[SOLVED] What is the best way to list friends in a paged format using opensocial?

What is the best way to list friends in a paged format using opensocial?

I have got the list of my friends and i can traverse through the collection to list them as per my requirement. But I want to show them in different pages as the people on myspace may have more than 100s of friends, so I dont want to show them on the same page. What would be the best way to show them in pages.

Currently i m doing the following:

    <script type="text/javascript">
    function init()
    {
    //creating request to get the friends
     var request = opensocial.newDataRequest();
             var params0 = {};
                params0[opensocial.IdSpec.Field.USER_ID] = opensocial.IdSpec.PersonId.OWNER;
                params0[opensocial.IdSpec.Field.NETWORK_DISTANCE] = 1;
                var idspec = opensocial.newIdSpec(params0);
                request.add(request.newFetchPeopleRequest(idspec), "friends");
                request.add(request.newFetchPersonRequest("OWNER"), "get_owner");
      request.send(got_myfriends);
}
function got_myfriends(response0)
{
//document.getElementById('message').innerHTML += "TESTING" + response0.data;
var friends=response0.get("friends").getData().asArray();
//document.getElementById('message').innerHTML += " TESTING0";
 var owner = response0.get('get_owner').getData();
document.getElementById('message').innerHTML += " TESTING1--ownerid: " + owner.getId();
document.getElementById('message').innerHTML +=" Total friends found: " + friends.length;
      var id, image, name;
        for(var i = 0; i < friends.length; i++){
          id = friends[i].getId();
            image = friends[i].getField(opensocial.Person.Field.THUMBNAIL_URL);
            name = friends[i].getDisplayName();
      document.getElementById('friendsdiv').innerHTML += "<div style='display:inline;float:left;width:150px;'><input type='radio' name='friendid' value='" + id + "'>" +  name  +"<BR><img src='" + image + "'></div>"; 
        }


}
</script>

Solution