我有这样的代码:
void OnCollisionEnter (Collision collision)
{
if (canCollide == true) {
canCollide = false;
ContactPoint contactPoint = collision.contacts [0];
GameObject cube = GameObject.CreatePrimitive (PrimitiveType.Cube);
cube.transform.localScale = new Vector3 (5, 5, 5);
cube.transform.position = contactPoint.point;
iTween.MoveTo (cube, iTween.Hash (
"y", 40,
"time", 0.5));
iTween.MoveTo (cube, iTween.Hash (
"position", new Vector3 (55, 79, 10),
"time", 0.5f,
"delay", 0.5f,
"oncompletetarget", GameObject.Find ("PlayerCar"),
"oncomplete", "IncrementGauge"));
Destroy (cube, 1.1f);
}
}好的,这两个动画都是顺序的。但是时间之和应该是1秒。但是破坏()不能延迟1秒,必须至少是1秒。为什么?我是否可以以某种方式更快地销毁多维数据集,并且仍然触发IncrementGauge() (在销毁时1秒,防止触发IncrementGauge())。
谢谢。
发布于 2015-08-09 21:50:49
这是因为如果超过1.0秒,iTween库会在每个更新调用中进行检查,这不可能是1秒,因为它依赖于框架,浮点算法不精确。
您的问题可以通过销毁多维数据集并在传递给参数DestroyCubeAndIncrementGauge()作为字符串传递给参数oncomplete的同一个回调方法中递增来解决。
public class CollisionObject : MonoBehaviour
{
[SerializeField]
private bool canCollide = true;
void OnCollisionEnter(Collision collision)
{
if (canCollide == true) {
canCollide = false;
ContactPoint contactPoint = collision.contacts [0];
GameObject cube = GameObject.CreatePrimitive(PrimitiveType.Cube);
cube.transform.localScale = new Vector3(5, 5, 5);
cube.transform.position = contactPoint.point;
iTween.MoveTo(cube, iTween.Hash(
"y", 40,
"time", 0.5));
iTween.MoveTo(cube, iTween.Hash(
"position", new Vector3 (55, 79, 10),
"time", 0.5f,
"delay", 0.5f,
"oncompletetarget", this.gameObject,
"oncompleteparams", cube,
"oncomplete", "DestroyCubeAndIncrementGauge"
));
}
}
public void DestroyCubeAndIncrementGauge(GameObject cube)
{
// destroy the cube
Destroy(cube);
// inform the PlayerCar to IncrementGauge
GameObject.Find("PlayerCar").SendMessage("IncrementGauge");
}
}https://stackoverflow.com/questions/31893015
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