我想知道是否有任何webgl纹理水平翻转的功能。
我已经谷歌了解这个问题。 但是我通过使用 gl.pixelStorei(gl.UNPACK_FLIP_Y_WEBGL,true); 来获得垂直翻转纹理的答案。
我还尝试编写一个简单的函数来切换像素值以获得我想要的翻转纹理。 但我仍然想知道有什么方法可以直接用于水平翻转webgl纹理吗?
答案 0 :(得分:3)
更好的问题是你想做什么。你说你想要翻转纹理。为什么?你想要完成什么?
有很多方法可以翻转纹理。绘制某个方向不需要翻转纹理的数据。假设您正在绘制一个带有纹理的矩形。你可以
所有这些都比试图实际翻转纹理数据更快,更有用。今天你问如何x翻转数据。明天你会问如何旋转数据。您通常不会通过x翻转或旋转纹理来操纵顶点和/或纹理坐标。
再说一遍,你认为x-flipping纹理数据的实际尝试是什么?
旧的固定函数OpenGL有一个纹理矩阵,对应于上面的最后一个想法。你可能有像这样的顶点着色器
attribute vec4 position;
attribute vec2 texcoord;
uniform mat4 matrix;
varying vec2 v_texcoord;
void main() {
gl_Position = matrix * position;
// pass through to fragment shader
v_texcoord = texcoord;
}
你可以添加像这样的纹理矩阵
attribute vec4 position;
attribute vec2 texcoord;
uniform mat4 matrix;
uniform mat4 textureMatrix; // ADDED!!
varying vec2 v_texcoord;
void main() {
gl_Position = matrix * position;
// pass through to fragment shader after
// multiplying by texture matrix
v_texcoord = (textureMatrix * vec4(texcoord, 0, 1)).xy; // CHANGED !!
}
这是一个使用纹理矩阵以多种方式绘制相同纹理的示例
var m4 = twgl.m4;
var gl = twgl.getWebGLContext(document.getElementById("c"));
var programInfo = twgl.createProgramInfo(gl, ["vs", "fs"]);
var arrays = {
position: {
numComponents: 2,
data: [
0, 0,
1, 0,
0, 1,
0, 1,
1, 0,
1, 1,
],
},
texcoord: [
0, 0,
1, 0,
0, 1,
0, 1,
1, 0,
1, 1,
],
};
var bufferInfo = twgl.createBufferInfoFromArrays(gl, arrays);
// make a texture from a 2d canvas. We'll make an F so we can see it's orientation
var ctx = document.createElement("canvas").getContext("2d");
ctx.width = 64;
ctx.height = 64;
ctx.fillStyle = "red";
ctx.fillRect(0, 0, ctx.canvas.width, ctx.canvas.height / 2);
ctx.fillStyle = "blue";
ctx.fillRect(0, ctx.canvas.height / 2, ctx.canvas.width, ctx.canvas.height / 2);
ctx.fillStyle = "yellow";
ctx.font = "100px sans-serif";
ctx.textAlign = "center";
ctx.textBaseline = "middle";
ctx.fillText("F", ctx.canvas.width / 2, ctx.canvas.height / 2);
var tex = twgl.createTexture(gl, { src: ctx.canvas });
var uniforms = {
matrix: m4.identity(),
textureMatrix: m4.identity(),
texture: tex,
};
gl.useProgram(programInfo.program);
twgl.setBuffersAndAttributes(gl, programInfo, bufferInfo);
for (var ii = 0; ii < 10; ++ii) {
var flipX = ii & 0x1;
var flipY = ii & 0x2;
var swap = ii & 0x4;
var rot = ii & 0x8;
var x = ii % 5;
var y = ii / 5 | 0;
var m = uniforms.matrix;
m4.ortho(0, gl.canvas.width, gl.canvas.height, 0, -1, 1, m);
m4.translate(m, [x * 59 + 2, y * 59 + 2, 0], m);
m4.scale(m, [58, 58, 1], m);
var tm = uniforms.textureMatrix;
m4.identity(tm);
if (flipX) {
m4.translate(tm, [1, 0, 0], tm);
m4.scale(tm, [-1, 1, 1], tm);
}
if (flipY) {
m4.translate(tm, [0, 1, 0], tm);
m4.scale(tm, [1, -1, 1], tm);
}
if (swap) {
m4.multiply(tm, [
0, 1, 0, 0,
1, 0, 0, 0,
0, 0, 1, 0,
0, 0, 0, 1,
], tm);
}
if (rot) {
m4.translate(tm, [0.5, 0.5, 0], tm);
m4.rotateZ(tm, Math.PI * 0.25, tm);
m4.translate(tm, [-0.5, -0.5, 0], tm);
}
twgl.setUniforms(programInfo, uniforms);
twgl.drawBufferInfo(gl, gl.TRIANGLES, bufferInfo);
}
canvas { border: 1px solid black; }
<script id="vs" type="notjs">
attribute vec4 position;
attribute vec2 texcoord;
uniform mat4 matrix;
uniform mat4 textureMatrix;
varying vec2 v_texcoord;
void main() {
gl_Position = matrix * position;
v_texcoord = (textureMatrix * vec4(texcoord, 0, 1)).xy;
}
</script>
<script id="fs" type="notjs">
precision mediump float;
uniform sampler2D texture;
varying vec2 v_texcoord;
void main() {
gl_FragColor = texture2D(texture, v_texcoord);
}
</script>
<script src="https://twgljs.org/dist/twgl-full.min.js"></script>
<canvas id="c"></canvas>
答案 1 :(得分:2)
实际上,最常用的方法是减去您要翻转的轴:
TexCoord.y = 1.0 - TexCoord.y;