ios – Swift Spritekit等距地图触摸位置

我已经开始做一个小的 swift / spritekit项目来教我自己的游戏开发.
它以等距地图开始,我设法绘制.
但是我在地图的不同瓷砖上获得精确的触摸位置时遇到了麻烦.
它有效,但有点不合适,似乎不一致.

这是我的功能：

class PlayScene: SKScene {

let map = SKNode()
    override func didMoveToView(view: SKView) {

        let origin = view.frame.origin
        let mapOrigin = CGPointMake(origin.x + self.frame.width / 4,origin.y - self.frame.height / 4)


        let mapConfig: Int[][] =

       [[0,1,0],[0,[2,2,2],0]]

    drawMap(mapConfig,mapOrigin: mapOrigin)
}

用：

func drawMap(mapConfig:Int[][],mapOrigin:CGPoint)
{
    let tileHeight:CGFloat = 25.5
    let numColumns:Int = 8
    let numRows:Int = 8

    var position = mapOrigin
    var column: Int = 0
    var row: Int = 0

    for column = 0; column < numColumns; column++
    {
        for row = 0; row < numRows; row++
        {
            position.x = mapOrigin.x + CGFloat(column) * tileHeight
            position.y = mapOrigin.y + CGFloat(row) * tileHeight
            let isoPosition = twoDToIso(position)
            placeTile(isoPosition,mapConfig: mapConfig[row][column])
        }
    }
    self.addChild(map)
}

func placeTile(position:CGPoint,mapConfig:Int)
{
 switch mapConfig
    {
    case 0:
        let sprite = SKSpriteNode(imageNamed:"grassTile")
        sprite.position = position
        sprite.setScale(0.1)
        sprite.name = "\(position)"
        self.map.addChild(sprite)
    case 1:
        let sprite = SKSpriteNode(imageNamed:"roadTile")
        sprite.position = position
        sprite.setScale(0.1)
        sprite.name = "\(position)"
        self.map.addChild(sprite)
    default:
        let sprite = SKSpriteNode(imageNamed:"roadTileLTR")
        sprite.position = position
        sprite.setScale(0.1)
        sprite.name = "\(position)"
        self.map.addChild(sprite)
    }
}

然后我想隐藏我触摸的瓷砖(用于测试)：

override func touchesBegan(touches: NSSet,withEvent event: UIEvent)
{
    for touch: AnyObject in touches
    {
        let locationNode = touch.locationInNode(self)
        nodeAtPoint(locationNode).hidden = true
    }
}

但它并不总是隐藏正确的瓷砖.
那我该怎么解决呢？我的代码是否根本错误(可能)？或者我需要以某种方式将位置转换为iso坐标？或者玩瓷砖位面罩？

无论如何,谢谢你的帮助！

解决方法

我在等轴测图中遇到了类似的问题.

问题是您单击的节点大于显示的节点(它具有透明部分).有关该问题的更好解释,请参阅my question here.

以下是我如何解决它(抱歉代码在Objective-C中)：

1.在tile的边缘之后创建一个CGPathRef(tileSize是纹理的大小).此代码适用于常规等距切片,而不是六边形,但想法是相同的.

// ObjC
-(CGPathRef)createTextureFrame:(CGSize)tileSize
{
    CGMutablePathRef path = CGPathCreateMutable();
    CGPathMoveToPoint(path,NULL,-(self.tileSize.height / 2));
    CGPathAddLineToPoint(path,(self.tileSize.width / 2),0);
    CGPathAddLineToPoint(path,(self.tileSize.height / 2));
    CGPathAddLineToPoint(path,-(self.tileSize.width / 2),0);
    CGPathCloseSubpath(path);
    return path;
}

// Swift
func createTextureFrame(tileSize:CGSize) -> CGPathRef {
    CGMutablePathRef path = CGPathCreateMutable()
    CGPathMoveToPoint(path,nil,-(self.tileSize.height / 2))
    CGPathAddLineToPoint(path,0)
    CGPathAddLineToPoint(path,(self.tileSize.height / 2))
    CGPathAddLineToPoint(path,0)
    CGPathCloseSubpath(path)
    return path
}

2.创建一个函数,检查给定的点是否在CGPathRef(textureFrame)中.

// ObjC
-(BOOL)isPointOnNode:(CGPoint)point
{
    return CGPathContainsPoint(textureFrame,point,YES);
}

// Swift
func isPointOnNode(point:CGPoint) -> Bool {
    return CGPathContainsPoint(textureFrame,YES)
}

3.对于每个触摸的节点,检查我们所在的textureFrame.

// ObjC
UITouch *touch = [touches anyObject];
NSArray *nodes = [self nodesAtPoint:[touch locationInNode:self]];
for (SKNode *node in nodes)
{
    CGPoint locationInNode = [touch locationInNode:node];
    if ([node isPointOnNode:locationInNode])
    {
        node.hidden = YES;
    }
}

// Swift
var touch = touches.anyObject() as UITouch
var nodes = self.nodesAtPoint(touch.locationInNode(self))
for node in nodes as [SKNode] {
    var locationInNode = touch.locationInNode(node)

    if node.isPointOnNode() {
        node.hidden = true
    }
}

我不知道它是否是最好的解决方案,但效果很好.
我希望它有帮助:)

编辑：添加了Swift版本的代码

ios – Swift Spritekit等距地图触摸位置

解决方法

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