Unity中，设定指定Menu的状态作为功能开关

redhat9i · 发表于 2022-1-26 09:41

例如，做一个这样的开关：勾选时，显示Log；不勾选时，不显示Log。

image.png

勾选时

image.png

不勾选时

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#if UNITY_EDITORusing System.Collections;using System.Collections.Generic;using UnityEngine;using UnityEditor;namespace GM{ public class GMTools { private const string ShowLogMenuPath = "Tools/GM/显示Log"; [MenuItem("Tools/GM/显示Log")] public static void ShowLogTool() { Menu.SetChecked(ShowLogMenuPath, !Menu.GetChecked(ShowLogMenuPath)); } public static bool ShouldShowLog{ get { return Menu.GetChecked(ShowLogMenuPath); } } }}#endifusing System.Collections;using System.Collections.Generic;using UnityEngine;public class Test : MonoBehaviour{ // Start is called before the first frame update void Start() { } // Update is called once per frame void Update() { if (UnityEngine.Input.GetKeyDown(KeyCode.Space)) { ShowLog("显示log111"); Debug.Log("显示log222"); } } public void ShowLog(string str) {#if UNITY_EDITOR if (GM.GMTools.ShouldShowLog) { Debug.Log(str); } }#endif}

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Unity中，设定指定Menu的状态作为功能开关

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