layui根据属性隐藏表格按钮

 

 根据状态显示按钮

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<script type="text/html" id="currentTableBar">              <a class="layui-btn layui-btn-normal layui-btn-xs data-count-edit" lay-event="edit">编辑</a>            {{# if(d.state==0){}}            <a class="layui-btn layui-btn-xs layui-btn data-count-delete" lay-event="start">启用</a>            {{# } if(d.state==1){}}            <a class="layui-btn layui-btn-xs layui-btn-danger data-count-delete" lay-event="delete">禁用</a>            {{#}}}            <a class="layui-btn layui-btn-xs layui-btn-warm data-count-delete" lay-event="reset">重置密码</a>          </script>

　　d代表当前行的对象，layui模板引擎写在{{}}中，d为内置对象不需要声明

 

 

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if(d.state == 0){   *************  }else if(d.state == 1){   *************  }