fix: sort warehouse tree by name, fix tree batch delete cascade, and implement safe history location autofill

inventory-backend/app/services/inbound/buy_service.py

@@ -1,6 +1,7 @@
 # inventory-backend/app/services/inbound/buy_service.py
 from app.extensions import db
 from app.models.inbound.buy import StockBuy
+from app.models.inbound.product import StockProduct
 from app.models.base import MaterialBase
 from datetime import datetime, timedelta, timezone
 from sqlalchemy import or_, func, text, and_
@@ -525,3 +526,42 @@ class BuyInboundService:
     def get_history_locations(base_id):
         return [r[0] for r in
                 db.session.query(StockBuy.warehouse_location).filter(StockBuy.base_id == base_id).distinct().all()]
+
+    @staticmethod
+    def get_last_location_by_base_id(base_id):
+        """
+        获取指定物料最近一次入库的库位（跨表查询）
+        查询顺序：采购入库 -> 成品入库 -> 半成品入库，返回最新入库的库位
+        """
+        from app.models.inbound.semi import StockSemi
+
+        # 1. 查询采购入库最新记录
+        last_buy = StockBuy.query.filter(
+            StockBuy.base_id == base_id
+        ).order_by(StockBuy.in_date.desc()).first()
+
+        # 2. 查询成品入库最新记录
+        last_product = StockProduct.query.filter(
+            StockProduct.base_id == base_id
+        ).order_by(StockProduct.in_date.desc()).first()
+
+        # 3. 查询半成品入库最新记录
+        last_semi = StockSemi.query.filter(
+            StockSemi.base_id == base_id
+        ).order_by(StockSemi.in_date.desc()).first()
+
+        # 比较三个表中的最新入库时间，返回最新的库位
+        candidates = []
+        if last_buy and last_buy.warehouse_location:
+            candidates.append((last_buy.in_date, last_buy.warehouse_location))
+        if last_product and last_product.warehouse_location:
+            candidates.append((last_product.in_date, last_product.warehouse_location))
+        if last_semi and last_semi.warehouse_location:
+            candidates.append((last_semi.in_date, last_semi.warehouse_location))
+
+        if not candidates:
+            return ""
+
+        # 按时间倒序排序，返回最新的库位
+        candidates.sort(key=lambda x: x[0] if x[0] else datetime.min, reverse=True)
+        return candidates[0][1] if candidates[0][1] else ""