feat(20241226-fieldpermission): 优化列权限逻辑

1、后端优化,当多个角色的时候,合并列权限配置
2、前端优化,有多级表头时,列权限设置无效的bug
This commit is contained in:
李小涛
2024-12-26 08:56:19 +08:00
parent 15c87ddd26
commit dddafa4826
3 changed files with 38 additions and 60 deletions

View File

@@ -404,7 +404,7 @@ PLUGINS_URL_PATTERNS = []
# ********** 一键导入插件配置开始 **********
# 例如:
# from dvadmin_upgrade_center.settings import * # 升级中心
from dvadmin3_celery.settings import * # celery 异步任务
# from dvadmin3_celery.settings import * # celery 异步任务
# from dvadmin_third.settings import * # 第三方用户管理
# from dvadmin_ak_sk.settings import * # 秘钥管理管理
# from dvadmin_tenants.settings import * # 租户管理

View File

@@ -1,7 +1,5 @@
# -*- coding: utf-8 -*-
from itertools import groupby
from django.db.models import F
from rest_framework.decorators import action
from rest_framework.permissions import IsAuthenticated
@@ -12,7 +10,7 @@ from dvadmin.utils.models import get_custom_app_models
class FieldPermissionMixin:
@action(methods=['get'], detail=False,permission_classes=[IsAuthenticated])
@action(methods=['get'], detail=False, permission_classes=[IsAuthenticated])
def field_permission(self, request):
"""
获取字段权限
@@ -27,45 +25,32 @@ class FieldPermissionMixin:
if finded is False:
return []
user = request.user
if user.is_superuser==1:
data = MenuField.objects.filter( model=model['model']).values('field_name')
# 创建一个默认字典来存储最终的结果
result = {}
if user.is_superuser == 1:
data = MenuField.objects.filter(model=model['model']).values('field_name')
for item in data:
item['is_create'] = True
item['is_query'] = True
item['is_update'] = True
else:
roles = request.user.role.values_list('id', flat=True)
data= FieldPermission.objects.filter(
field__model=model['model'],role__in=roles
).values( 'is_create', 'is_query', 'is_update',field_name=F('field__field_name'))
field_name = item.pop('field_name')
result[field_name] = {}
result[field_name]['is_create'] = True
result[field_name]['is_query'] = True
result[field_name]['is_update'] = True
else:
roles = request.user.role.values_list('id', flat=True)
data = FieldPermission.objects.filter(
field__model=model['model'], role__in=roles
).values('is_create', 'is_query', 'is_update', field_name=F('field__field_name'))
"""
合并权限
这段代码首先根据 field_name 对列表进行排序,
然后使用 groupby 按 field_name 进行分组。
对于每个组,它创建一个新的字典 merged
并遍历组中的每个字典将布尔值字段使用逻辑或or操作符进行合并如果 merged 中还没有该字段,则默认为 False
其他字段(如 field_name则直接取组的关键字即 key
"""
# 使用field_name对列表进行分组, # groupby 需要先对列表进行排序,因为它只能对连续相同的元素进行分组。
grouped = groupby(sorted(list(data), key=lambda x: x['field_name']), key=lambda x: x['field_name'])
data = []
# 遍历分组,合并权限
for key, group in grouped:
# 初始化一个空字典来存储合并后的结果
merged = {}
for item in group:
# 合并权限, True值优先
merged['is_create'] = merged.get('is_create', False) or item['is_create']
merged['is_query'] = merged.get('is_query', False) or item['is_query']
merged['is_update'] = merged.get('is_update', False) or item['is_update']
merged['field_name'] = key
data.append(merged)
return DetailResponse(data=data)
# 遍历原始数据并填充结果字典
for item in data:
field_name = item.pop('field_name')
for key, value in item.items():
if field_name in result:
if value:
result[field_name][key] = True
else:
result[field_name] = {}
result[field_name][key] = value
return DetailResponse(data=result)