layui根据属性隐藏表格按钮
根据状态显示按钮
<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
if(d.state == 0){ ************* }else if(d.state == 1){ ************* }