Question

我正在使用序列来运行SKActions列表。然而，我想要做的是运行SKAction，然后一次运行两个，然后依次运行一个。

这是我的代码：

SKNode *ballNode = [self childNodeWithName:@"ball"];

    if (ballNode != Nil){
        ballNode.name = nil;

        SKAction *delay = [SKAction waitForDuration:3];
        SKAction *scale = [SKAction scaleTo:0 duration:1];
        SKAction *fadeOut = [SKAction fadeOutWithDuration:1];
        SKAction *remove = [SKAction removeFromParent];

        //put actions in sequence
        SKAction *moveSequence = [SKAction sequence:@[delay, (run scale and fadeout at the same time), remove]];

        //run action from node (child of SKLabelNode)
        [ballNode runAction:moveSequence];
    }

我怎样才能做到这一点？我假设我不能使用序列？

Answer 1

使用群组操作。

从sprite kit编程指南：

组操作是一组操作，一旦执行完组就会立即执行。如果希望同步操作，可以使用组

SKSpriteNode *wheel = (SKSpriteNode*)[self childNodeWithName:@"wheel"];
CGFloat circumference = wheel.size.height * M_PI;
SKAction *oneRevolution = [SKAction rotateByAngle:-M_PI*2 duration:2.0];
SKAction *moveRight = [SKAction moveByX:circumference y:0 duration:2.0];
SKAction *group = [SKAction group:@[oneRevolution, moveRight]];
[wheel runAction:group];

Answer 2

Swift中的一个例子是：

    let textLabel = SKLabelNode(text: "Some Text")

    let moveTo = CGPointMake(600, 20)

    let big = SKAction.scaleTo(3.0, duration: 0.1)
    let med = SKAction.scaleTo(1.0, duration: 0.3)
    let reduce = SKAction.scaleTo(0.2, duration: 1.0)
    let move = SKAction.moveTo(moveTo, duration: 1.0)
    let fade = SKAction.fadeOutWithDuration(2.0)
    let removeNode = SKAction.removeFromParent()
    let group = SKAction.group([fade, reduce])

    let sequence = SKAction.sequence([big, med, move, group, removeNode])

    self.addChild(textLabel)
    textLabel.runAction(sequence)

一次运行两个SKActions

2 个答案: