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修复sqlalchemy scalar_subquery()告警 fix #89
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4linuxfun committed Jan 3, 2023
commit 7f83fbf187fc6d501c77f0dc7b01c68ba02691f9
2 changes: 1 addition & 1 deletion server/crud/base.py
Original file line number Diff line number Diff line change
Expand Up @@ -99,7 +99,7 @@ def search(self, session: Session, search: Pagination, filter_type: Optional[Dic
else:
subquery = subquery.order_by(getattr(self.model, order_col))
subquery = subquery.offset(
(search.page - 1) * search.page_size).limit(1).subquery()
(search.page - 1) * search.page_size).limit(1).scalar_subquery()
if search.model == 'desc':
sql = sql.where(getattr(self.model, order_col) <= subquery).order_by(desc(getattr(self.model, order_col))).limit(
search.page_size)
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2 changes: 1 addition & 1 deletion server/crud/internal/dictonary.py
Original file line number Diff line number Diff line change
Expand Up @@ -12,7 +12,7 @@ class CRUDDict(CRUDBase[DataDict]):

class CRUDItem(CRUDBase[DictItem]):
def get_items_by_code(self, db: Session, code: str):
dict_id = select(DataDict.id).where(DataDict.code == code).subquery()
dict_id = select(DataDict.id).where(DataDict.code == code).scalar_subquery()
sql = select(self.model).where(self.model.dict_id == dict_id).where(self.model.enable == 1).order_by(
self.model.sort)
return db.exec(sql).all()
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