jQuery .offset（）獲取未定義的類var對象的頂部

Question

我正在處理下面的代碼，並有點難倒爲什麼我在我的調試器中得到一個錯誤。顯然在以var skillsLimit = ...開頭的行中，變量skillsOffset的類.top未定義。我是否在這裏錯誤地設置了skillsOffset的變量？

// Set Pie graph to position fixed during a specified range 
     var $window = $(window); 
     var windowHeight = $(window).height(); 
     var pos = $window.scrollTop(); //position of the scrollbar 
     var $this = $(this); 

     $window.bind('scroll', function(){ //when the user is scrolling... 
      var pos = $window.scrollTop(); //position of the scrollbar    
      var skillsOffset = $('#skills').offset(); 
      var skillsLimit = $('#skills').skillsOffset.top + $('#skills').outerHeight(); 
      if (pos > skillsOffset) { 
       $('.chartwell-pies').css({ 'position' : 'fixed' }); 
      }    
     }); 

Answer 1

從分配中刪除$('#skills')到skillsLimit，你已經得到了在上述線路的對象：

var pos = $window.scrollTop(); //position of the scrollbar    
var skillsOffset = $('#skills').offset(); 
var skillsLimit = skillsOffset.top + $('#skills').outerHeight();