控制和创建多个精灵阵列的Java Libgdx [英] Controlling and creating multiple sprites array Java Libgdx

问题描述

我想创建一个游戏，有精灵和每秒另一个是催生，我试图用这个作为基础：的 https://github.com/libgdx/libgdx/wiki/A-simple-game 然而，当新的派生于它破坏了旧的，他们开始产卵越来越快。 下面是相应和code：

ownCreate方式：

 阵列＆LT;矩形＆GT;鸡;
鸡=新的Array＆LT;矩形＆GT;（）;
spawnChicken（）;
 

新的产卵鸡的方法：

 私人无效spawnChicken（）{


        INT选择;
        选择= MathUtils.random（0，3）;
        开关（选择）{

        情况下0：
            chicken.x = MathUtils.random（0,1920-85）;
            chicken.y = 0;
            打破;
        情况1：
            chicken.x = MathUtils.random（0,1920-85）;
            chicken.y = 1080至85年;
            打破;
        案例2：
            chicken.x = 0;
            chicken.y = MathUtils.random（0,1080-66）;
            打破;
        案例3：
            chicken.x = 1920年至1985年;
            chicken.y = MathUtils.random（0,1080-66）;
            打破;
        }


    chicken.height = 66;
    chicken.width = 85;
    chickens.add（鸡）;
    runningChickens ++;
    lastSpawnTime = TimeUtils.nanoTime（）;
 

Render方法：

 如果（TimeUtils.nanoTime（） -  lastSpawnTime＆GT; 10亿）
    spawnChicken（）;

迭代器＆LT;矩形＆GT; ITER = chickens.iterator（）;

而（iter.hasNext（））{
    矩形鸡= iter.next（）;
    //运动
 

非常相似，维基：

 私人阵列＆LT;矩形＆GT;雨滴;
   专用长lastDropTime;私人无效spawnRaindrop（）{
      矩形雨滴=新的Rectangle（）;
      raindrop.x = MathUtils.random（0，800-64）;
      raindrop.y = 480;
      raindrop.width = 64;
      raindrop.height = 64;
      raindrops.add（雨滴）;
      lastDropTime = TimeUtils.nanoTime（）;
   }
 雨滴=新的Array＆LT;矩形＆GT;（）;
   spawnRaindrop（）;
   如果（TimeUtils.nanoTime（） -  lastDropTime＆GT;十亿）spawnRaindrop（）;
 迭代器＆LT;矩形＆GT; ITER = raindrops.iterator（）;
   而（iter.hasNext（））{
      矩形雨滴= iter.next（）;
      raindrop.y  -  = 200 * Gdx.graphics.getDeltaTime（）;
      如果（raindrop.y + 64℃的）iter.remove（）;
   }
 

解决方案

把

 矩形鸡=新的Rectangle（）;
 

在 spawnChicken（）方法的开始。

---

原因

• 如果您保持鸡一员，它会在每次 spawnChicken（）方法被调用时更换。
• 此外，相同的对象每次尝试产卵鸡时添加到阵列中。

I am trying to create a game which has sprites and every second another is spawned, i tried using this as a base: https://github.com/libgdx/libgdx/wiki/A-simple-game However when the new one spawns in it destroys the old one and they start to spawn faster and faster. Here is the relevent code:

ownCreate method:

Array<Rectangle> chickens;
chickens = new Array<Rectangle>();
spawnChicken();

New Spawn Chicken method:

private void spawnChicken() {


        int choice;
        choice = MathUtils.random(0, 3);
        switch (choice){

        case 0:
            chicken.x = MathUtils.random(0,1920-85);
            chicken.y = 0;          
            break;
        case 1:
            chicken.x = MathUtils.random(0,1920-85);
            chicken.y = 1080-85;    
            break;
        case 2:
            chicken.x = 0;
            chicken.y = MathUtils.random(0,1080-66);
            break;
        case 3:
            chicken.x = 1920-85;
            chicken.y = MathUtils.random(0,1080-66);
            break;
        }


    chicken.height = 66;
    chicken.width = 85;
    chickens.add(chicken);
    runningChickens ++;
    lastSpawnTime = TimeUtils.nanoTime();

Render method:

if(TimeUtils.nanoTime() - lastSpawnTime > 1000000000)
    spawnChicken();

Iterator<Rectangle> iter = chickens.iterator();

while(iter.hasNext()){      
    Rectangle chicken = iter.next();
    //movement

Very similar to the wiki:

private Array<Rectangle> raindrops;
   private long lastDropTime; private void spawnRaindrop() {
      Rectangle raindrop = new Rectangle();
      raindrop.x = MathUtils.random(0, 800-64);
      raindrop.y = 480;
      raindrop.width = 64;
      raindrop.height = 64;
      raindrops.add(raindrop);
      lastDropTime = TimeUtils.nanoTime();
   }
 raindrops = new Array<Rectangle>();
   spawnRaindrop();
   if(TimeUtils.nanoTime() - lastDropTime > 1000000000) spawnRaindrop();
 Iterator<Rectangle> iter = raindrops.iterator();
   while(iter.hasNext()) {
      Rectangle raindrop = iter.next();
      raindrop.y -= 200 * Gdx.graphics.getDeltaTime();
      if(raindrop.y + 64 < 0) iter.remove();
   }

解决方案

Put

Rectangle chicken = new Rectangle();

at the start of spawnChicken() method.

---

Reasons

• If you keep chicken a member, it will be replaced every time spawnChicken() method is called.
• Also, the same object is added to the array every time you try to spawn a chicken.

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