Question

说明

我正在构建一个我面临一些问题的游戏。我有两个按照相同规则运行的对象（A和B）：从上到下，它们会产生，向下移动并在到达边缘时被移除。它们中的每一个在产卵之前都有其特定的持续时间，这样，有时它们会被覆盖（如下所示）。

问题

问题很简单：有没有办法让A和B产生不同的持续时间，但仍然是随机的？

代码

如果需要，可以下载此项目here。

import SpriteKit

class GameScene: SKScene {

    //Declarations
    var A = SKSpriteNode()
    var B = SKSpriteNode()
    var durationA = CGFloat()
    var durationB = CGFloat()

    //Setup
    func setupA(){
        A = SKSpriteNode(imageNamed: "A")
        A.position = CGPoint(x: self.frame.width / 2, y: self.frame.height)
    }
    func setupB(){
        B = SKSpriteNode(imageNamed: "B")
        B.position = CGPoint(x: self.frame.width / 2, y: self.frame.height)
    }

    //Actions
    func actionsA(){

        //Start spawning, moving and removing
        let spawnA = SKAction.runBlock({
            () in

            //Spawn A
            self.setupA()
            self.addChild(self.A)

            //Move left and remove when go off screenk
            let frameHeight = CGFloat(self.frame.height)
            let moveA = SKAction.moveByX(0, y: -frameHeight, duration: NSTimeInterval(0.0028 * frameHeight))  //duration: faster or slower
            let removeA = SKAction.removeFromParent()

            let moveAndRemoveA = SKAction.sequence([moveA, removeA])

            self.A.runAction(moveAndRemoveA)
        })

        //Spawn A each 1.75~2.25 seconds
        durationA = (CGFloat(arc4random_uniform(51)) + 175.0) / 100.0
        let spawnAfterDurationA = SKAction.repeatActionForever(SKAction.sequence([SKAction.waitForDuration(Double(durationA)), spawnA]))
        runAction(spawnAfterDurationA)
    }
    func actionsB(){

        //Start spawning, moving and removing
        let spawnB = SKAction.runBlock({
            () in

            //Spawn B
            self.setupB()
            self.addChild(self.B)

            //Move left and remove when go off screen
            let frameHeight = CGFloat(self.frame.height)
            let moveB = SKAction.moveByX(0, y: -frameHeight, duration: NSTimeInterval(0.0028 * frameHeight))   //duration: faster or slower
            let removeB = SKAction.removeFromParent()

            let moveAndRemoveB = SKAction.sequence([moveB, removeB])

            self.B.runAction(moveAndRemoveB)
        })

        //Spawn B each 0.5~1.0 seconds
        durationB = (CGFloat(arc4random_uniform(51)) + 50.0) / 100.0
        let spawnAfterDurationB = SKAction.repeatActionForever(SKAction.sequence([SKAction.waitForDuration(Double(durationB)), spawnB]))
        runAction(spawnAfterDurationB)
    }

    override func didMoveToView(view: SKView) {
        /* Setup your scene here */
        actionsA()
        actionsB()
    }
}

<2>蒂莫西史密斯的尝试

如果需要，可以下载here。

import SpriteKit

class GameScene: SKScene {

    //Declarations
    var A = SKSpriteNode()
    var B = SKSpriteNode()

    //Setup
    func setupA(){
        A = SKSpriteNode(imageNamed: "A")
        A.position = CGPoint(x: self.frame.width / 2, y: self.frame.height)
    }
    func setupB(){
        B = SKSpriteNode(imageNamed: "B")
        B.position = CGPoint(x: self.frame.width / 2, y: self.frame.height)
    }

    //Actions
    func actionsA(){

        //Spawn, move and remove
        let spawnA = SKAction.runBlock({
            () in

            //Spawn A
            self.setupA()
            self.addChild(self.A)

            //Move left and remove when go off screenk
            let frameHeight = CGFloat(self.frame.height)
            let moveA = SKAction.moveByX(0, y: -frameHeight, duration: NSTimeInterval(0.0028 * frameHeight))  //duration: faster or slower
            let removeA = SKAction.removeFromParent()

            let moveAndRemoveA = SKAction.sequence([moveA, removeA])

            self.A.runAction(moveAndRemoveA)
        })

        runAction(spawnA)
    }
    func actionsB(){

        //Spawn, move and remove
        let spawnB = SKAction.runBlock({
            () in

            //Spawn B
            self.setupB()
            self.addChild(self.B)

            //Move left and remove when go off screen
            let frameHeight = CGFloat(self.frame.height)
            let moveB = SKAction.moveByX(0, y: -frameHeight, duration: NSTimeInterval(0.0028 * frameHeight))   //duration: faster or slower
            let removeB = SKAction.removeFromParent()

            let moveAndRemoveB = SKAction.sequence([moveB, removeB])

            self.B.runAction(moveAndRemoveB)
        })

        runAction(spawnB)
    }

    //Spawn
    func spawn(){

        let random = Int(arc4random_uniform(10)+1) //pick a number from 1 to 10

        if random <= 8{
            actionsB()
        }
        else{
            actionsA()
        }
    }

    override func didMoveToView(view: SKView) {
        /* Setup your scene here */
        runAction(SKAction.repeatActionForever(SKAction.sequence([SKAction.waitForDuration(0.5, withRange: 0.25), SKAction.runBlock(self.spawn)])))
    }
}

提前谢谢，
路易斯。

Answer 1

一种可能的解决方案是使用单个函数spawn，每0.25到1.00秒随机调用一次。在函数内部，随机选择A或B，并生成它。看起来你想要B比A更频繁出现，所以你可以使用加权随机性（从0.45到1选择一个数字并舍入到最接近的整数）。

这是否有效取决于B分离是否超过1秒是否可以接受。

Answer 2

每次运行(arc4random_uniform(50)+51)/100时，它都会为您提供0.5到1.0秒之间的随机持续时间。您可以使用以下代码生成A＆amp;的随机延迟B

durationA = (arc4random_uniform(50)+51)/100 durationB = (arc4random_uniform(50)+51)/100

此外，您可以运行for循环以将durationB与durationA进行比较，并且仅在durationB与durationA不同时才接受$(function(){ var hasBeenTrigged = false; $(window).scroll(function() { if ($(this).scrollTop() >= 100) { // if scroll is greater/equal then 100 and hasBeenTrigged is set to false. $("#sticky_nav").show(); hasBeenTrigged = true; } else { $("#sticky_nav").hide(); } }); })。

在任意持续时间内生成对象而不覆盖

说明

问题

代码

2 个答案: