triangle, etc.):
Live demo https://jsfiddle.net/Kolosovsky/tdqv6pk2/
let points = [
{ x: 125, y: 50 },
{ x: 250, y: 65 },
{ x: 300, y: 125 },
{ x: 175, y: 175 },
{ x: 100, y: 125 },
];
let minX = Math.min(...points.map(point => point.x));
let minY = Math.min(...points.map(point => point.y));
let maxX = Math.max(...points.map(point => point.x));
let maxY = Math.max(...points.map(point => point.y));
let pivot = {
x: maxX - ((maxX - minX) / 2),
y: maxY - ((maxY - minY) / 2)
};
let degrees = 90;
let radians = degrees * (Math.PI / 180);
let cos = Math.cos(radians);
let sin = Math.sin(radians);
function rotatePoint(pivot, point, cos, sin) {
return {
x: (cos * (point.x - pivot.x)) - (sin * (point.y - pivot.y)) + pivot.x,
y: (sin * (point.x - pivot.x)) + (cos * (point.y - pivot.y)) + pivot.y
};
}
let boundingBox = {
x1: Number.POSITIVE_INFINITY,
y1: Number.POSITIVE_INFINITY,
x2: Number.NEGATIVE_INFINITY,
y2: Number.NEGATIVE_INFINITY,
};
points.forEach((point) => {
let rotatedPoint = rotatePoint(pivot, point, cos, sin);
boundingBox.x1 = Math.min(boundingBox.x1, rotatedPoint.x);
boundingBox.y1 = Math.min(boundingBox.y1, rotatedPoint.y);
boundingBox.x2 = Math.max(boundingBox.x2, rotatedPoint.x);
boundingBox.y2 = Math.max(boundingBox.y2, rotatedPoint.y);
});
Bounding box of an ellipse:
Live demo https://jsfiddle.net/Kolosovsky/sLc7ynd1/
let centerX = 350;
let centerY = 100;
let radiusX = 100;
let radiusY = 50;
let degrees = 200;
let radians = degrees * (Math.PI / 180);
let radians90 = radians + Math.PI / 2;
let ux = radiusX * Math.cos(radians);
let uy = radiusX * Math.sin(radians);
let vx = radiusY * Math.cos(radians90);
let vy = radiusY * Math.sin(radians90);
let width = Math.sqrt(ux * ux + vx * vx) * 2;
let height = Math.sqrt(uy * uy + vy * vy) * 2;
let x = centerX - (width / 2);
let y = centerY - (height / 2);

straight line between two points

On a HTML canvas I have multiple points starting from 1 to N, this is basically a connect the dots application and is activated on touchstart.
There is validation so that they can only connect the dots from 1 and go to 2 (.. n). The issue is that right now is there is no validation that the line is a straight line and I am looking for an algorithm to do this, Here is what I have thought so far
For 2 points (x1,y1) to (x2,y2) get all the coordinates by finding the slope and using the formula y = mx + b
on touchmove get the x,y co-oridnates and make sure it is one of the points from the earlier step and draw a line else do not draw the line.
Is there a better way to do this or are there any different approaches that I can take ?

Edit: I originally misunderstood the question, it seems.
As far as validating the path: I think it would be easier just to have a function that determines whether a point is valid than calculating all of the values beforehand. Something like:
function getValidatorForPoints(x1, y1, x2, y2) {
var slope = (y2 - y1) / (x2 - x1);
return function (x, y) {
return (y - y1) == slope * (x - x1);
}
}
Then, given two points, you could do this:
var isValid = getValidatorForPoints(x1, y1, x2, y2);
var x = getX(), y = getY();// getX and getY get the user's new point.
if (isValid(x, y)) {
// Draw
}
This approach also gives you some flexibility—you could always modify the function to be less precise to accommodate people who don't quite draw a straight line but are tolerably close.
Precision:
As mentioned in my comment, you can change the way the function behaves to make it less exacting. I think a good way to do this is as follows:
Right now, we are using the formula (y - y1) == slope * (x - x1). This is the same as (slope * (x - x1)) - (y - y1) == 0. We can change the zero to some positive number to make it accept points "near" the valid line as so:
Math.abs((slope * (x - x1)) - (y - y1)) <= n
Here n changes how close the point has to be to the line in order to count.
I'm pretty sure this works as advertised and helps account for people's drawing the line a little crooked, but somebody should double check my math.

function drawGraphLine(x1, y1, x2, y2, color) {
var dist = Math.ceil(Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)));
var angle = Math.atan2(y2-y1, x2-x1)*180/Math.PI;
var xshift = dist - Math.abs(x2-x1);
var yshift = Math.abs(y1-y2)/2;
var div = document.createElement('div');
div.style.backgroundColor = color;
div.style.position = 'absolute';
div.style.left = (x1 - xshift/2) + 'px';
div.style.top = (Math.min(y1,y2) + yshift) + 'px';
div.style.width = dist+'px';
div.style.height = '3px';
div.style.WebkitTransform = 'rotate('+angle+'deg)';
div.style.MozTransform = 'rotate('+angle+'deg)';
}
// By Tomer Almog