識別兩個骷髏Kinect

Question

我用一個kinect做了一個遊戲（pong），我可以識別一個骷髏，並且我可以做出手勢來控制左邊的杖，但是當我試圖和兩個玩家一起玩時，不會識別這個骷髏其他玩家。這是我到目前爲止：

private void SensorSkeletonFrameReady(object sender, SkeletonFrameReadyEventArgs e) 
     { 
      Skeleton[] skeletons = new Skeleton[0]; 

      using (SkeletonFrame skeletonFrame = e.OpenSkeletonFrame()) 
      { 
       if (skeletonFrame != null) 
       { 
        skeletons = new Skeleton[skeletonFrame.SkeletonArrayLength]; 
        skeletonFrame.CopySkeletonDataTo(skeletons); 
       } 
       if (skeletons.Length != 0) 
       { 
        foreach (Skeleton skel in skeletons) 
        { 
         if (skel.TrackingState == SkeletonTrackingState.Tracked) 
         { 
          this.tracked(skel); 
          this.trackedLeft(skel); 
         } 
        } 
       } 
      } 
     } 

     public void tracked(Skeleton skeleton) 
     { 
      Joint jHandRight = skeleton.Joints[JointType.HandRight]; 
      Joint jHipCenter = skeleton.Joints[JointType.HipCenter]; 
      if ((jHipCenter.Position.Z - jHandRight.Position.Z) > 0.2) 
      { 
       //Consider hand raised in front of them 
       //System.Diagnostics.Debug.WriteLine("Hand: Raised"); 
       //MessageBox.Show("POR FAVORRRRRRRR"); 
       //movement[0] = false; 
       movement[0] = true; 
       movement[1] = false; 
      } 
      else 
      { 
       //Hand is lowered by the users side 
       //System.Diagnostics.Debug.WriteLine("Hand: Lowered"); 
       //MessageBox.Show("A SERRRRIIIIIOOOOOOOOOOOOOO"); 
       //movement[1] = false; 
       movement[1] = true; 
       movement[0] = false; 
      } 
     } 

有人可以幫助我。

Answer 1

您需要一種方法來區分骨架一和二。有關如何執行此操作，請參見Kinect user Detection。然後，您可以將玩家一和二的骨骼傳遞給您兩種不同的方法。我使用玩家ID，因爲如果一幀或兩幀的骨架丟失，它們的ID保持不變

int id1 = 0, id2 = 0; 

... 

if (skeletons.Length != 0) 
{ 
    foreach (Skeleton skel in skeletons) 
    { 
     if (skel.TrackingState == SkeletonTrackingState.Tracked) 
     { 
       if (skel.TrackingID == id1) 
        this.tracked(skel); 
       else if (skel.TrackingID == id2) 
        this.trackedLeft(skel); 
       else 
       { 
        if (id1 != 0 && id2 == 0) 
         id2 = skel.TrackingID; 
        else if (id2 != 0 && id1 == 0) 
         id1 = skel.TrackingID; 
       } 
     } 
    } 
    }