273 lines
9.2 KiB
Python
273 lines
9.2 KiB
Python
from app.extensions import db
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from app.models.bom import BomTable
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from app.models.base import MaterialBase
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from app.models.inbound.buy import StockBuy
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from sqlalchemy import func, distinct, or_
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import uuid
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from datetime import datetime
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class BomService:
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# ====================== 新版 BOM 逻辑（基于 bom_no） ======================
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@staticmethod
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def generate_bom_no():
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"""生成唯一的 BOM 编号 (作为默认备选)"""
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timestamp = datetime.now().strftime('%Y%m%d%H%M%S')
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unique = str(uuid.uuid4())[:8]
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return f'BOM-{timestamp}-{unique}'
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@staticmethod
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def get_bom_list(keyword=None):
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"""
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获取所有 BOM 配方（按 bom_no 分组）
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支持模糊搜索：BOM编号、父件名称/规格、子件名称/规格
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"""
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# 1. 如果有搜索关键词，先筛选出符合条件的 bom_no 集合
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filtered_bom_nos = None
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if keyword:
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kw = f'%{keyword}%'
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# 条件A: 匹配 BOM编号 或 父件信息
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# 需要 join 父件表
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q1 = db.session.query(BomTable.bom_no).join(
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MaterialBase, BomTable.parent_id == MaterialBase.id
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).filter(
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or_(
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BomTable.bom_no.ilike(kw),
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MaterialBase.name.ilike(kw),
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MaterialBase.spec_model.ilike(kw)
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)
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)
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# 条件B: 匹配 子件信息
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# 需要 join 子件表
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q2 = db.session.query(BomTable.bom_no).join(
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MaterialBase, BomTable.child_id == MaterialBase.id
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).filter(
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or_(
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MaterialBase.name.ilike(kw),
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MaterialBase.spec_model.ilike(kw)
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)
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)
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# 取并集 (Union)
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filtered_bom_nos = q1.union(q2).distinct().all()
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filtered_bom_nos = [row[0] for row in filtered_bom_nos]
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# 如果搜不到任何结果，直接返回空
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if not filtered_bom_nos:
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return []
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# 2. 原有的分组聚合查询
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subq_query = db.session.query(
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BomTable.bom_no,
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BomTable.parent_id,
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BomTable.version,
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func.count(BomTable.child_id).label('child_count')
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)
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# 应用筛选
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if filtered_bom_nos is not None:
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subq_query = subq_query.filter(BomTable.bom_no.in_(filtered_bom_nos))
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subq = subq_query.group_by(BomTable.bom_no, BomTable.parent_id, BomTable.version).subquery()
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query = db.session.query(
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subq.c.bom_no,
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subq.c.parent_id,
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subq.c.version,
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subq.c.child_count,
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MaterialBase.name.label('parent_name'),
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MaterialBase.spec_model.label('parent_spec')
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).join(MaterialBase, subq.c.parent_id == MaterialBase.id)
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# 按 bom_no 倒序排列
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query = query.order_by(subq.c.bom_no.desc())
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results = query.all()
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return [{
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'bom_no': row.bom_no,
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'parent_id': row.parent_id,
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'parent_name': row.parent_name,
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'parent_spec': row.parent_spec or '',
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'version': row.version,
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'child_count': row.child_count
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} for row in results]
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@staticmethod
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def get_bom_detail(bom_no):
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"""
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根据 bom_no 获取配方详情
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返回包含父件信息和子件列表的对象
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"""
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rows = db.session.query(
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BomTable,
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MaterialBase.name.label('child_name'),
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MaterialBase.spec_model.label('child_spec')
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).join(
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MaterialBase, BomTable.child_id == MaterialBase.id
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).filter(
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BomTable.bom_no == bom_no
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).all()
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if not rows:
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return None
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first = rows[0]
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parent_id = first.BomTable.parent_id
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# 获取父件的名称和规格
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parent_material = MaterialBase.query.filter(MaterialBase.id == parent_id).first()
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if parent_material:
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parent_name = parent_material.name
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parent_spec = parent_material.spec_model or ''
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else:
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parent_name = ''
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parent_spec = ''
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children = []
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for bom, child_name, child_spec in rows:
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children.append({
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'child_id': bom.child_id,
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'child_name': child_name,
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'child_spec': child_spec or '',
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'dosage': float(bom.dosage) if bom.dosage else 0.0,
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'remark': bom.remark or ''
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})
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return {
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'bom_no': bom_no,
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'parent_id': parent_id,
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'parent_name': parent_name,
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'parent_spec': parent_spec,
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'version': first.BomTable.version,
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'children': children
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}
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@staticmethod
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def save_bom(data):
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"""
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保存或更新一个 BOM 配方
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data 结构：
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{
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"bom_no": "用户输入或自动生成",
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"version": "版本号，默认v1",
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"parent_id": 父件ID,
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"children": [...]
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}
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"""
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bom_no = data.get('bom_no')
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version = data.get('version', 'v1')
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parent_id = data['parent_id']
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children = data['children']
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# 校验父件不能与子件相同
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for child in children:
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if child['child_id'] == parent_id:
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raise ValueError('父件与子件不能是同一物料')
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# 如果未提供 bom_no，则生成一个新的 (兼容旧逻辑，但现在前端会传值)
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if not bom_no or not bom_no.strip():
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# 如果前端没传，抛出异常要求用户填写，或者自动生成
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# 这里选择自动生成作为兜底，但推荐前端校验必填
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bom_no = BomService.generate_bom_no()
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# 删除该 bom_no 下所有现有记录 (全量更新模式)
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BomTable.query.filter_by(bom_no=bom_no).delete()
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# 插入新记录
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for child in children:
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bom = BomTable(
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bom_no=bom_no,
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version=version,
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parent_id=parent_id,
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child_id=child['child_id'],
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dosage=child.get('dosage', 0),
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remark=child.get('remark', '')
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)
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db.session.add(bom)
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db.session.commit()
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return bom_no
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@staticmethod
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def get_bom_with_stock_by_bom_no(bom_no):
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"""
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根据 bom_no 获取配方详情，并计算每个子件的库存和最大可生产数量
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"""
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detail = BomService.get_bom_detail(bom_no)
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if not detail:
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return None
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for child in detail['children']:
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# 查询该子件在 StockBuy 中的可用库存总量
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stock_qty = db.session.query(
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func.coalesce(func.sum(StockBuy.available_quantity), 0)
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).filter(
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StockBuy.base_id == child['child_id']
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).scalar() or 0
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child['current_stock'] = float(stock_qty)
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dosage = child['dosage']
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child['max_producible'] = int(stock_qty // dosage) if dosage > 0 else 0
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return detail
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# ====================== 兼容旧接口（基于 parent_id） ======================
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@staticmethod
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def get_bom_no_by_parent(parent_id):
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"""
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根据父件 ID 获取其最新的 BOM 编号（用于兼容旧接口）
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"""
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row = BomTable.query.filter_by(parent_id=parent_id) \
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.order_by(BomTable.version.desc()).first()
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return row.bom_no if row else None
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@staticmethod
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def create_or_update_bom(parent_id, child_list, bom_no=None, version='v1'):
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"""
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兼容旧接口的保存方法（保留原有调用方式）
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如果提供了 bom_no，则更新该 bom_no；否则为父件创建新 BOM。
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"""
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# 校验父件不能与子件相同
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for item in child_list:
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if item['child_id'] == parent_id:
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raise ValueError('父件与子件不能是同一物料')
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# 如果未提供 bom_no，尝试查找现有的，否则新建
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if not bom_no:
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existing = BomTable.query.filter_by(parent_id=parent_id).first()
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bom_no = existing.bom_no if existing else BomService.generate_bom_no()
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# 删除该 bom_no 下所有记录
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BomTable.query.filter_by(bom_no=bom_no).delete()
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# 插入新的
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for item in child_list:
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bom = BomTable(
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bom_no=bom_no,
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version=version,
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parent_id=parent_id,
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child_id=item['child_id'],
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dosage=item.get('dosage', 0),
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remark=item.get('remark', '')
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)
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db.session.add(bom)
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db.session.commit()
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return True
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@staticmethod
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def get_bom_with_stock(parent_id):
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"""
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兼容旧接口：根据父件 ID 获取 BOM 及库存信息
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（实际会找到对应的 bom_no 再调用新方法）
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"""
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bom_no = BomService.get_bom_no_by_parent(parent_id)
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if not bom_no:
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return []
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detail = BomService.get_bom_with_stock_by_bom_no(bom_no)
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if not detail:
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return []
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return detail['children'] |