swift - 对象定位已关闭

Question

嗨，我正在尝试让一个对象(在本例中是绿色 Frog )在与场景一样宽的平台上与玩家 Sprite (红色 Frog )对齐生成，我的意思是，正在让对象生成，以便当玩家前进时它不会与对象重叠。 (图中显示绿色 Frog 位于两只红色 Frog 之间且与其中一只红色 Frog 不在一条直线上)

我的对象定位代码如下

obstacle.position = CGPointMake(-(backgroundSprite.size.width / 2) + CGFloat(randomX) + (spacing * CGFloat(i)), 0)

目前，它在屏幕左侧、场景一半处生成。背景 Sprite 是对象被添加到的对象，其定义如下:

let theSize:CGSize = CGSizeMake(levelUnitWidth, levelUnitHeight)       
let tex:SKTexture = SKTexture(imageNamed: imageName)          
backgroundSprite = SKSpriteNode(texture: tex, color: SKColor.blackColor(), size: theSize)

随机 x 也是在背景 Sprite 的 x 轴上随机生成它们的原因(我也尝试过调整，但没有成功)

let randomX = arc4random_uniform(  UInt32 (backgroundSprite.size.height) )

最后，间距是同一级别单位中对象之间的距离。

let spacing:CGFloat = 250

我尝试过实现玩家 Sprite 宽度作为引用，但没有成功。有人可以告诉我我在这里做错了什么吗？

如果您需要查看全部代码，这里是完整的代码:

if (theType == LevelType.road) {

for (var i = 0; i < Int(numberOfObjectsInLevel); i++) {

     let obstacle:Object = Object()


     obstacle.theType = LevelType.road


     obstacle.createObject()
     addChild(obstacle)


     let spacing:CGFloat = 250
     obstacle.position = CGPointMake((backgroundSprite.size.width / 4) + CGFloat(randomX) + (spacing * CGFloat(i)), 0)

    }

编辑:

我尝试用我已有的代码来实现您在编辑帖子中编写的代码，这就是我得到的。

  if (theType == LevelType.road) {


        let xAxisSpawnLocations: [CGFloat] = {

            var spawnLocations:[CGFloat] = []

            //Create 5 possible spawn locations
            let numberOfNodes = 5

            for i in 0...numberOfNodes - 1 {

                /*
                Spacing between nodes will change if:

                1) number of nodes is changed,
                2) screen width is changed,
                3) node's size is changed.
                */

                var xPosition = (frame.maxX - thePlayer.size.width) / CGFloat((numberOfNodes - 1)) * CGFloat(i)

                //add a half of a player's width because node's anchor point is (0.5, 0.5) by default
                xPosition += thePlayer.size.width/2.0

                spawnLocations.append( xPosition )
            }

            return spawnLocations
        }()

        print(xAxisSpawnLocations)

        let yAxisSpawnLocations: [CGFloat] = [0]

        let obstacle:Object = Object()

        obstacle.theType = LevelType.road


        obstacle.createObject()
        addChild(obstacle)

        let randx = xAxisSpawnLocations[Int(arc4random_uniform(UInt32(xAxisSpawnLocations.count)))]

        obstacle.position = CGPoint(x: randx, y: yAxisSpawnLocations[0] )
        obstacle.zPosition = 200

    }

编辑2:

所以这次再次以正确的方式实现了代码，我得到了这个:

所以玩家仍然没有与对象对齐，并且由于某种原因它只在屏幕的右侧生成。我认为这是因为我有一个包含所有内容的 worldNode。

worldNode 保存在 worldNode 中起始点为 (0,0) 的玩家，并且还保存保存对象的关卡单位。相机位置以玩家节点为中心，我不确定这是否是问题所在，但我将提供下面的代码，以便您查看。

let startingPosition:CGPoint = CGPointMake(0, 0)

woldNode代码:

let worldNode:SKNode = SKNode()


//creates the world node point to be in the middle of the screen
self.anchorPoint = CGPointMake(0.5, 0.5)
addChild(worldNode)

//adds the player as a child node to the world node 
worldNode.addChild(thePlayer)
thePlayer.position = startingPosition
thePlayer.zPosition = 500   

相机定位代码:

 override func didSimulatePhysics() {

    self.centerOnNode(thePlayer) 
}

//centers the camera on the node world.
func centerOnNode(node:SKNode) {

    let cameraPositionInScene:CGPoint = self.convertPoint(node.position, fromNode: worldNode)
    worldNode.position = CGPoint(x: worldNode.position.x , y:worldNode.position.y - cameraPositionInScene.y  )

}

我很确定这是我的问题，但请告诉我你的想法。

Answer 1

就像我在评论中所说，关键是预定义 x(和 y)轴的坐标并基于此生成节点。首先，让我们在 GameScene 类中定义一个玩家:

 let player = SKSpriteNode(color: .redColor(), size: CGSize(width: 50, height: 50))

现在，预定义生成位置(x 轴和 y 轴):

let xAxisSpawnLocations: [CGFloat] = [50.0, 125.0, 200.0, 275.0, 350.0, 425.0]
let yAxisSpawnLocations: [CGFloat] = [50.0, 125.0,  200.0, 275.0]

现在，当我们知道可能的位置时，首先定位我们的玩家并将其添加到场景中:

player.position = CGPoint(x: xAxisSpawnLocations[0], y: yAxisSpawnLocations[0])
player.zPosition = 10
addChild(player)

您可以根据玩家的宽度、高度以及屏幕尺寸创建这些位置，但为了简单起见，我对所有内容进行了硬编码。

所以，让我们用绿色 Frog 填充玩家正上方的一行:

for xLocation in xAxisSpawnLocations {

     let greenFrog = SKSpriteNode(color: .greenColor(), size: player.size)
     greenFrog.position = CGPoint(x: xLocation, y: yAxisSpawnLocations[1])
      addChild(greenFrog)
}

结果会是这样的:

或者，例如，将玩家向右移动一个位置，并在他的正上方制作一列绿色 Frog :

player.position = CGPoint(x: xAxisSpawnLocations[1], y: yAxisSpawnLocations[0])


 for yLocation in yAxisSpawnLocations {

      let greenFrog = SKSpriteNode(color: .greenColor(), size: player.size)
       greenFrog.position = CGPoint(x: xAxisSpawnLocations[1], y: yLocation)
      addChild(greenFrog)
  }

它应该看起来像这样:

编辑:

根据您的评论，您可以根据节点数量、屏幕宽度和节点大小在屏幕上分布节点:

let xAxisSpawnLocations: [CGFloat] = {

    var spawnLocations:[CGFloat] = []

    //Create 5 possible spawn locations
    let numberOfNodes = 5

    for i in 0...numberOfNodes - 1 {

        /*
            Spacing between nodes will change if:

            1) number of nodes is changed,
            2) screen width is changed,
            3) node's size is changed.
        */

        var xPosition = (frame.maxX - player.size.width) / CGFloat((numberOfNodes - 1)) * CGFloat(i)

        //add a half of a player's width because node's anchor point is (0.5, 0.5) by default
        xPosition += player.size.width/2.0

        spawnLocations.append( xPosition )
    }

    return spawnLocations
}()

print(xAxisSpawnLocations)

您应该处理添加太多节点或节点太大时发生的情况，但这可以让您了解如何沿 x 轴分布节点并保持它们之间的相同距离。