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Examples of python menu recursive query and data conversion to json

Y2J

Release： 2017-04-21 16:18:21

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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 &#39;唯一标识&#39;,
 `menu_name` varchar(40) DEFAULT NULL COMMENT &#39;菜单名称&#39;,
 `menu_url` varchar(100) DEFAULT NULL COMMENT &#39;菜单链接&#39;,
 `type` varchar(1) DEFAULT NULL COMMENT &#39;类型&#39;,
 `parent` varchar(32) DEFAULT NULL COMMENT &#39;父级目录id&#39;,
 `del_flag` varchar(1) NOT NULL DEFAULT &#39;0&#39; COMMENT &#39;删除标志 0：不删除 1：已删除&#39;,
 `create_time` datetime DEFAULT CURRENT_TIMESTAMP COMMENT &#39;创建时间&#39;,
 `update_time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP COMMENT &#39;更新时间&#39;,
 PRIMARY KEY (`id`) USING BTREE
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT=&#39;菜单表&#39;;

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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

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Test Method

  def test_set_subMenus(self):
    # 一级菜单
    rootMenus = queryByParent(&#39;&#39;)
    for menu in rootMenus:
      subMenus = queryByParent(menu.id)
      menu.subMenus = set_subMenus(menu.id, subMenus)

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Note: The basic process is: first query the first-level menu, and then pass the id of the menu at this level and the submenu list of this level menu to the set_subMenus method, and recursively perform the submenu list. Lower-level menu settings;

supports passing the menu ID to query all submenus under the menu. 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 &#39;%s(id=%s,menu_name=%s,menu_url=%s,type=%s,parent=%s)&#39; % (
      self.class.name, self.id, self.menu_name, self.menu_url, self.type, self.parent)

  repr = str

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So, the method of recursively setting submenus is 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=&#39;&#39;, menu_name=&#39;&#39;, menu_url=&#39;&#39;, type=&#39;&#39;, parent=&#39;&#39;, subMenus=&#39;&#39;)")
    # 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(&#39;subMenus&#39;)

  # 子菜单列表不为空
  if len(subMenuDtos):
    return subMenuDtos
  else: # 没有子菜单了
    return None

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Remarks:

When a menu has no submenu, delete the "subMenus" attribute, otherwise a null value will appear when converting to Json
## The model_list_2_dto_list method can convert the Menu list into a MenuDto list
to_pinyin is a method of converting Chinese characters into pinyin. You don’t need to pay attention here.

The method of View layer returning Json

  def get(self):
    param = request.args
    id = param[&#39;id&#39;]
    # 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=&#39;&#39;, menu_name=&#39;&#39;, menu_url=&#39;&#39;, type=&#39;&#39;, parent=&#39;&#39;, subMenus=&#39;&#39;)")
    # 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=&#39;&#39;, menu_name=&#39;&#39;, menu_url=&#39;&#39;, type=&#39;&#39;, parent=&#39;&#39;, subMenus=&#39;&#39;)")
        menu.subMenus = set_subMenuDtos(menu.id, subMenuDtos)
      else:
        menu.delattr(&#39;subMenus&#39;)

    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=&#39;&#39;, msg=&#39;&#39;, code=0):
  if msg == &#39;&#39;:
    return jsonify({&#39;code&#39;: code, &#39;data&#39;: data})
  elif data == &#39;&#39;:
    return jsonify({&#39;code&#39;: code, &#39;msg&#39;: msg})
  else:
    return jsonify({&#39;code&#39;: code, &#39;msg&#39;: msg, &#39;data&#39;: data})

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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 it to dict type first, otherwise the returned string will have "\"

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