I created a plane by using
this.plane = BABYLON.MeshBuilder.CreatePlane("plane", { width: 0.5, height: 10 }, scene, true);
and then I try to modify the plane height in rendering, but the height is not changed.
I think you are looking for scaling. Here is the link to Babylon.js Scaling. Essentially, you do
objectName.scaling.x = 1.50 //This is a scaling multiplier (you can substitute x, y, and z)
//Or, if you want to scale in more than one direction, you can use a vector3
objectName.scaling = new Babylon.Vector3(1.5,.5, 2)
This should let you dynamically change the dimensions of your object. If you wanted the make the change visibly noticeable (instead of jumping straight to the correct size) you could also add an animation to it.
Related
For example, if you go to Twitter and click on an image, you can see they have a nice color that is close to what you see on the image. I tried looking up ways to achieve this as well as trying to figure it out on my own but no luck. I'm not sure if there's a color: relative property or not.
if you want to use the a colour that exists in your image and set it as a background colour you need to use the canvas element in the following manner:
HTML (this is your image)
<img src="multicolour.jpg" id="mainImage">
JS
window.onload = function() {
// get the body element to set background (this can change dependending of your needs)
let body = document.getElementsByTagName("body")
// get references to the image element that contains the picture you want to match with background
let referenceImage = document.getElementById("mainImage");
// create a canvas element (but don't add it to the page)
let canvas = document.createElement("canvas");
// make the canvas size the same as your image
canvas.width = referenceImage.offsetWidth
canvas.height = referenceImage.offsetHeight
// create the canvas context
let context = canvas.getContext('2d')
// usage your image reference to draw the image in the canvas
context.drawImage(referenceImage,0,0);
// select a random X and Y coordinates inside the drawn image in the canvas
// (you don't have to do this one, but I did to demonstrate the code)
let randomX = Math.floor(Math.random() * (referenceImage.offsetWidth - 1) + 1)
let randomY = Math.floor(Math.random() * (referenceImage.offsetHeight - 1) + 1)
// THIS IS THE MOST IMPORTANT LINE
// getImageData takes 4 arguments: coord x, coord y, sample size w, and sample size h.
// in our case the sample size is going to be of 1 pixel so it retrieves only 1 color
// the method gives you the data object which constains and array with the r, b, g colour data from the selected pixel
let color = context.getImageData(randomX, randomY, 1, 1).data
// use the data to dynamically add a background color extracted from your image
body[0].style.backgroundColor = `rgb(${color[0]},${color[1]},${color[2]})`
}
here is a gif of the code working... hopefully this helps
UPDATE
Here is the code to select two random points and create a css3 background gradient
window.onload = function() {
// get the body element to set background (this can change dependending of your needs)
let body = document.getElementsByTagName("body")
// get references to the image element that contains the picture you want to match with background
let referenceImage = document.getElementById("mainImage");
// create a canvas element (but don't add it to the page)
let canvas = document.createElement("canvas");
// make the canvas size the same as your image
canvas.width = referenceImage.offsetWidth
canvas.height = referenceImage.offsetHeight
// create the canvas context
let context = canvas.getContext('2d')
// usage your image reference to draw the image in the canvas
context.drawImage(referenceImage,0,0);
// select a random X and Y coordinates inside the drawn image in the canvas
// (you don't have to do this one, but I did to demonstrate the code)
let randomX = Math.floor(Math.random() * (referenceImage.offsetWidth - 1) + 1)
let randomY = Math.floor(Math.random() * (referenceImage.offsetHeight - 1) + 1)
// THIS IS THE MOST IMPORTANT LINE
// getImageData takes 4 arguments: coord x, coord y, sample size w, and sample size h.
// in our case the sample size is going to be of 1 pixel so it retrieves only 1 color
// the method gives you the data object which constains and array with the r, b, g colour data from the selected pixel
let colorOne = context.getImageData(randomX, randomY, 1, 1).data
// THE SAME TO OBTAIN ANOTHER pixel data
let randomX2 = Math.floor(Math.random() * (referenceImage.offsetWidth - 1) + 1)
let randomY2 = Math.floor(Math.random() * (referenceImage.offsetHeight - 1) + 1)
let colorTwo = context.getImageData(randomX2, randomY2, 1, 1).data
// use the data to dynamically add a background color extracted from your image
//body[0].style.backgroundColor = `rgb(${allColors[0]},${allColors[1]},${allColors[2]})`
body[0].style.backgroundImage = `linear-gradient(to right, rgb(${colorOne[0]},${colorOne[1]},${colorOne[2]}),rgb(${colorTwo[0]},${colorTwo[1]},${colorTwo[2]}))`;
}
The following are your options.
1. Use an svg.
As far as I know there's no way to have javascript figure out what color is being used in a png and set it as a background color. But you can work the other way around. You can have javascript set the background color and an svg image to be the same color.
See this stackoverflow answer to learn more about modifying svgs with javascript.
2. Use a custom font.
There are fonts out there that provide a bunch of icons instead of letters, you can also create your own font if you feel so inclined to do so. With css you just have to set the font-color of that icon to be the same as the background-color of your other element.
Font Awesome provides a bunch of useful custom icons. If the image you need to use happens to be similar to one of theirs, you can just go with them.
3. Use canvas
If you really want to spend the time to code it up you can use a html <canvas/> element and put the image into it. From there you can inspect certain details about the image like its color, then apply that color to other elements. I won't go into too much detail about using this method as it seems like it's probably overkill for what you're trying to do, but you can read up more about from this stackoverflow answer.
4. Just live with it.
Not a fun solution, but this is usually the option I go with. You simply have to hard-code the color of the image into your css and live with it. If you ever need to modify the color of the image, you have to remember to update your css also.
I am trying to convert a mouse event to pixel coordinates within a video. By pixel coordinates, I mean coordinates relative to the original video resolution.
My video element has object-fit: contain, which means that the top left corner of the video is not necessarily located at position (0,0), as this picture shows:
If I click on the top-left corner of the white section in this video then I want to get (0,0), but in order to do this I need to discover the offset of the video content (white area) relative to the video element (black border).
How can I recover this offset?
I am already aware of width, height, videoWidth, and videoHeight, but these only let me account for the scaling, not the offset.
The offset can be deduced. I think this kind of code should do the trick:
if(videoHeight/height > videoWidth/width){
scale = videoHeight/height;
offsetX = (videoWidth - width*scale)/2;
offsetY = 0;
}
else{
scale = videoWidth/width;
offsetY = (videoHeight - height*scale)/2;
offsetX = 0;
}
I was also interested in getting the actual pixel positions from mouse or touch events when using object-fit, and this is the only result I found when searching. Although I suspect it is probably too late to be helpful to you, I thought I'd answer in case anybody else comes across this in future like I did.
Because I'm working on code with other people, I needed a robust solution that would work even if someone changed or removed the object-fit or object-property in the css
The approach that I took was:
Implement the cover, contain etc algorithms myself, just functions doing math, not dependent on the DOM
Use getComputedStyle to get the element's objectFit and objectPosition properties
Use .getBoundingClientRect() to get the DOM pixel size of the element
Pass the element's current objectFit, objectPosition, its DOM pixel size and it's natural pixel size to my function to figure out where the fitted rectangle sat within the element
You then have enough information to transform the event point to a pixel location
There's more code than would comfortably fit here, but getting the size of the fitted rectangle for cover or contain is something like:
if ( fitMode === 'cover' || fitMode === 'contain' ) {
const wr = parent.width / child.width
const hr = parent.height / child.height
const ratio = fitMode === 'cover' ? Math.max( wr, hr ) : Math.min( wr, hr )
const width = child.width * ratio
const height = child.height * ratio
const size = { width, height }
return size
}
// handle other object-fit modes here
Hopefully this gives others a rough idea of how to solve this problem themselves, alternately I have published the code at the link below, it supports all object-fit modes and it includes examples showing how to get the actual pixel point that was clicked:
https://github.com/nrkn/object-fit-math
I'm trying to draw scaled image on canvas in javafx. Using this code:
Image image = ...;
canvas.setWidth(scale * width);
canvas.setHeight(scale * height);
GraphicsContext gc = canvas.getGraphicsContext2D();
gc.drawImage(image, 0, 0, scale * width, scale * height);
// this gives same result
// gc.scale(scale, scale);
// gc.drawImage(editableImage, 0, 0, width, height);
It works really fast but makes blurred images like this:
This is not what I'd like to see. Instead I want to get this picture:
Which can be drawn by manually setting each pixel color with such code:
PixelReader reader = image.getPixelReader();
PixelWriter writer = gc.getPixelWriter();
for (int y = 0; y < scale * height; ++y) {
for (int x = 0; x < scale * width; ++x) {
writer.setArgb(x, y, reader.getArgb(x / scale, y / scale));
}
}
But I cannot use this approach as it's too slow. It took couple of seconds to draw 1Kb image scaled 8 times. So I ask if there's any way to disable this blurry effect for drawing on canvas?
UPD 10/07/2019:
Looks like the issue is fixed! Now GraphicsContext should have property "image smoothing" controlling this behavior.
INITIAL ANSWER
I guess I've found answer to my question. As this issue says that there's no way to specify filtering options in graphics context.
Description:
When drawing an image in a GraphicsContext using the drawImage()
method to enlarge a small image to a larger canvas, the image is being
interpolated (possibly using a bilinear or bicubic algorithm). But
there are times like when rendering color maps (temperature,
zooplancton, salinity, etc.) or some geographical data (population
concentration, etc.) where we want to have no interpolation at all
(ie: use the nearest neighbor algorithm instead) in order to represent
accurate data and shapes.
In Java2D, this is possible by setting the appropriate
RenderingHints.KEY_RENDERING on the Graphics2D at hand. Currently on
JavaFX's GraphicsContext there is no such way to specify how the image
is to be interpolated.
The same applies when shrinking images too.
This could be expanded to support a better form of smoothing for the
"smooth" value that is available in both Image and ImageView and that
does not seem to work very well currently (at least on Windows).
The issue was created in 2013 but it's still untouched so unlikely it will be resolved soon.
i wrote the following code for drawing a rotate rectangle
var s:UIComponent = new UIComponent();
s.graphics.lineStyle(1, 0x0000FF);
s.graphics.drawRect(50, 50, 200, 200);
s.rotation = 30;
template.addChild(s);
where template is a canvas. Its rotate nicely but the problem is the position is not in right place. i.e. it is not in (50,50) after rotate. How can i solve this problem?
Thanks in advance.
What you're seeing is the default rotation around the origin in Flex (in your case the X,Y coordinates of 50,50) I am assuming that you want to rotate around the center (as I recently had to do). There is the jury rig way to do it, by adjusting the origin point based on rotation angle. Then there is the rotate effect:
import mx.effects.Rotate;
var s:UIComponent = new UIComponent();
var rotator:Rotate = new Rotate(s);
s.graphics.lineStyle(1, 0x0000FF);
s.graphics.drawRect(50, 50, 200, 200);
rotator.angleFrom = 0;
rotator.angleTo = 30;
rotator.originX = s.width/2;
rotator.originY = s.height/2;
template.addChild(s);
rotator.play();
Now I've noticed some problems with this that required me to set the width and height of the rotated object again after the play() method, but other than that I get the ideal rotation situation.
Another downside to this is that it's an effect, which means it visually rotates the object. I will post back when I correct that, if you don't want to see the object rotate.
The Answer is duration
just by adding rotator.duration = 1 before play it happens so quick the user won't see it. 1 being 1 millisecond. I tried 0, but that resulted in no rotation occurring. Obviously if you want to see the effect in action you can increase that length of time by using any value in milliseconds.
Ok, so I've been trying to get this concept to work for the day now and have had zero luck. I am not so good with math, so any help will be appreciated. I am trying to rotate a centered container from it's center. The problem with this code is when I use the rotatePicture method, it doesn't rotate from the center, instead it rotates from the box's top-left corner. Here's the code...
import mx.effects.Rotate;
private function init():void
{
calculateCoordinates();
}
private function rotateBox():void
{
var m:Matrix = myBox.transform.matrix;
var centerX:Number = myBox.width / 2;
var centerY:Number = myBox.height / 2;
var centerPoint:Point = new Point(centerX, centerY);
var transformPoint:Point= m.transformPoint(centerPoint);
m.translate(-transformPoint.x, -transformPoint.y);
m.rotate(90 * Math.PI / 180);
m.translate(transformPoint.x, transformPoint.y);
myBox.transform.matrix = m;
this.callLater(calculateCoordinates);
//calculateCoordinates();
}
private function calculateCoordinates():void
{
var x : Number = (myBox.parent.width - myBox.width) / 2;
x = x < 0 ? 0 : x;
var y : Number = (myBox.parent.height - myBox.height) / 2;
y = y < 0 ? 0 : y;
myBox.move(x, y);
}
Ok, this was a bit tricky and i'm working out a couple of details, but in case anyone had a similar issue, I found the general answer. Just took a movie break to refresh the brain...
I had to place a variable for how many turns the canvas had rotated which was easy since I was restricting the degrees to only 90. Then I place in a switch statement that tested the turns variable and recalculated the (x,y) coordinates based off of the turns. Since I knew that the Rotate class would create a cooler effect and end with the same result, I ended up using that instead.
Cheers!
I don't know what your background is, but in my experience this is a classic case of "out-thinking yourself".
You've written quite a bit of code to do something that is actually native to the Flash display API's.
Since you seem to be using Flex I'll just tell you that the simple way to achieve this is to dynamically reposition your display clip contents so that the center of your contents is at the 0,0 point of your clip.
This gets harder the more content you have, but if you just have something like an image or a button or what have you, it's really easy to just calculate the height and width, then divide by 2 and subtract.
Then the rotation property will work just fine.
In Flash it's even easier because you can just make a new clip, bind your class to the clip, and place all yours contents in the Flash authoring tool positioned properly for rotation to work as expected.
Yeah, what Back2Dos said.
Personally, I'd wrap the container in another sprite, position it so its center is at (0,0) in that sprites coordinate space, and then just rotate the sprite ... it's simple and robust ...
I'd like to use <s:Rotate> to rotate center. Hope useful to you.