I have a rounded square texture, drawn in paint, with black border. How can I change only the inner color without changing the border color?
EDIT: I did it, but I want to ask if it can be done without using another texture? I also wonder if there is a connection between this and 9patch as in scaling. Can we use that or is there something like that?
You can tint whatever it is what you are drawing, using SpriteBatch. This means that color you specify is multiplied with the color of the image. For example if your image is a white rounded rectangle with a black border and you tint it using the color purple then the inside will be purple and the border will remain black. You didn't provide enough information to be more specific. But if you are for example using the Sprite class then you can use the setColor method to tint it. Likewise if you are using the Image class. If you are drawing the "texture" directly using SpriteBatch then you can use the setColor method of the SpriteBatch.
Related
I am refering to a older question saying color blending with GDI+
Using GDI+ with Windows Forms, I want to be able to draw with a pen and blend color based on the destination pixel color.
For example, if I draw a line and it passes over black pixels, I want it to be a lighter color (like white for example) so that it's visible. When that same line passes over white pixels, it should be a darker color (black for example) so that it's still clearly visible.
the answers says to use a color matrix for transformation
so i started implementing it..
My image is present in raw data format in rgb48
Gdiplus::Bitmap image(input.width,input.height,input.width*6,PixelFormat48bppRGB,(unsigned char*)rgb48);
Gdiplus::Image *images= image.GetThumbnailImage(input.width,input.height);
Gdiplus::TextureBrush brush(images);
Gdiplus::Pen pen(&brush);
Gdiplus::ColorMatrix matrix={
-1.0f,0.0f,0.0f,0.0f,0.0f,
0.0f,-1.0f,0.0f,0.0f,0.0f,
0.0f,0.0f,-1.0f,0.0f,0.0f,
0.0f,0.0f,0.0f,1.0f,0.0f,
1.0f,1.0f,1.0f,0.0f,1.0f,
};
Gdiplus::Graphics gfx(&image1);
Gdiplus::ImageAttributes imageAttr;
imageAttr.SetColorMatrix(&matrix);
gfx.DrawImage(images,Gdiplus::Rect(0,0,input.width,input.height),0,0,1024,1024,Gdiplus::UnitPixel,&imageAttr);
I am not getting what i expect..Can some one help me in finding the mistake i m doing.
You can use the alpha component of a color to specify transparency, so that colors can be combined. However, if you want the combination to be something other than the alpha blend, you must draw the pixels yourself. You could first draw into a transparent bitmap, then render that onto the destination pixel by pixel.
I'm using texture and textureAtlas in libgdx.
This textures loads images of white circle.
How can I overlay its color with another color?
Or maybe their is another approach for this?
private final static TextureAtlas textureAtlas = new TextureAtlas(Gdx.files.internal("spritesheet.atlas"));
private final static TextureAtlas.AtlasRegion texture = textureAtlas.findRegion("Bubble.001");
EDIT:
As I said I have a white circle and I want to make it red (without the need for another image with a red circle)
You can use the lerp method in Color class.
actor.setColor(Color.WHITE.cpy().lerp(tintingColor, .5f));
lerp changes your Color object, so use the cpy command before it to preserve the original color. In this case I use WHITE as my original color, which preserves the color of your Actor's texture. lerp will overlay the tintingColor with a strength of 50% in this case.
When rendering textures that have an alpha-channel, a white border appears around the non-transparent part (the border seems to be the pixels that have an alpha > 0 and < 1):
The original texture is created in illustrator and exported as a png. here it is:
(well, seems stackoverflow altered the image, adjusting pixels that are not completely opaque/transparent, so here is a link)
it is probably the blending, though i dont know what is wrong with the setup:
gl.enable(gl.BLEND);
gl.blendFunc(gl.SRC_ALPHA, gl.ONE_MINUS_SRC_ALPHA);
[Update]
Here is a rendered version, where i added a alpha-gradient to the left part of the texture (so it is getting from 0 opacity to 1 until the half)
this texture is the only texture rendered at this position. it seems to be whitest around a=0.5. really weird. the background is just a cleared color:
gl.clearColor(0.603, 0.76, 0.804, 1.0);
gl.clear(gl.COLOR_BUFFER_BIT | gl.DEPTH_BUFFER_BIT);
// render objects here
the depth-function looks like:
gl.enable(gl.DEPTH_TEST);
gl.depthFunc(gl.LEQUAL);
any ideas? thanks a lot.
[Update 2]
Answering my own question: the effect occurs when the background-color of the canvas or the body of the html-page is white. I don't have an explanation, though.
Use premultiplied alpha and this problem will go away.
See: http://home.comcast.net/~tom_forsyth/blog.wiki.html#%5B%5BPremultiplied%20alpha%5D%5D
This is problem related to texturing linear interpolation. On borders, some interpolated pixels will take half white half green, and 0.5 alpha. You should modify your texture to extend your borders with one more green pixel, even if it is totally transparent.
What's your draw order? This looks like a depth buffering issue to me — you start with a white background, draw the thing with the border so that it's composited on the white, then draw the thing behind the thing with the border. Those areas where the border was blended with the original white background will have stored a value in the depth buffer equal to the depth of their plane, so when the object behind is subsequently drawn, its pixels are discarded in that area.
The general rule is to draw transparent objects after opaque objects, usually from back to front. If you're using additive blending then it's often good enough to disable the depth buffer after the opaque draw and draw them in any order.
When setting the FragColor in the shader, try multiplying the image RGB with the image alpha.
I'm working with SVG using the Raphael library. I can apply a fill colour to an object like so:
circle.attr({fill: "#ff0000"});
And this also works (though the Raphael documentation doesn't mention it):
circle.attr({fill: "url(pattern.png)"});
I am able to use transparent PNGs as fill patterns, and transparency works as expected. The svg object is completely transparent where the fill pattern image is transparent. But what I would like to do is specify both a fill pattern image and a fill colour, so that the colour would show through where the pattern image is transparent - similar to the 'background' property using CSS, for example. Is this possible with SVG?
You can define a pattern that has a rect with a fill, and an image that is your png on top of that rect. Then use the pattern as fill for the circle (or whatever element you want).
This means stepping outside of Raphaël, or extending it to do what you want. Note that what ({fill: "url(pattern.png)"}) does is to create a pattern element and and append an image element pointing to the given url. It's quite possible to hack Raphaël to allow you to pass a color too, and then you deal with that in the code that creates the pattern by creating a rect of the same dimensions as the image with the given fill color.
I should say that if you want it to work with IE<9 then you probably need to implement it in VML too.
Other options include drawing two shapes, one with color fill and the other with the raster image fill. Yet another is to make the png include the background color so that it's not transparent.
In XAML there is the concept of a [Border] object that behaves much like an SVG rectangle except that a XAML [Border] can contain a child element.
So I could create a [Border] with a black stroke and inside it have another border that had a white stroke and a fill of Green.
This gives the the look of a single rectangle with both a black border & then an inner white border.
How can you create this in SVG?
You need multiple paths. In this particular case you could 'cheat' by having the same path twice, with the lower path having a larger border that appears 'outside' the upper object; in general, though you will need to create paths with offsets if you want this behavior.