Question

所以下面你会找到一小段代码。这段代码的作用是允许玩家点击'p'键暂停游戏，当发生这种情况时，会弹出一个gui，玩家看起来和动作控制被禁用。我的问题是停用和重新激活gui，因为它是一个游戏对象。它让我停用它，但是当我尝试激活它时，我收到错误。

代码：

    UnityEngine.Component walkScriptOld = GameObject.FindWithTag ("Player").GetComponent ("CharacterMotor");
    UnityEngine.Behaviour walkScript = (UnityEngine.Behaviour)walkScriptOld;
    UnityEngine.GameObject guiMenu = GameObject.FindWithTag ("Canvas");
    if ((Input.GetKey ("p")) && (stoppedMovement == true)) {
        stoppedMovement = false;
        walkScript.enabled = true;
        guiMenu.SetActive(true);
    } else if ((Input.GetKey ("p")) && (stoppedMovement == false)) {
        stoppedMovement = true;
        walkScript.enabled = false;
        guiMenu.SetActive(false);
    }

错误：

NullReferenceException: Object reference not set to an instance of an object MouseLook.Update () (at Assets/Standard Assets/Character Controllers/Sources/Scripts/MouseLook.cs:44)

Answer 1

您在此处提供的代码似乎在更新中。因此，每个帧都会找到并存储guiMenu对象。

您要做的是将对象缓存在Awake或Start函数中，其余代码将正常工作。另请注意，缓存始终是一种很好的做法。

    //This is the Awake () function, part of the Monobehaviour class
    //You can put this in Start () also
    UnityEngine.GameObject guiMenu;  
    void Awake () {
        guiMenu = GameObject.FindWithTag ("Canvas");
    }

    // Same as your code
    void Update () {
        if ((Input.GetKey ("p")) && (stoppedMovement == true)) {
            stoppedMovement = false;
            walkScript.enabled = true;
            guiMenu.SetActive(true);
        } else if ((Input.GetKey ("p")) && (stoppedMovement == false)) {
            stoppedMovement = true;
            walkScript.enabled = false;
            guiMenu.SetActive(false);
        }
    }

统一激活和停用游戏对象

1 个答案: