javascript - jQuery .offset() 未定义 var 对象的类“top”

Question

我正在处理以下代码，并且有点难过为什么我在调试器中遇到错误。显然在以varvar skillsLimit = ...类开头的行中是未定义的。我在这里错误地设置了var吗？.topskillsOffsetskillsOffset

// 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

$('#skills')从对 的赋值中删除skillsLimit，您已经在上面的行中获得了对象：

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