Question

我想生成一个客户端和服务器都可以看到的弹跳球。当客户连接并且下一个场景（球在弹跳的地方）被加载时，必须产生球。

问题是OnPlayerConnected（）方法没有被调用，有没有人知道任何替代方案以及它为什么不起作用？

public class MainMenu : MonoBehaviour {

public string connectionIP = "127.0.0.1";
public int connectionPort = 25001;
bool isClientConnected = false;

void Awake()
{
    // Make sure list is empty and request a new list
    MasterServer.RequestHostList("MyUniqueGameType");
}



// Use this for initialization
void OnGUI()
{
    HostData[] data = MasterServer.PollHostList ();
    // Go through all the hosts in the host list
    foreach (var element in data) 
    {
        //Debug.Log ("server found");
        GUILayout.BeginHorizontal ();    
        var name = element.gameName + " " + element.connectedPlayers + " / " + element.playerLimit;
        GUILayout.Label (name);  
        GUILayout.Space (5);
        string hostInfo;
        hostInfo = "[";
        foreach (var host in element.ip)
            hostInfo = hostInfo + host + ":" + element.port + " ";
        hostInfo = hostInfo + "]";
        GUILayout.Label (hostInfo);  
        GUILayout.Space (5);
        GUILayout.Label (element.comment);
        GUILayout.Space (5);
        GUILayout.FlexibleSpace ();
        if (GUI.Button (new Rect (10, 30, 120, 20), "Client Connect")) 
        {
            Network.Connect (connectionIP, connectionPort);
            isClientConnected = true;
            Application.LoadLevel ("scene1");

            //  Debug.Log ("spawning ball");
            //  SpawnBall();


        }
        GUILayout.EndHorizontal ();  

    }


    if (GUI.Button (new Rect (10, 50, 120, 20), "Initialize Server"))
    {
        Network.InitializeServer (2, connectionPort, !Network.HavePublicAddress ());
        //  Network.InitializeServer (2, 25002, !Network.HavePublicAddress ());
        MasterServer.RegisterHost ("MyUniqueGameType", "JohnDoes game", "l33t game for all");
                    /*
                    while(!isClientConnected)
                    {
                        Debug.Log("waiting for client");
                    }   
                    */

        Application.LoadLevel("scene1");

    }
}

void OnPlayerConnected(NetworkPlayer player)
{
    Debug.Log ("OnPlayerConnected");
    SpawnBall();
}


void SpawnBall()
{
    GameObject objPrefab = Resources.Load("Ball") as GameObject;
    GameObject obj = Instantiate(objPrefab) as GameObject;
}

Answer 1

加载新场景时，此脚本将停止执行。因此，永远不会在客户端或服务器上调用消息处理程序OnPlayerConnected。

您可以做的是使用RPC（远程过程调用）让客户端在连接并加载场景时通知服务器（这将在第二个场景中）：

void Start() {
    if(Network.isClient) {
        networkView.RPC("SpawnBall", RPCMode.AllBuffered)
    }
}

[RPC]
void SpawnBall(NetworkViewID viewID, Vector3 location) {
    GameObject objPrefab = Resources.Load("Ball") as GameObject;
    GameObject obj = Instantiate(objPrefab) as GameObject;
}

现在当客户端加载场景时，他会向服务器发送一个RPC，告诉他产生一个球。请注意Start函数中我使用Network.isClient指定只在客户端加载场景时才执行此代码。 RPCMode.AllBuffered会将RPC发送给所有连接的播放器，包括发送它的客户端。

RPC是一种有价值的工具，如果在Unity中进行网络连接，您应该熟悉它们。

当客户连接时产卵球

1 个答案: