ios - 无法调用非函数类型 SKShapeNode 的值

Question

我一直在尝试修复此错误，并尝试使该字符包含在将是 int 的行中。

 func isRightTileAt(location:CGPoint) ->Bool {
    //as shape node so we can get fill
    var currentRect = self.atPoint(location) as! SKShapeNode
    //get the 10th character which will contain the row and make it an int
   // let rowOfNode = Int(currentRect.name![10]) //error(tried both of these)
    var rowOfNode = Int(currentRect(name[10])) //error 
    //flip position is used for the row index below the screen to flip it to the top.
    var currentRow = self.flipPosition + 1
    var currentRowOfClick = self.flipPosition

    //we reuse the flip position because it hasn't flipped yet but it normally contains the right row.
    //because flip position happens after this check so it won't be sent back around yet
    if self.flipPosition == 5 {
        currentRowOfClick = 0
    }
    //if they are at least on the right row
    if rowOfNode == currentRowOfClick && currentRect.fillColor.hash == 65536{
        return true
    }
    return false
}

Answer 1

访问name的字符存在一些挑战。 SKNode的属性(property)或SKNode子类(例如 SKShapeNode )。

首先，从name开始是 String? ，需要将其拆开。

guard let string = self.name else {
    return
}

其次，你无法访问 String 的字符与 Int下标；您需要使用 String.Index .

// Since Swift is zero based, the 10th element is at index 9; use 10 if you want the 11th character.
let index = string.index(string.startIndex, offsetBy: 9)
// The 10th character of the name
let char = string[index]

第三，你不能直接转换 Character到Int 。您需要将字符转换为 String然后将字符串转换为 Int .

let rowString = String(char)

// Unwrap since Int(string:String) returns nil if the string is not an integer
guard let row = Int(rowString) else {
    return
}

此时，row是name的第10个字符转换为Int .

或者，您可以将上述内容实现为扩展

extension String {
    func int(at index:Int) -> Int? {
        let index = self.index(self.startIndex, offsetBy: index)
        let string = String(self[index])
        return Int(string)
    }
}

并使用它

guard let name = self.name, let row = name.int(at:9) else {
    return
}