好的我有两个脚本，第一个使用invokerepeating以给定的速率产生障碍物。第二个脚本每20秒增加障碍物的速度。我的问题是，一旦速度增加，障碍物之间的差距就会增大，这导致每个障碍之间存在很大差距。关于如何解决这个问题的任何想法？
using UnityEngine;
using System.Collections;
public class Generate : MonoBehaviour
{
public Vector3 startPosition;
public GameObject object1;
public GameObject object2;
public GameObject object3;
public int random;
int score = 0;
public int count = 0;
private float timeLeft = 20.0f;
private int num1 =3;
private int num2 =3;
// Use this for initialization
void Start()
{
startPosition = new Vector3(5, 0, 0);
random = Random.Range(1, 4);
InvokeRepeating("CreateObstacle", num1, num2);
}
// Update is called once per frame
/*void OnGUI ()
{
GUI.color = Color.black;
GUILayout.Label(" Score: " + score.ToString());
}*/
void Update(){
fast ();
}
void CreateObstacle()
{
if (random == 1) {
GameObject go = Instantiate(object1, startPosition, Quaternion.identity) as GameObject;
go.transform.parent = GameObject.Find("4 - Obstacles").transform;
random = Random.Range(1, 4);
//Destroy(gameObject, 10);
}
else if (random == 2) {
GameObject go = Instantiate(object2, startPosition, Quaternion.identity) as GameObject;
go.transform.parent = GameObject.Find("4 - Obstacles").transform;
random = Random.Range(1, 4);
//Destroy(gameObject, 10);
}
else if(random == 3) {
GameObject go = Instantiate(object3, startPosition, Quaternion.identity) as GameObject;
go.transform.parent = GameObject.Find("4 - Obstacles").transform;
random = Random.Range(1, 4);
//Destroy(gameObject, 10);
}
}
void fast(){
//Timer
timeLeft -= Time .deltaTime;
if (timeLeft <= 0.0f) {
count ++;
timeLeft += 20.0f;
num1 += 1;
num2 += 1;
}
if (count == 1) {
num1 += 1;
num2 += 1;
count ++;
}
else if (count == 3) {
num1 += 1;
num2 += 1;
count ++;
}
}
void OnGUI () {
// Make a background box
GUI.Box (new Rect (10, 10, 100, 90), "Number" + num1);
}
}
第二个脚本
using System.Collections.Generic;
using System.Linq;
using UnityEngine;
/// <summary>
/// Parallax scrolling script that should be assigned to a layer
/// </summary>
public class Scrolling : MonoBehaviour
{
public int count = 0;
private float timeLeft = 20.0f;
/// <summary>
/// Scrolling speed
/// </summary>
private Vector2 speed = new Vector2(2, 2);
/// <summary>
/// Moving direction
/// </summary>
public Vector2 direction = new Vector2(-1, 0);
/// <summary>
/// Movement should be applied to camera
/// </summary>
public bool isLinkedToCamera = false;
/// <summary>
/// 1 - Background is infinite
/// </summary>
public bool isLooping = false;
/// <summary>
/// 2 - List of children with a renderer.
/// </summary>
private List<Transform> backgroundPart;
// 3 - Get all the children
void Start()
{
// For infinite background only
if (isLooping)
{
// Get all the children of the layer with a renderer
backgroundPart = new List<Transform>();
for (int i = 0; i < transform.childCount; i++)
{
Transform child = transform.GetChild(i);
// Add only the visible children
if (child.renderer != null)
{
backgroundPart.Add(child);
}
}
// Sort by position.
// Note: Get the children from left to right.
// We would need to add a few conditions to handle
// all the possible scrolling directions.
backgroundPart = backgroundPart.OrderBy(
t => t.position.x
).ToList();
}
}
void Update()
{
//InvokeRepeating("fast", 1, 1);
fast();
// Movement
Vector3 movement = new Vector3(
speed.x * direction.x,
speed.y * direction.y,
0);
movement *= Time.deltaTime;
transform.Translate(movement);
// Move the camera
if (isLinkedToCamera)
{
Camera.main.transform.Translate(movement);
}
// 4 - Loop
if (isLooping)
{
// Get the first object.
// The list is ordered from left (x position) to right.
Transform firstChild = backgroundPart.FirstOrDefault();
if (firstChild != null)
{
// Check if the child is already (partly) before the camera.
// We test the position first because the IsVisibleFrom
// method is a bit heavier to execute.
if (firstChild.position.x < Camera.main.transform.position.x)
{
// If the child is already on the left of the camera,
// we test if it's completely outside and needs to be
// recycled.
if (firstChild.renderer.IsVisibleFrom(Camera.main) == false)
{
// Get the last child position.
Transform lastChild = backgroundPart.LastOrDefault();
Vector3 lastPosition = lastChild.transform.position;
Vector3 lastSize = (lastChild.renderer.bounds.max - lastChild.renderer.bounds.min);
// Set the position of the recyled one to be AFTER
// the last child.
// Note: Only work for horizontal scrolling currently.
firstChild.position = new Vector3(lastPosition.x + lastSize.x, firstChild.position.y, firstChild.position.z);
// Set the recycled child to the last position
// of the backgroundPart list.
backgroundPart.Remove(firstChild);
backgroundPart.Add(firstChild);
}
}
}
}
}
void fast(){
//Timer
timeLeft -= Time .deltaTime;
if (timeLeft <= 0.0f) {
count ++;
timeLeft += 20.0f;
speed.x += 0.4f;
speed.y += 0.4f;
}
if (count == 1) {
speed.x += 0.4f;
speed.y += 0.4f;
count ++;
}
else if (count == 3) {
speed.x += 0.4f;
speed.y += 0.4f;
count ++;
}
/* else if (count == 5) {
speed.x ++;
speed.y ++;
count ++;
}*/
else if (count == 5) {
speed.x += 5;
speed.y += 5;
count ++;
}
}
void OnGUI ()
{
GUI.color = Color.black;
GUILayout.Label(" Speed: " + speed.ToString() + " Count: " + count.ToString ());
GUILayout.Label(" Time: " + timeLeft.ToString());
}
}
答案 0 :(得分：0)
您还必须减少障碍产生的时间（您在fast()方法中尝试但不会起作用）。您正在更新num1和num2值，但这些值永远不会分配给InvokeRepeating。
更新num1和num2之后还取消上一次调用并创建一个新的调用（我相信这次你只需要num2）。在fast()方法的末尾添加这两行。
CancelInvoke("CreateObstacle");
InvokeRepeating("CreateObstacle", num2, num2);
InvokeRepeating中的第一个时间值定义在为第一个方法调用方法之前等待的时间。第二时间值定义将重复方法调用的持续时间。 （简而言之，第一个值仅用于第一次，第二个值用于重复的方法调用）。