runBlock 发生后延迟 SKAction.sequence 中的下一个操作（Swift）？

The duration财产为moveTo在 a 内时不被跟随runBlock，允许序列中的后续操作立即执行，而该操作应该只在之后执行duration秒。

代码A（正确执行的序列）：

let realDest = CGPointMake(itemA.position.x, itemA.position.y)
let moveAction = SKAction.moveTo(realDest, duration: 2.0)

itemB.runAction(SKAction.sequence([SKAction.waitForDuration(0.5), moveAction, SKAction.runBlock {
    itemB.removeFromParent()
}]))

代码 B（序列未正确执行）：

let badMoveAction = SKAction.runBlock {
    let realDest = CGPointMake(itemA.position.x, itemA.position.y)
    let moveAction = SKAction.moveTo(realDest, duration: 2.0)
    itemB.runAction(moveAction)
}

itemB.runAction(SKAction.sequence([SKAction.waitForDuration(0.5), badMoveAction, SKAction.runBlock {
    itemB.removeFromParent()
}]))

In Code A, itemB后被删除moveAction完成（约 2 秒）。这是正确的顺序。

In Code B, itemB之前被删除badMoveAction结束，意义itemB永远不会离开原来的位置。就好像持续时间属性没有得到尊重Code B.

我们怎样才能移动itemB as in Code B但确保序列中的下一个操作不会开始，直到badMoveAction完成？

这应该做你想做的。我只是稍微重新安排了代码。

itemB.runAction(SKAction.sequence([

    // wait for half a second
    SKAction.waitForDuration(0.5),
    SKAction.runBlock({

        // after waiting half a second, get itemA's position
        let realDest = CGPointMake(itemA.position.x, itemA.position.y)
        let moveAction = SKAction.moveTo(realDest, duration: 2.0)

        // move to that position, after we get there, remove itemB from scene
        itemB.runAction(moveAction, completion: {
            itemB.removeFromParent()
        })

    })
]))