perf: 优化菜单管理模块service层及异常处理

This commit is contained in:
insistence
2024-07-10 11:08:07 +08:00
parent 4a17ffbebe
commit 013ea25731
4 changed files with 129 additions and 96 deletions

View File

@@ -1,4 +1,4 @@
from sqlalchemy import select, update, delete, and_
from sqlalchemy import select, update, delete, and_, func
from sqlalchemy.ext.asyncio import AsyncSession
from module_admin.entity.do.menu_do import SysMenu
from module_admin.entity.do.user_do import SysUser, SysUserRole
@@ -156,3 +156,35 @@ class MenuDao:
delete(SysMenu)
.where(SysMenu.menu_id.in_([menu.menu_id]))
)
@classmethod
async def has_child_by_menu_id_dao(cls, db: AsyncSession, menu_id: int):
"""
根据菜单id查询菜单关联子菜单的数量
:param db: orm对象
:param menu_id: 菜单id
:return: 菜单关联子菜单的数量
"""
menu_count = (await db.execute(
select(func.count('*'))
.select_from(SysMenu)
.where(SysMenu.menu_id == menu_id)
)).scalar()
return menu_count
@classmethod
async def check_menu_exist_role_dao(cls, db: AsyncSession, menu_id: int):
"""
根据菜单id查询菜单关联角色数量
:param db: orm对象
:param menu_id: 菜单id
:return: 菜单关联角色数量
"""
role_count = (await db.execute(
select(func.count('*'))
.select_from(SysRoleMenu)
.where(SysRoleMenu.menu_id == menu_id)
)).scalar()
return role_count