ios - “SKNode？” 没有名为“职位”的成员

Question

我究竟做错了什么？我似乎无法弄清楚这一点。我试过在后面加一个感叹号var thisBlock = self.childNodeWithName(block)： ,

这给了我一个新的错误说法。type () does not confirm to protocol 'BooleanType'.

func blockRunner() {
        for(block, blockStatus) in self.blockStatuses {
            var thisBlock = self.childNodeWithName(block)
            if blockStatus.shouldRunBlock() {
                blockStatus.timeGapForNextRun = random()
                blockStatus.currentInterval = 0
                blockStatus.isRunning = true
            }

            if blockStatus.isRunning {
                if thisBlock.position.x = blockMaxX{
                    thisBlock.position.x -= CGFloat(self.groundSpeed)
                } else {
                    thisBlock.position.x = self.origBlockPositionX
                    blockStatus.isRunning = false
                    self.score++
                    if ((self.score % 5) == 0) {
                        self.groundSpeed++
                    }
                    self.scoreText.text = String(self.score)
                }
            } else {
                blockStatus.currentInterval++
            }
        }
    }

Answer 1

childNodeWithName()确实返回一个可选的SKNode?，您必须打开它才能使用。不知道为什么var thisBlock = self.childNodeWithName(block)!没有解决你的问题。我建议使用可选绑定( if let) 语法：

if let thisBlock = self.childNodeWithName(block) {
    if blockStatus.shouldRunBlock() {
        blockStatus.timeGapForNextRun = random()
        blockStatus.currentInterval = 0
        blockStatus.isRunning = true
    }

    if blockStatus.isRunning {
        ... rest of your code
}

如果没有子节点，这具有额外的优势，即不会崩溃。它只是不会进入区块。