将几何添加到THREE.Object3D

Question

我编辑了这篇文章,以使其更加清晰.

好吧,我在这里要做的是表示由8个不同三角形组成的图块.每个三角形应该能够独立地改变它的颜色.

所以,我遇到的问题是,当我改变单个三角形的颜色时,它会改变线条的颜色,正如你在第二张图片中看到的那样.

这是创建磁贴的代码:

var tile=[];
var n=0;
for(var i=0; i<4; i++){
    for(var j=0; j<2; j++){
        var triangle = new THREE.Object3D();
        var lineMaterial = new THREE.LineBasicMaterial({color:0xffffff, transparent:true, opacity:0.5});
        var triangleMaterial = new THREE.MeshPhongMaterial({color:COLOR_OFF ,emissive:EMISSIVE_OFF ,side:THREE.DoubleSide, shading:THREE.FlatShading});

        var geometry = new THREE.Geometry();
        geometry.vertices.push( triangleVectors[j][0], triangleVectors[j][1], triangleVectors[j][2] );
        var face = new THREE.Face3(0, 1, 2);
        geometry.faces.push(face);

        triangle.add(new THREE.LineSegments(new THREE.Geometry(), lineMaterial));
        triangle.add( new THREE.Mesh( new THREE.Geometry(), triangleMaterial));

        triangle.children[ 0 ].geometry = new THREE.WireframeGeometry(geometry);
        triangle.children[ 1 ].geometry = geometry;

        triangle.rotation.z = Math.PI*i/2;
        triangle.position.x = TILE_LENGTH*x;
        triangle.position.y = TILE_LENGTH*y;

        n++;                        
        tile.push({'triangle':triangle,'number':n,'state':"OFF"});                      
        scene.add(triangle);
     }
}

要更新Tile三角形的状态我正在使用此代码:

for(var j=0;j<tile.length; j++){

    tile[j].triangle.children[0].material.color.set(COLOR_OFF);
    tile[j].triangle.children[1].material.color.set(COLOR_OFF);
    tile[j].state="OFF";

    for(var k=0; k<pathUpdates[step].length; k++){
        if(pathUpdates[step][k].number == tile[j].number){
           tile[j].triangle.children[0].material.color.set(COLOR_ON);               
           tile[j].triangle.children[1].material.color.set(COLOR_ON);
           floor[i].tile[j].state="ON";
        }
    }
}

这是改变材料的正确方法吗？

我简化了一些代码以更好地展示我的问题.但是,真正的代码在此存储库https://github.com/tul1/Tile.git中.如果你想看一下它的工作原理我在http://tul1.github.io/examples/dale.html中进行了部署.

Answer 1

所以,我遇到的问题是,当我改变单个三角形的颜色时,它会改变两者之间的线条颜色,正如您在第二个图像中看到的那样.

我看着你的源代码,如果你想要做的是不影响之间使它们保持白色,简单地删除,改变颜色的线线的颜色lineMaterial:

for(var j=0;j<tile.length; j++){
    // The line below changes all line colors to 0x156289:
    // tile[j].triangle.children[0].material.color.set(COLOR_OFF);

    tile[j].triangle.children[1].material.color.set(COLOR_OFF);
    tile[j].state="OFF";

    for(var k=0; k<pathUpdates[step].length; k++){
        if(pathUpdates[step][k].number == tile[j].number){
           // The line below changes the active line color to 0x891528
           // tile[j].triangle.children[0].material.color.set(COLOR_ON); 

           tile[j].triangle.children[1].material.color.set(COLOR_ON);
           floor[i].tile[j].state="ON";
        }
    }
}

结果

如果您没有触摸线条上的颜色,它们将像您第一次创建时一样保持白色:

分组

如果要将所有切片分组到一个对象中,则应在全局范围内(在所有循环之外)声明一个THREE.Group对象for().然后,您可以在循环的每次迭代中单独添加每个三角形for().这是关于如何实现这个的一些伪代码:

// Create tileGroup outside the loop
var tileGroup = new THREE.Group();

for(var x=0; x<nX; x++){
    for(var y=0; y<nY; y++){
        var tile=[];
        var n=0;
        for(var i=0; i<4; i++){
            for(var j=0; j<2; j++){
                // All your existing triangle-creation code goes in here

                // ...

                // After triangle has been created, we add triangle
                // to group, instead of adding it to scene
                tileGroup.add(triangle);
            }
        }
    }
}

// Finally, we add tileGroup to scene
scene.add(tileGroup);

// Now we can manipulate all tiles at once within the group
tileGroup.position.set(0, 15, 0);