jQuery实现动态给table赋值的方法示例

<table class="layui-table" lay-skin="line" id="datas">
  <colgroup>
    <col width="150">
    <col width="200">
    <col>
  </colgroup>
  <thead>
  <tr>
    <th>昵称</th>
    <th>加入时间</th>
    <th>签名</th>
  </tr>
  </thead>
  <tbody>
  <tr id="template">
    <td id="id"></td>
    <td id="url"></td>
    <td id="title"></td>
  </tr>
  </tbody>
</table>

<script>
  var data = [{
      "id": 1,
      "url": "http://www.jqcool.net",
      "switch": 1,
      "order": 1,
      "pid": 0,
      "title": "Online Program knowledge share and study platform"
    },
      {
        "id": 2,
        "url": "http://www.baidu.com",
        "switch": 0,
        "order": 2,
        "pid": 2,
        "title": "这是测试的数数据这是测试的数数据"
      },
      {
        "id": 3,
        "url": "http://www.taobao.com",
        "switch": 0,
        "order": 3,
        "pid": 2,
        "title": "淘宝购物"
      },
      {
        "id": 4,
        "url": "http://www.jqcool.net1",
        "switch": 1,
        "order": 4,
        "pid": 2,
        "title": "Online Program knowledge share and study platform2"
      },
      {
        "id": 5,
        "url": "http://www.baidu.com1",
        "switch": 0,
        "order": 5,
        "pid": 2,
        "title": "这是测试的数数据2"
      },
      {
        "id": 6,
        "url": "http://www.taobao.com1",
        "switch": 1,
        "order": 6,
        "pid": 0,
        "title": "淘宝购物2"
      }];
  $.each(data, function (i, n) {
    var row = $("#template").clone();
    row.find("#id").text(n.id);
    row.find("#url").text(n.url);
    row.find("#title").text(n.title);
    row.appendTo("#datas");//添加到模板的容器中
  });
</script>