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VistasHow to render different dimensioned rectangles with the same border texture and have the corners not stretch?
This is my texture. I would like to have a bunch of differently sized rectangle, that means they can all have different aspect-ratios.
If I use one texture
on a Plane, then the textures would get stretched, especially around the corners. (I don't care if the middle gets scratched since its just a gradient).
It looks like this.
If possible, I would like to have differently sized rectangles all use one texture and have the corners not stretched. Like this.
·
Juan Pablo Isaza
1 Respuestas
Responde la pregunta0
I made a BufferGeometry implementation of 9-slicing. It works by creating 9 planes with respective normal UVs. When you stretch it, only the edges and center plane is scaled, so the corners don't get distorted.
const basePlaneVertices = [0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0];
class NineSlicePlane extends THREE.Mesh {
constructor(material, dimensions = { width: 100, height: 100, border: 25 }, uvBorder = 0.25) {
let geometry = new THREE.BufferGeometry();
let vertices = [];
let uvs = [];
const vertexOffsetsX = [0, dimensions.border, dimensions.width - dimensions.border, dimensions.width];
const vertexOffsetsY = [0, dimensions.border, dimensions.height - dimensions.border, dimensions.height];
const uvOffsets = [0, uvBorder, 1 - uvBorder, 1];
for (let i = 0; i < 3; i++) {
for (let j = 0; j < 3; j++) {
for (let k = 0; k < 6; k++) {
vertices.push(
basePlaneVertices[k * 2] * (vertexOffsetsX[i + 1] - vertexOffsetsX[i]) + vertexOffsetsX[i] - dimensions.width * 0.5,
basePlaneVertices[k * 2 + 1] * (vertexOffsetsY[j + 1] - vertexOffsetsY[j]) + vertexOffsetsY[j] - dimensions.height * 0.5,
0
);
uvs.push(uvOffsets[i + basePlaneVertices[k * 2]], uvOffsets[j + basePlaneVertices[k * 2 + 1]]);
}
}
}
geometry.setAttribute('position', new THREE.BufferAttribute(new Float32Array(vertices), 3));
geometry.setAttribute('uv', new THREE.BufferAttribute(new Float32Array(uvs), 2));
super(geometry, material);
}
}
·
Juan Pablo Isaza
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Responde la pregunta