ajax成功后如何刷新表体

Question

$.ajax({
            type:"POST",
            url:"abc.php",
            data:dataString,
            success:function(response){ 

            //alert(response);
            // Here I want to write code to refresh table body          
            }
    });

Answer 1

通过在成功函数中替换表 ID 来试试这个

        success:function(response){ 

          $("#table_id").load(window.location + " #table_id");

        }

Space in important.