Why is offset().top throwing an error?

Question

JavaScript

Hello! There is this code:

var elem = jQuery('.table-price-wrapper tr').eq(find);
            var ypos = jQuery(elem).offset().top;
           jQuery('.table-price-wrapper tr').removeClass('active');
        	jQuery('.table-price-wrapper').animate({scrollTop: jQuery('.table-price-wrapper').scrollTop() - 20 + ypos + 'px'}, 700);
            jQuery('.table-price-wrapper tr:eq(' + find + ')').addClass('active');

In the console, the error immediately - Uncaught TypeError: Cannot read property 'top' of undefined
What could be the error? the find variable is a digit that always exists.

Answer 1

Butter oil...

//eq возвращает jQuery-объект, а не обычный элемент
//И получается у вас такое действо:
$($(something)).offset().top;

-->
hmm....check what happens in elem ?

Answer 2

Check the find variable, if it is greater than the number of elements, there will be an error like this