ios - Swift 中 SKAction 中 performSelector 的替代方法

Question

我正在将 Sprikit 应用程序转换为 Swift。但是我在转换这种方法时遇到了问题:

SKAction *releaseBalls = [SKAction sequence:@[[SKAction performSelector:@selector(createMyNode) onTarget:self],[SKAction waitForDuration:1]    ]];

Swift 中是否有替代代码？谢谢

Answer 1

试试这个

class MyScene: SKScene {

    func doAction() {
        let releaseBalls = SKAction.sequence([
            SKAction.runBlock(self.createMyNode),
            SKAction.waitForDuration(1)
            ])
        // run action
    }

    func createMyNode() {
        // create the nodes
    }
}