python递归查询菜单并转换成json实例

CREATE TABLE `tb_menu` ( `id` varchar(32) NOT NULL COMMENT '唯一标识', `menu_name` varchar(40) DEFAULT NULL COMMENT '菜单名称', `menu_url` varchar(100) DEFAULT NULL COMMENT '菜单链接', `type` varchar(1) DEFAULT NULL COMMENT '类型', `parent` varchar(32) DEFAULT NULL COMMENT '父级目录id', `del_flag` varchar(1) NOT NULL DEFAULT '0' COMMENT '删除标志 0：不删除 1：已删除', `create_time` datetime DEFAULT CURRENT_TIMESTAMP COMMENT '创建时间', `update_time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP COMMENT '更新时间', PRIMARY KEY (`id`) USING BTREE) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT='菜单表';

def set_subMenus(id, menus):  """  根据传递过来的父菜单id，递归设置各层次父菜单的子菜单列表  :param id: 父级id  :param menus: 子菜单列表  :return: 如果这个菜单没有子菜单，返回None;如果有子菜单，返回子菜单列表  """  # 记录子菜单列表  subMenus = []  # 遍历子菜单  for m in menus:    if m.parent == id:      subMenus.append(m)  # 把子菜单的子菜单再循环一遍  for sub in subMenus:    menus2 = queryByParent(sub.id)    # 还有子菜单    if len(menus):      sub.subMenus = set_subMenus(sub.id, menus2)  # 子菜单列表不为空  if len(subMenus):    return subMenus  else: # 没有子菜单了    return None

  def test_set_subMenus(self):    # 一级菜单    rootMenus = queryByParent('')    for menu in rootMenus:      subMenus = queryByParent(menu.id)      menu.subMenus = set_subMenus(menu.id, subMenus)

class MenuDto():  def __init__(self, id, menu_name, menu_url, type, parent, subMenus):    super().__init__()    self.id = id    self.menu_name = menu_name    self.menu_url = menu_url    self.type = type    self.parent = parent    self.subMenus = subMenus  def __str__(self):    return '%s(id=%s,menu_name=%s,menu_url=%s,type=%s,parent=%s)' % (      self.__class__.__name__, self.id, self.menu_name, self.menu_url, self.type, self.parent)  __repr = __str__

def set_subMenuDtos(id, menuDtos):  """  根据传递过来的父菜单id，递归设置各层次父菜单的子菜单列表  :param id: 父级id  :param menuDtos: 子菜单列表  :return: 如果这个菜单没有子菜单，返回None;如果有子菜单，返回子菜单列表  """  # 记录子菜单列表  subMenuDtos = []  # 遍历子菜单  for m in menuDtos:    m.name = to_pinyin(m.menu_name)    if m.parent == id:      subMenuDtos.append(m)  # 把子菜单的子菜单再循环一遍  for sub in subMenuDtos:    menus2 = queryByParent(sub.id)    menusDto2 = model_list_2_dto_list(menus2,                     "MenuDto(id='', menu_name='', menu_url='', type='', parent='', subMenus='')")    # 还有子菜单    if len(menuDtos):      if len(menusDto2):        sub.subMenus = set_subMenuDtos(sub.id, menusDto2)      else: # 没有子菜单，删除该节点        sub.__delattr__('subMenus')  # 子菜单列表不为空  if len(subMenuDtos):    return subMenuDtos  else: # 没有子菜单了    return None

  def get(self):    param = request.args    id = param['id']    # 如果id为空，查询的是从根目录开始的各级菜单    rootMenus = queryByParent(id)    rootMenuDtos = model_list_2_dto_list(rootMenus,                       "MenuDto(id='', menu_name='', menu_url='', type='', parent='', subMenus='')")    # 设置各级子菜单    for menu in rootMenuDtos:      menu.name = to_pinyin(menu.menu_name)      subMenus = queryByParent(menu.id)      if len(subMenus):        subMenuDtos = model_list_2_dto_list(subMenus,                          "MenuDto(id='', menu_name='', menu_url='', type='', parent='', subMenus='')")        menu.subMenus = set_subMenuDtos(menu.id, subMenuDtos)      else:        menu.__delattr__('subMenus')    menus_json = json.dumps(rootMenuDtos, default=lambda o: o.__dict__, sort_keys=True, allow_nan=false,                skipkeys=true)    # 需要转字典，否则返回的字符串会带有“/”    menus_dict = json_dict(menus_json)    return fullResponse(menus_dict)fullResponsefrom flask import jsonifydef fullResponse(data='', msg='', code=0):  if msg == '':    return jsonify({'code': code, 'data': data})  elif data == '':    return jsonify({'code': code, 'msg': msg})  else:    return jsonify({'code': code, 'msg': msg, 'data': data})