上一篇已经对绕非定轴转动有所了解,这篇郭先生继续说一说逻辑转体游戏的制作,这部分我们同样会遇到一些小问题,首先是根据数据渲染陷阱和目标区域,然后是对可以转动的判定,最后是获胜的判定。

1. 根据数据渲染陷阱和目标区域

首先我们P一张底图和陷阱图,如下图

three.js 制作逻辑转体游戏(下)-LMLPHP

three.js 制作逻辑转体游戏(下)-LMLPHP

就像这样,然后就是根据数据渲染陷阱和目标区域了,首先陷阱的个数是固定的,而目标区域是随小方块的数量而定,先看数据

end: [[-1, -4], [-1, -5]],
trap: [[-1, -7], [-6, -2]],

这里我们看一下Shader怎么写的

let texture1 = new THREE.TextureLoader().load('/static/images/base/luojizhuanti.png');
let texture2 = new THREE.TextureLoader().load('/static/images/base/stack.png');
let trapArray = [];
let targetArray = new Array(7).fill('').map(() => new THREE.Vector2(0,0));
square[this.game].trap.forEach(d => {
    trapArray.push(new THREE.Vector2(d[0], d[1]));
})
square[this.game].end.forEach((d,i) => {
    targetArray[i] = new THREE.Vector2(d[0], d[1]);
})
uniforms = {
    texture1: {
        value: texture1
    },
    texture2: {
        value: texture2
    },
    point0: {
        value: trapArray[0]
    },
    point1: {
        value: trapArray[1]
    },
    target: {
        value: targetArray
    }
}
uniforms[ "texture2" ].value.wrapS = uniforms[ "texture2" ].value.wrapT = THREE.RepeatWrapping;
let planeMate = new THREE.ShaderMaterial({
    side: THREE.DoubleSide,
    uniforms: uniforms,
    vertexShader: `
        varying vec2 vUv;
        varying vec3 pos;
        void main() {
            vUv = uv;
            pos = position;
            gl_Position = projectionMatrix * modelViewMatrix * vec4( position, 1.0 );
        }
    `,
    fragmentShader: `
        varying vec2 vUv;
        varying vec3 pos;
        uniform vec2 point0;
        uniform vec2 point1;
        uniform vec2 target[7];
        uniform sampler2D texture1;
        uniform sampler2D texture2;
        void main() {
            int index = 0;
            vec2 newUv = vec2(vUv.x * 7.0, vUv.y * 8.0);
            vec4 tcolor1 = texture2D( texture1, vUv );
            vec4 tcolor2 = texture2D( texture2, newUv );
            vec4 resultColor = tcolor1;
            if (pos.x < point0.x * 10.0 + 45.0 && pos.x > point0.x * 10.0 + 35.0 && pos.y < - point0.y * 10.0 - 40.0 && pos.y > - point0.y * 10.0 - 50.0) {
                resultColor = tcolor2;
            } else if(pos.x < point1.x * 10.0 + 45.0 && pos.x > point1.x * 10.0 + 35.0 && pos.y < - point1.y * 10.0 - 40.0 && pos.y > - point1.y * 10.0 - 50.0) {
                resultColor = tcolor2;
            } else {
                for(int i=0; i < 7; i++) {
                    if (pos.x < target[i].x * 10.0 + 45.0 && pos.x > target[i].x * 10.0 + 35.0 && pos.y < - target[i].y * 10.0 - 40.0 && pos.y > - target[i].y * 10.0 - 50.0) {
                        resultColor = vec4(1.0, 0.5, 0.0, 1.0);
                    }
                }
            }
            gl_FragColor = resultColor;
        }
    `
})

texture1和texture2是两个纹理图,trapArray是盛放陷阱的数组,targetArray是目标区域,默认长度是7,且默认值都是new THREE.Vector2(0,0),然后我们将二维向量加到以上两个数组中,最后添加到uniforms中,最后传到ShaderMaterial中,顶点着色器我们只需要将position和ui传到片元着色器中,关键是片元着色器,首先我们先得到一个新uv,这个新uv是沿x方向重复7次,沿y方向重复8次,然后tcolor1和tcolor2分别是底图的颜色和重复了7*8的陷阱的颜色。if中是渲染第一个陷阱,else if是渲染第二个陷阱,else中循环target数组,渲染target区域,具体的判断其实很简单。这样我们就根据关卡渲染了陷阱。

2. 对是否可以旋转进行判定

因为小方块是不可以超过底图的边缘的,而且也不可以直接覆盖到陷阱上面,因为这个操作是在点击上下左右的时候就要先判断可行性,但是此时我们还没有转,所以我们就要先拷贝一个boxes,先进行旋转看看出没出界或者压没压到陷阱,我们是这样实现的。

judge(num) {
    judgeGroup = new THREE.Group();
    boxesCopy = [];
    for(let i=0; i<boxes.length; i++) {
        let geom = new THREE.BoxGeometry(ratio, ratio, ratio);
        let mate = new THREE.MeshBasicMaterial({color: 0xffaa00, transparent: true, opacity: .8});
        let mesh = new THREE.Mesh(geom, mate);
        mesh.position.copy(boxes[i].position);
        boxesCopy[i] = mesh;
    }
    if(num == 1) {
        var offset = new THREE.Vector3(box3.max.x, 0, 0);
        judgeGroup.position.copy(offset);
        boxesCopy.forEach(d => {
            d.position.sub(offset);
            judgeGroup.add(d);
        })
        judgeGroup.rotation.z = - Math.PI / 2;
    } else if(num == 2) {
        var offset = new THREE.Vector3(box3.min.x, 0, 0);
        judgeGroup.position.copy(offset);
        boxesCopy.forEach(d => {
            d.position.sub(offset);
            judgeGroup.add(d);
        })
        judgeGroup.rotation.z = Math.PI / 2;
    } else if(num == 3) {
        var offset = new THREE.Vector3(0, 0, box3.min.z);
        judgeGroup.position.copy(offset);
        boxesCopy.forEach(d => {
            d.position.sub(offset);
            judgeGroup.add(d);
        })
        judgeGroup.rotation.x = - Math.PI / 2;
    } else if(num == 4) {
        var offset = new THREE.Vector3(0, 0, box3.max.z);
        judgeGroup.position.copy(offset);
        boxesCopy.forEach(d => {
            d.position.sub(offset);
            judgeGroup.add(d);
        })
        judgeGroup.rotation.x = Math.PI / 2;
    }
    judgeGroup.updateMatrixWorld();
    let canPass = true;
    boxesCopy.forEach(d => {
        var trans = new THREE.Vector3();
        d.matrixWorld.decompose(trans, new THREE.Quaternion(), new THREE.Vector3());
        let x = Math.round((trans.x - 5) / 10);
        let z = Math.round((trans.z - 5) / 10);
        let y = Math.round((trans.y + 5) / 10);
        if(x > -1 || x < -7 || z > -1 || z < -8) {
            canPass = false;
        } else {
            square[this.game].trap.forEach(d => {
                if(d[0] == x && d[1] == z && y == 1) {
                    canPass = false;
                }
            })
        }
    })
    return canPass;
},

boxesCopy就是对boxes进行的拷贝,num就是我们的上下左右操作,最后一个循环就是判断是否可翻转,x,y,z值分别对应我们的格子,if判断时候出界,因为x的界限就是[-1,-7],z的界限就是[-1,-8]。else是判断是否压到陷阱,只要有一个成立,canPass就会变成false。这就完成了简单的旋转判断。

3. 获胜的判定

获胜的判定很简单,在每一个旋转之后,比较boxes和end数组,如果两个数组一样,那么就说明胜利了,代码如下

computedWin() {
    let win = true;
    let temp = [];
    boxes.forEach(d => {
        let x = Math.round((d.position.x - 5) / 10);
        let z = Math.round((d.position.z - 5) / 10);
        temp.push([x, z]);
    })
    square[this.game].end.forEach(d => {
        if(!temp.some(dd => dd[0] == d[0] && dd[1] == d[1])) {
            win = false;
        }
    })
    if(win) {
        this.win();
    }
},

最后加上一点tween动画,这样我们就完成了一个逻辑转体的游戏,游戏玩起来还是比较有意思的。

转载请注明地址:郭先生的博客

08-20 22:31