使可放置對象捕捉到網格

Question

我在我的游戲中有一個構建機制，我想知道我是如何做到這一點的，所以每當我放下對象時它都會對齊網格。 每當我按0時，它會將對象移動到我的鼠標上，每當我點擊它時，都會將對象放下。 我想讓它在放置時對齊網格，這樣物體就不會碰撞。

 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

您可以通過簡單地舍入transform.position每個軸來捕捉代碼中的網格：

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

Answer 2

一個更加數學的方法，你可以輕松設置網格大小（除了Lews Therin的答案）將是：

1. 取position mod yourGridSize （比如，你的網格大小是64，位置是144. 那么 ：144 mod 64 = 16）
2. 取mod並從位置減去：144 - 16 = 128
3. 最后，將上面的結果除以yourGridSize ： yourGridSize = 2

現在你知道你的位置是進入網格的2個街區。 對三軸應用此操作：

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);

令人費解？ 也許。 有效？ 哎呀！