Alex
Alex

Reputation: 2335

Draw a grid with SpriteKit

What would be the best way to draw a grid like this by using the SpriteKit 2D game engine? Requirements:

I'm planning to use a SKSpriteNode (from an image) as a player moving in different squares in this grid.
So, I'll save in a 2 dimensional array the central point (x,y) of each square and then move from the player's current position to that position. If you have a better suggestion for this too, I'd like to hear it.

I would appreciate a solution in Swift (preferably 2.1), but Objective-C would do too. Planning on using this only on iPhone devices.
My question is close to this one. Any help is appreciated.

Upvotes: 22

Views: 9168

Answers (1)

0x141E
0x141E

Reputation: 12773

I suggest you implement the grid as a texture of an SKSpriteNode because Sprite Kit will renders the grid in a single draw call. Here's a example of how to do that:

class Grid:SKSpriteNode {
    var rows:Int!
    var cols:Int!
    var blockSize:CGFloat!

    convenience init?(blockSize:CGFloat,rows:Int,cols:Int) {
        guard let texture = Grid.gridTexture(blockSize: blockSize,rows: rows, cols:cols) else {
            return nil
        }
        self.init(texture: texture, color:SKColor.clear, size: texture.size())
        self.blockSize = blockSize
        self.rows = rows
        self.cols = cols
    }

    class func gridTexture(blockSize:CGFloat,rows:Int,cols:Int) -> SKTexture? {
        // Add 1 to the height and width to ensure the borders are within the sprite
        let size = CGSize(width: CGFloat(cols)*blockSize+1.0, height: CGFloat(rows)*blockSize+1.0)
        UIGraphicsBeginImageContext(size)

        guard let context = UIGraphicsGetCurrentContext() else {
            return nil
        }
        let bezierPath = UIBezierPath()
        let offset:CGFloat = 0.5
        // Draw vertical lines
        for i in 0...cols {
            let x = CGFloat(i)*blockSize + offset
            bezierPath.move(to: CGPoint(x: x, y: 0))
            bezierPath.addLine(to: CGPoint(x: x, y: size.height))
        }
        // Draw horizontal lines
        for i in 0...rows {
            let y = CGFloat(i)*blockSize + offset
            bezierPath.move(to: CGPoint(x: 0, y: y))
            bezierPath.addLine(to: CGPoint(x: size.width, y: y))
        }
        SKColor.white.setStroke()
        bezierPath.lineWidth = 1.0
        bezierPath.stroke()
        context.addPath(bezierPath.cgPath)
        let image = UIGraphicsGetImageFromCurrentImageContext()
        UIGraphicsEndImageContext()

        return SKTexture(image: image!)
    }

    func gridPosition(row:Int, col:Int) -> CGPoint {
        let offset = blockSize / 2.0 + 0.5
        let x = CGFloat(col) * blockSize - (blockSize * CGFloat(cols)) / 2.0 + offset
        let y = CGFloat(rows - row - 1) * blockSize - (blockSize * CGFloat(rows)) / 2.0 + offset
        return CGPoint(x:x, y:y)
    }
}

And here's how to create a grid and add a game piece to the grid

class GameScene: SKScene {
    override func didMove(to: SKView) {
        if let grid = Grid(blockSize: 40.0, rows:5, cols:5) {
            grid.position = CGPoint (x:frame.midX, y:frame.midY)
            addChild(grid)

            let gamePiece = SKSpriteNode(imageNamed: "Spaceship")
            gamePiece.setScale(0.0625)
            gamePiece.position = grid.gridPosition(row: 1, col: 0)
            grid.addChild(gamePiece)
        }
    }
}

Update:

To determine which grid square was touched, add this to init

self.isUserInteractionEnabled = true

and this to the Grid class:

override func touchesBegan(_ touches: Set<UITouch>, withEvent event: UIEvent?) {
    for touch in touches {
        let position = touch.location(in:self)
        let node = atPoint(position)
        if node != self {
            let action = SKAction.rotate(by:CGFloat.pi*2, duration: 1)
            node.run(action)
        }
        else {
            let x = size.width / 2 + position.x
            let y = size.height / 2 - position.y
            let row = Int(floor(x / blockSize))
            let col = Int(floor(y / blockSize))
            print("\(row) \(col)")
        }
    }
}

Upvotes: 43

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