Make repository list() reject unrecognised filter names
Repository.list(**filters) let each implementation pick out the filter names it understood and silently ignore the rest, so a typo or an unsupported filter returned every row in the table instead of an error. This was hit for real: LeaveWishRepository.list(parent_personal_data_uuid =...) returns every wish in the database, because that implementation only recognises workspace_id and child_id. A guard written on top of it would have passed while checking nothing. All ten implementations now validate against their supported set and raise ValueError naming the offending key. ValueError rather than a domain error because passing a filter that does not exist is a programming bug to fix, not a runtime condition to handle -- it must not become an HTTP 4xx. Co-Authored-By: Claude Opus 5 <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_01RXMhc9nT6sNz1SL4j8dznL
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@@ -6,6 +6,7 @@ from uuid import UUID
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import aiosqlite
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from app.entities.user import User
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from app.repositories.filter_validator import validate_filters
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from app.repositories.user_repository import UserRepository
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from app.utils.errors import NotFoundError
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@@ -72,6 +73,7 @@ class SQLiteUserRepository(UserRepository):
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raise NotFoundError(f"User {id} not found")
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async def list(self, **filters: object) -> list[User]:
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validate_filters(filters, set())
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cursor = await self._conn.execute('SELECT * FROM "user"')
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rows = await cursor.fetchall()
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return [self._row_to_entity(r) for r in rows]
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