构造多级菜单树(来源mini分支)

applications/common/admin.py

@@ -1,3 +1,5 @@
+import copy
+from collections import OrderedDict
 from io import BytesIO
 from flask import session, make_response, current_app
 from flask_login import current_user
@@ -22,16 +24,8 @@ def add_auth_session():
 
 # 生成菜单树
 def make_menu_tree():
-    # power0 = Power.query.filter(
-    #     Power.type == 0,
-    # ).all()
-    # power1 = Power.query.filter(
-    #     Power.type == 1
-    # ).all()
-    # 获取当前用户的角色
     role = current_user.role
-    power0 = []
-    power1 = []
+    powers = []
     for i in role:
         # 如果角色没有被启用就直接跳过
         if i.enable == 0:
@@ -41,31 +35,28 @@ def make_menu_tree():
             # 如果权限关闭了就直接跳过
             if p.enable == 0:
                 continue
-            # 一级菜单
-            if int(p.type) == 0:
-                power0.append(p)
-            # 二级菜单
-            else:
-                power1.append(p)
+            # 一二级菜单
+            if int(p.type) == 0 or int(p.type) == 1:
+                powers.append(p)
 
     power_schema = PowerSchema(many=True)  # 用已继承 ma.ModelSchema 类的自定制类生成序列化类
-    power0_dict = power_schema.dump(power0)  # 生成可序列化对象
-    power1_dict = power_schema.dump(power1)  # 生成可序列化对象
-    power0_dict = sorted(power0_dict, key=lambda i: i['sort'])
-    power1_dict = sorted(power1_dict, key=lambda i: i['sort'])
-    # print(power0)
-    # print(power1)
+    power_dict = power_schema.dump(powers)  # 生成可序列化对象
+    power_dict.sort(key=lambda x: x['id'], reverse=True)
 
-    menu = []
+    menu_dict = OrderedDict()
+    for _dict in power_dict:
+        if _dict['id'] in menu_dict:
+            # 当前节点添加子节点
+            _dict['children'] = copy.deepcopy(menu_dict[_dict['id']])
+            _dict['children'].sort(key=lambda item: item['sort'])
+            # 删除子节点
+            del menu_dict[_dict['id']]
 
-    for p0 in power0_dict:
-        for p1 in power1_dict:
-            if p0.get('id') == p1.get('parent_id'):
-                if p0.get("children") is None:
-                    p0['children'] = []
-                p0['children'].append(p1)
-        menu.append(p0)
-    return menu
+        if _dict['parent_id'] not in menu_dict:
+            menu_dict[_dict['parent_id']] = [_dict]
+        else:
+            menu_dict[_dict['parent_id']].append(_dict)
+    return sorted(menu_dict.get(0), key=lambda item: item['sort'])
 
 
 # 生成验证码