android - how to set button position -

i'm beginner android developer have skills in c#. have made sliding puzzle c# , want made on android.

this code :

  tablelayout layout = new tablelayout(this);   setcontentview(layout);   layout.setbackgroundcolor(color.gray);   final textview text = new textview(this);   layout.addview(text);    layout.setstretchallcolumns(true);       layout.setshrinkallcolumns(true);     int rows = 3; final int colomuns = 3;     ( int = 0 ; < rows ; ++)   {        tablerow tr = new tablerow(this);        ( int k = 0 ; k < colomuns ; k++)       {           final int indexk = k+1;               final button btn = new button(this);               final int indexi = i;               final int index = indexi*colomuns+indexk ;                 btn.setid(index);                 btn.settextsize(36);                 btn.setpadding(20, 20, 20, 20);                 btn.settag(new point(indexi, indexk));                 btn.settext(""+ index);                  //layout params                 tablerow.layoutparams params = new tablerow.layoutparams();                 params.setmargins(5, 5, 5, 5);                 btn.setlayoutparams(params);                 //----------------                 btn.setbackgroundcolor(color.white);                 btn.setonclicklistener(new view.onclicklistener()                  {                     public void onclick(view v) {                           tablerow.layoutparams leftviewparams = new tablerow.layoutparams();                         leftviewparams.column = 1;                           btn.setlayoutparams(leftviewparams);                         point p = (point) btn.gettag();                        }                 });                 tr.addview(btn);        }        layout.addview(tr);   } 

but, have problem because haven't found "position" property, , want how layout ? , choose tablelayout right ?

thanks,

you should add button layout , offset position using margins/padding, shouldn't positioning based on x,y

use left margin , top margins try , align require.