Swift / SpriteKit: evenly distribute SKShapeNodes horizontally and vertically into a grid - ios

Question

I want to equally distribute an n amount of square SKShapeNodes over the screen to create a square grid. I tried searching for other examples of this, but I cannot find the solution I'm looking for as most solutions only focus on only distributing views horizontally.
I'm creating the squares in a for loop, however I can only get them to move on either the x or the y axes, not both. The below example gives me the best result, but as expected from the code only the first two squares are visible, the third one is created off screen:
amountOfTiles is a CGFloat set to 4
tile is an SKShapeNode
func addTiles() {
let screenWidth = self.view?.frame.size.width
tileWidth = screenWidth! / (amountOfTiles / 2)
tileHeight = tileWidth
tileX = 0
tileY = 0
tileRect = CGRect(x: tileX, y: tileY, width: tileWidth, height: tileHeight)
for _ in 0...Int(amountOfTiles) {
tile = SKShapeNode(rect: tileRect)
tile.position = CGPoint(x: tileX, y: tileY)
tile.fillColor = .red
addChild(tile)
tileX += tileWidth
tileY += tileHeight
}
}
My guess is that I should not update the Y and X positions of the squares at the same time, but I'm not sure how to solve this.

Answer 1

Fixed it myself. I removed amountOfTiles and added two separate CGFloats for x and y, which I update in a separate loop:
func addTiles() {
let xAmount:CGFloat = 4
let yAmount:CGFloat = 4
let screenWidth = self.view?.frame.size.width
tileWidth = screenWidth! / (xAmount)
tileHeight = tileWidth
tileY = 0
tileRect = CGRect(x: tileX, y: tileY, width: tileWidth, height: tileHeight)
for _ in 0..<Int(xAmount) {
tileX = 0
for _ in 0..<Int(yAmount) {
tile = SKShapeNode(rect: tileRect)
tile.fillColor = .red
addChild(tile)
tile.position = CGPoint(x: tileX, y: tileY)
tileX += tileWidth
}
tileY += tileHeight
}
}