使用layui+ajax实现简单的菜单权限管理及排序的方法

//菜单循环遍历打印拼接操作
  $(function () {
  $.ajax( {
   url : projectName+'/sys/menu/findMenuByRole',
   type : 'post',
   dataType : 'json',
   success : function(data) {
   var menu = ""; //定义变量存储
   for(var i = 0;i<data.length;i++){
    menu += "<li class='layui-nav-item '>"
    if(data[i].pareMenuId == 0){ //取出父元素的菜单，拼进页面
    menu += "<a href='javascript:;'>"+data[i].menuName+"</a>"
    for(var j = 0;j<data.length;j++){ //继续遍历这几条数据
     if(data[j].mid > data[i].mid && data[j].mid < (data[i].mid)+1000){ //取出这个父元素所对应的子元素
     menu += "<dl class='layui-nav-child'>"
     menu += "<dd>"
     menu +=  "<a href='"+data[j].menuPath+"' target='option'>"+'&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'+data[j].menuName+"</a>"
     menu += "</dd>"
     menu += "</dl>"
     }
    }
    }
    menu += "</li>";
   }
   $("#nav").html(menu);
   var element = layui.element;
   element.init()//初始化element事件，使菜单展开
   }
  });
  })

 SELECT
  t1.*
  FROM
  (
  SELECT
   sys_menu.id,
   sys_menu.mid,
   sys_menu.menu_name,
   sys_menu.menu_path,
   sys_menu.pare_menu_id,
   sys_menu.`status`,
   sys_menu.`order`,
   abc.`status` AS status2
  FROM
   sys_menu
  INNER JOIN (
   SELECT
   *
   FROM
   role_menu
   WHERE
   id IN (
    SELECT
    min(id)
    FROM
    role_menu
    GROUP BY
    role_id,
    menu_id
   )
   AND role_menu.`status` = 1
  ) abc ON abc.menu_id = sys_menu.id
  INNER JOIN sys_role ON sys_role.id = abc.role_id
  INNER JOIN user_role ON user_role.role_id = sys_role.id
  INNER JOIN sys_user ON sys_user.id = user_role.user_id
  WHERE
   sys_user.id = #{id}
  AND sys_menu.`status` = 1
  ORDER BY
   mid
  ) t1
 ORDER BY
  t1.pare_menu_id,
  t1.`order`;