Kinect user Detection

I am developing an application When an kinect sensor detects an skeleton that person can work on it if other person comes near to the existing user it detects the second person.
I want to restrict to the user the kinect sensor first detects it if other user comes this should not detect the other one.
thanks in advance

Also see Jurgeon D's answer on Kinect SDK player detection, as it deals with skeleton index. @Fixus is also right in that you could use a ID. But if you mean more than 2 people are detected, then only one is detected, that is not programming, that is in the Kinect's hardware and @FelixK. was right.

Skeletal Index

void nui_SkeletonFrameReady(object sender, SkeletonFrameReadyEventArgs e) 
{
    SkeletonFrame sf = e.SkeletonFrame;
    //check which skeletons in array are active and
    // use that array indexes for player index
    SkeletonData player1 = sf.Skeletons[playerIndex1];
    SkeletonData player2 = sf.Skeletons[playerIndex2];
}

Skeletal IDs

void nui_SkeletonFrameReady(object sender, SkeletonFrameReadyEventArgs e)
{
    SkeletonFrame sf = e.SkeletonFrame;

    if (sf.TrackingState == SkeletalTrackingState.Tracked)
     {
          int ID1 = sf.TrackingID;
     }

Also the code for detecting humans

 DepthImageFrame depthFrame;
 short[] rawDepthData = new short[depthFrame.PixelDataLength];
 depthFrame.CopyPixelDataTo(rawDepthData); 
 Byte[] pixels = new byte[depthFrame.Height * depthFrame.Width * 4];     
 int player = rawDepthData[depthIndex] & DepthImageFrame.PlayerIndexBitmask;

 if (player > 0)
 {
     //do something
 }      

Kinect will detec the new user cause it is his job :) BUT remember that every user has his own ID so you always know that the first user is first and second is second. That way you can work only on the skeleton of the user that you want to work with

If i understand your question correctly this is not possible, you can't modify the Kinects behaviour and how it detects users ( If there is nothing in the Framework; i don't think there is something ).

You have to solve this in your code.