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)
fullResponse

from flask import jsonify


def 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})