如果玩家5秒不动，Unity就会弹出显示。

Question

我在Unity 3d上有一个任务，如果玩家5秒不动，屏幕中央就会出现一个弹出窗口，如果玩家移动，弹出窗口就会消失。请教我如何编写这个任务的逻辑，谢谢。

Answer 1

下面是代码，会检查用户的鼠标位置，看其在最近5秒内是否有移动。如果没有，那么弹出窗口就会出现。如果这里的注释很难读懂（我认为是这样），请将这段代码复制并粘贴到Visual Studio中，这样颜色将有助于区分代码和注释。

[SerializeField] GameObject popupWindow = null;

float totTime;
float timeBeforePause = 5f;
Vector3 updatedMousePosition;

private void Update()
{
    //  Add the time delta between frames to the totTime var
    totTime += Time.deltaTime;

    //  Check to see if the current mouse position input is equivalent to updateMousePosition from the previous update
    //  If they are equivalent, this means that the user hasn't moved the mouse
    if (Input.mousePosition == updatedMousePosition)
    {
        //  Since the user hasn't moved the mouse, check to see if the total Time is greater than the timeBeforePause
         if (totTime >= timeBeforePause)
        {
            //  Set the popup window to true in order to show the window (instantiate instead it if if doesn't exist already)
            popupWindow.SetActive(true);
        }
    }

    //  If the user has moved the mouse, set the totTime back to 0 in order to restart the totTime tracking variable
    else
    {
       totTime = 0;
    }

    //  Check to see if the popup window is visible (active)
    if (popupWindow.activeSelf == true)
    {
        //  Check to see if the user has pressed the Esc button
        if (Input.GetKeyDown(KeyCode.Escape))
        {
            //  Hide the window
            popupWindow.SetActive(false);
        }
    }

    //  Update the updatedMousePosition before the next frame/update loop executes
    updatedMousePosition = Input.mousePosition;
}

如果你想跟踪不同的用户输入（按键），你可以使用类似的方法。此外，你还必须在弹出窗口上实现某种按钮，允许用户在返回后从弹出窗口中退出。希望这能帮到你