控制和创建多个精灵阵列的Java Libgdx

本文介绍了控制和创建多个精灵阵列的Java Libgdx的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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 github/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: github/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.