c# - 我怎样才能生成楼梯？

Question

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class GenerateStairs : MonoBehaviour
{
    public GameObject stairsPrefab;
    public int delay = 3;
    public int stairsNumber = 5;
    public int stairsHeight = 0;
    public Vector3 stairsPosition;
    public Vector2 stairsSize;

    // Use this for initialization
    void Start ()
    {
        StartCoroutine(BuildStairs());
    }

    // Update is called once per frame
    void Update ()
    {

    }

    private IEnumerator BuildStairs()
    {
        for (float i = 0; i <= stairsSize.x; i++)
        {
            for (float k = 0; k <= stairsSize.y; k++)
            {
                stairsPosition = new Vector3(i, stairsHeight, k);
                GameObject stairs = Instantiate(stairsPrefab, stairsPosition, Quaternion.identity);
                stairs.transform.localScale = new Vector3(stairsSize.x, 1 , stairsSize.y);

                stairsHeight += 1;

                yield return new WaitForSeconds(delay);
            }
        }
    }

    private void CalculateNextStair()
    {

    }
}

我搞砸了。例如，我想 build 5 个楼梯，但循环超过了楼梯的大小，而不是楼梯的数量。

其次，它创建了 10 组楼梯而不是 5 组楼梯:

另一个问题是我怎样才能让每个楼梯都慢慢 build ？现在它只是缓慢地延迟实例化，但我如何延迟生成每个楼梯？

脚本检查器的屏幕截图:

我当前的代码:

using System.Collections;
using System.Collections.Generic;
using UnityEngine;

public class GenerateStairs : MonoBehaviour
{
    public GameObject stairsPrefab;
    public float delay = 0.3f;
    public int stairsNumber = 5;
    public int stairsPositions = 0;
    public int stairsStartPositionHeight = 0;
    public float stairsScalingHaight = 1;
    public Vector2 stairsPosition;
    public Vector2 stairsSize;

    // Use this for initialization
    void Start()
    {
        StartCoroutine(BuildStairs());
    }

    // Update is called once per frame
    void Update()
    {

    }

    private IEnumerator BuildStairs()
    {
        for (float i = 0; i <= stairsNumber; i++)
        {
            // x=0f, y=z=stairsHeight
            stairsPosition = new Vector3(0f, stairsPositions, stairsPositions);
            GameObject stairs = Instantiate(
                    stairsPrefab,
                    stairsPosition,
                    Quaternion.identity);
            stairs.transform.localScale = new Vector3(
                    stairsSize.x,
                    stairsScalingHaight,
                    stairsSize.y);

            stairsStartPositionHeight += 1;

            yield return new WaitForSeconds(delay);
        }
    }

    private void CalculateNextStair()
    {

    }
}

Answer 1

根本没有理由遍历楼梯的大小；您想要遍历 stairsNumber，它在您的代码中尚未使用。

此外，您不需要更改楼梯位置的 x 组件。将其保持在 0f(或您需要的任何值)。

楼梯位置的 y 和 z 分量(相对于起点)都应该是 stairHeight 的因子。在这种特殊情况下，您希望它们等于 stairHeight，以便获得“方形”阶梯形状。

private IEnumerator BuildStairs()
{
    for (int i = 0; i <= stairsNumber ; i++) {
        // x=0f, y=z=stairsHeight
        stairsPosition = new Vector3(0f, stairsHeight, stairsHeight);
        GameObject stairs = Instantiate(
                stairsPrefab, 
                stairsPosition, 
                Quaternion.identity);
        stairs.transform.localScale = new Vector3(
                stairsSize.x, 
                1f , 
                stairsSize.y);

        stairsHeight += 1f;

        yield return new WaitForSeconds(delay);
    }
}

如果您将 stairSize 更改为 Vector3，那么您可以直接使用 stairSize 作为 localScale ，并将 stairsHeight 增加 stairsSize.y 而不是仅仅增加 1f。

如果您想偏移楼梯的起始位置，则需要包含一个偏移量。我建议将它与高度计数器分开，直到您需要添加它们。

此外，如果您想要矩形大小的步长，请将 widthFactor 乘以高度以计算每个步长水平移动的距离。

组合这些更改可能如下所示:

Vector3 stairSize;
float stepWidthFactor=1f;
Vector3 stairsStartPosition;

private IEnumerator BuildStairs() {
    for (int i = 0; i <= stairsNumber ; i++) {

        stairsPosition = new Vector3(
                stairsStartPosition.x, 
                stairsStartPosition.y + stairsHeight,
                stairsStartPosition.z + stairsHeight*stepWidthFactor);

        GameObject stairs = Instantiate(
                stairsPrefab, 
                stairsPosition, 
                Quaternion.identity);

        stairsHeight += stairsSize.y;
        stairs.transform.localScale = stairSize;

        yield return new WaitForSeconds(delay);
    }
}