点击提交之后，如何只保留下面一行，把第一行表格中的的input框给去除，是remove的对象有问题？

Question

1. $("#submit").on('click', function (){
           let tableObj = document.getElementById("tableId");
   
           let rows = tableObj.rows;
           for (let i = 1; i < rows.length; i++) {
               let cells = rows[i].cells;
               let id = cells[0].innerHTML;
               let Vessel = cells[1].innerHTML;
               let Voyage = cells[2].innerHTML;
               let CtnWegtCla = cells[3].innerHTML;
               let CtnSize = cells[4].innerHTML;
               let CtnWegtRanLeft = cells[5].getElementsByTagName("input")[0].value;
               let CtnWegtRanRight = cells[6].getElementsByTagName("input")[0].value;
   
               let data = {
                   'id': id,
                   'Vessel': Vessel,
                   'Voyage': Voyage,
                   'CtnWegtCla': CtnWegtCla,
                   'CtnSize': CtnSize,
                   'CtnWegtRanLeft': CtnWegtRanLeft,
                   'CtnWegtRanRight': CtnWegtRanRight,
   
               };
               $.ajax({
                   url: "/berth/add_port_grade/",
                   type: "POST",
                   data: JSON.stringify(data),
   
                   success: function () {
                       alert('提交成功');
                       $(this).remove();
                       let tbodyObj = document.getElementsByTagName('tbody');
                       let newTr = document.createElement('tr');
                       let newTd = `
                       <td>${id}</td>
                       <td>${Vessel}</td>
                       <td>${Voyage}</td>
                       <td>${CtnWegtCla}</td>
                       <td>${CtnSize}</td>
                       <td>${CtnWegtRanLeft}</td>
                       <td>${CtnWegtRanRight}</td>
                       <td><input type="checkbox" class="delweight"/></td>
   
                       `;
                       newTr.innerHTML = newTd;
                       tbodyObj[0].appendChild(newTr);
   
                   }
               })
           }})
   

 

Answer 1

ajax请求已经跳出了当前对象，所以不能识别$(this)，可以在ajax之前申明一个$(this)的变量再使用