Question

我的游戏中有一个构建机制，我想知道如何做到这一点，所以每当我将对象放下时，它都会捕捉到网格。每当我按0时，它将对象移动到鼠标上，而当我单击时，将对象向下放置。我想在放置时使其对齐网格，以免对象碰撞。

using UnityEngine;
using UnityEngine.AI;

public class GroundPlacementController : MonoBehaviour
{
    [SerializeField]
    private GameObject placeableObjectPrefab;
    public NavMeshObstacle nav;


    [SerializeField]
    private KeyCode newObjectHotkey = KeyCode.A;
    
    private GameObject currentPlaceableObject;


    private float mouseWheelRotation;

    



    private void Update()
    {
        HandleNewObjectHotkey();
        nav = GetComponent<NavMeshObstacle>();
        if (currentPlaceableObject != null)
        {
            MoveCurrentObjectToMouse();
            RotateFromMouseWheel();
            ReleaseIfClicked();



        }
    }

    private void HandleNewObjectHotkey()
    {
        if (Input.GetKeyDown(newObjectHotkey))
        {
            if (currentPlaceableObject != null)
            {
                Destroy(currentPlaceableObject);
               
            }
            else
            {
                currentPlaceableObject = Instantiate(placeableObjectPrefab);


            }
        }

    }
    

    private void MoveCurrentObjectToMouse()
    {
        Ray ray = Camera.main.ScreenPointToRay(Input.mousePosition);

        RaycastHit hitInfo;
        if (Physics.Raycast(ray, out hitInfo))
        {
            currentPlaceableObject.transform.position = hitInfo.point;
            currentPlaceableObject.transform.rotation = Quaternion.FromToRotation(Vector3.up, hitInfo.normal);
            currentPlaceableObject.GetComponent<NavMeshObstacle>().enabled = false;


        }
    }

    private void RotateFromMouseWheel()
    {
        Debug.Log(Input.mouseScrollDelta);
        mouseWheelRotation += Input.mouseScrollDelta.y;
        currentPlaceableObject.transform.Rotate(Vector3.up, mouseWheelRotation * 90f);
    }

    private void ReleaseIfClicked()
    {
        if (Input.GetMouseButtonDown(0))
        {
            
            currentPlaceableObject.GetComponent<NavMeshObstacle>().enabled = true;
            print("disabled");
            currentPlaceableObject = null;
            print("removed prefab");
        }
    }
}

Answer 1

您只需将transform.position的每个轴四舍五入即可在代码中对齐网格：

var currentPosition = transform.position;
transform.position = new Vector3(Mathf.Round(currentPosition.x),
                                 Mathf.Round(currentPosition.y),
                                 Mathf.Round(currentPosition.z));

Answer 2

一种更数学的方法，并且您可以轻松设置栅格大小的方法（除了Lews Therin的答案）将是：

采用position mod yourGridSize（例如，您的网格大小为64，位置为144。然后：144 mod 64 = 16）
取mod结果并从位置中减去144-16 = 128
最后，将以上结果除以yourGridSize：128/64 = 2

现在您知道自己的位置是2格到网格中。对三个轴应用此操作：

var yourGridSize = 64;

var currentPosition = currentPlaceableObject.transform.position;
currentPlaceableObject.transform.position = new Vector3(((currentPosition.x - (currentPosition.x % yourGridSize)) / yourGridSize) * yourGridSize,
                                                        ((currentPosition.y - (currentPosition.y % yourGridSize)) / yourGridSize) * yourGridSize,
                                                        ((currentPosition.z - (currentPosition.z % yourGridSize)) / yourGridSize) * yourGridSize);

令人费解？也许。有效？ 哎呀！

使可放置对象对齐网格

2 个答案: