


Detailed explanation of python recursive query menu and conversion into json example code
This article mainly introduces the python recursive query menu and converts it into a json instance. It has certain reference value. Interested friends can refer to it.
I recently needed to write a menu in python, and it took me two or three days to get it done. Now I record it here, and friends who need it can learn from it.
Note: The article quotes the complete non-executable code and only excerpts the key parts of the code
Environment
Database: mysql
python:3.6
Table structure
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';
Python code
In the Menu object, there is a reference to the submenu list "subMenus", the type is list
Core code
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
Test method
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)
Note: The basic process is: first query the first-level menu, Then pass the id of the menu at this level and the submenu list of this level menu to the set_subMenus method to recursively set the lower-level menus of the submenu list;
supports passing the menu ID and querying the menu below. All submenus. If you pass a null character, the query starts from the root directory
In the "rootMenus" object, you can see the complete menu tree structure
Convert to Json
The ORM framework I use is: sqlalchemy. The Menu object queried directly from the database will report an error when it is converted to Json. A DTO class needs to be redefined to convert the Menu object into a Dto object.
MenuDto
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__
So, the method of recursively setting submenus was redefined
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
Note:
When a menu has no submenus, delete the "subMenus" attribute, otherwise a null value will appear when converting to Json
model_list_2_dto_list method can convert Menu list into MenuDto list
to_pinyin is a method to convert Chinese characters into pinyin, no need to pay attention here
View layer's method of returning Json
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})
Note: The meanings of json and dictionary in python are similar. When json is finally returned to the page, You need to use the json_dict method to convert to dict type first, otherwise the returned string will contain "\"
Query results
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