177 lines
6.0 KiB
Python
177 lines
6.0 KiB
Python
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"""
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共享 CTE 任务树加载器 — 使用 PostgreSQL Recursive CTE 一次性拉取完整任务树
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解决问题:原 _load_task_tree / _get_task_with_children_recursive 使用
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Python 递归逐层 SELECT,N 个节点产生 N+1 次数据库查询。
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现在无论树深度多大,仅执行 2 条查询(CTE + records selectinload)。
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"""
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from __future__ import annotations
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import uuid
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from fastapi import HTTPException, status
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from sqlalchemy import select
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from sqlalchemy.ext.asyncio import AsyncSession
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from sqlalchemy.orm import noload, selectinload
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from sqlalchemy.orm.attributes import set_committed_value
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from app.models.task import Task
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# ============================================================
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# 内存树组装(O(N) 时间 / O(N) 空间)
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# ============================================================
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def _build_tree_in_memory(tasks: list[Task]) -> dict[uuid.UUID, Task]:
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"""
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给定扁平 Task ORM 列表,在内存中通过哈希表组装嵌套树结构。
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关键安全设计:
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- 使用临时字典 temp_children_map 暂存父子关系,绝对不直接操作 ORM 的 child_tasks。
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- 通过 set_committed_value 注入最终列表,告诉 SQLAlchemy 这是"已提交数据",
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避免 add_task_record 等场景中 db.commit() 时触发级联 UPDATE 污染数据库。
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时间复杂度: O(N),空间复杂度: O(N)。
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"""
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if not tasks:
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return {}
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# ── Pass 1: 临时字典存储关系(不触碰 ORM 属性)──
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temp_children_map: dict[uuid.UUID, list[Task]] = {t.id: [] for t in tasks}
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task_map: dict[uuid.UUID, Task] = {t.id: t for t in tasks}
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# ── Pass 2: 挂载到临时字典 ──
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for t in tasks:
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pid = t.parent_task_id
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if pid is not None and pid in temp_children_map:
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temp_children_map[pid].append(t)
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# ── Pass 3: 排序 + set_committed_value 安全注入 ──
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for t in tasks:
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children = temp_children_map[t.id]
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if children:
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children.sort(key=lambda x: x.created_at)
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# 关键:标记为已提交数据,SQLAlchemy 不会对其生成 UPDATE
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set_committed_value(t, 'child_tasks', children)
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return task_map
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# ============================================================
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# 公开 API:按单一任务 ID 加载子树
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# ============================================================
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async def load_task_tree_by_root(
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db: AsyncSession, task_id: uuid.UUID
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) -> Task:
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"""
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使用 Recursive CTE 加载以 task_id 为根的完整任务子树。
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返回: 根 Task ORM 对象(child_tasks 已递归填充)。
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Raises:
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HTTPException(404): 根任务不存在。
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"""
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# ── Step 1: Recursive CTE — 收集所有子孙节点 ID ──
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# WITH RECURSIVE task_tree AS (
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# SELECT tasks.* FROM tasks WHERE tasks.id = :tid
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# UNION ALL
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# SELECT tasks.* FROM tasks
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# JOIN task_tree ON tasks.parent_task_id = task_tree.id
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# )
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anchor = (
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select(Task)
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.where(Task.id == task_id)
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.cte(name="task_tree", recursive=True)
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)
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task_tree_cte = anchor.union_all(
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select(Task).join(anchor, Task.parent_task_id == anchor.c.id)
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)
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# ── Step 2: 批量加载所有任务 + 关联数据 ──
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stmt = (
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select(Task)
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.options(
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noload(Task.child_tasks), # 禁掉模型默认 selectinload,由内存树接管
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noload(Task.parent_task), # 组装树不需要 parent 引用
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selectinload(Task.records), # 🔥 一次性预加载所有进度记录
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selectinload(Task.product), # 🔥 一次性预加载产品引用
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)
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.where(Task.id.in_(select(task_tree_cte.c.id)))
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)
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result = await db.execute(stmt)
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all_tasks = result.unique().scalars().all()
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if not all_tasks:
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raise HTTPException(
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status_code=status.HTTP_404_NOT_FOUND,
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detail=f"任务不存在: {task_id}",
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)
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# ── Step 3: 内存组装 ──
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task_map = _build_tree_in_memory(all_tasks)
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# 根任务一定在 map 中(CTE anchor 保证了这一点)
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return task_map[task_id]
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# ============================================================
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# 公开 API:按产品 ID 加载所有任务树
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# ============================================================
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async def load_task_trees_by_product(
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db: AsyncSession, product_id: uuid.UUID
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) -> list[Task]:
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"""
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使用 Recursive CTE 加载指定产品下的所有任务树。
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返回: 顶层任务列表(parent_task_id IS NULL),每项的 child_tasks 已递归填充。
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若无任务则返回空列表。
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"""
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# ── Step 1: Recursive CTE ──
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# WITH RECURSIVE product_task_tree AS (
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# SELECT tasks.* FROM tasks
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# WHERE tasks.product_id = :pid AND tasks.parent_task_id IS NULL
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# UNION ALL
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# SELECT tasks.* FROM tasks
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# JOIN product_task_tree ON tasks.parent_task_id = product_task_tree.id
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# )
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anchor = (
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select(Task)
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.where(
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Task.product_id == product_id,
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Task.parent_task_id.is_(None),
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)
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.cte(name="product_task_tree", recursive=True)
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)
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task_tree_cte = anchor.union_all(
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select(Task).join(anchor, Task.parent_task_id == anchor.c.id)
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)
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# ── Step 2: 批量加载 ──
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stmt = (
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select(Task)
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.options(
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noload(Task.child_tasks),
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noload(Task.parent_task),
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selectinload(Task.records),
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selectinload(Task.product),
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)
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.where(Task.id.in_(select(task_tree_cte.c.id)))
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)
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result = await db.execute(stmt)
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all_tasks = result.unique().scalars().all()
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if not all_tasks:
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return []
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# ── Step 3: 内存组装 ──
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_build_tree_in_memory(all_tasks)
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# ── Step 4: 返回排序后的顶层任务 ──
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roots = [t for t in all_tasks if t.parent_task_id is None]
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roots.sort(key=lambda t: t.created_at)
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return roots
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