使用 (where user_id=1) 时,我可以获得结果,但是当注入参数时,我得到了这样的回溯:“用户”对象不可迭代,任何将参数注入原始查询的想法,我的意见.py: class TransactionViews(viewsets.ViewSet): def list(self, request): user = get_object_or_404(User, pk=request.user.id) queryset = Transaction.objects.raw("SELECT product.name, transaction.* from product inner join variant on product.id = variant.product_id inner join transactions_variants on variant.id = transactions_variants.variant_id inner join transaction on transactions_variants.transaction_id = transaction.id where user_id=%s",user) serializer = TransactionSerializer(queryset, many=True) return Response(serializer.data, status=200)还有一个问题:queryset = Transaction.objects.filter(user_id=user).exclude(deleted_at__isnull=False)我如何将此“.exclude(deleted_at__isnull=False)”转换为我的原始查询(此排除是软删除的条件)Traceback:File "/usr/local/lib/python3.7/site-packages/django/core/handlers/exception.py" in inner 34. response = get_response(request)File "/usr/local/lib/python3.7/site-packages/django/core/handlers/base.py" in _get_response 126. response = self.process_exception_by_middleware(e, request)File "/usr/local/lib/python3.7/site-packages/django/core/handlers/base.py" in _get_response 124. response = wrapped_callback(request, *callback_args, **callback_kwargs)File "/usr/local/lib/python3.7/site-packages/django/views/decorators/csrf.py" in wrapped_view 54. return view_func(*args, **kwargs)File "/usr/local/lib/python3.7/site-packages/rest_framework/viewsets.py" in view 116. return self.dispatch(request, *args, **kwargs)File "/usr/local/lib/python3.7/site-packages/rest_framework/views.py" in dispatch 495. response = self.handle_exception(exc)File "/usr/local/lib/python3.7/site-packages/rest_framework/views.py" in handle_exception 455. self.raise_uncaught_exception(exc)
2 回答
ibeautiful
TA贡献1993条经验 获得超5个赞
正如清楚明确地记录的那样,您需要将查询参数作为元组或列表传递 - 即使只有一个参数。IOW,你想要:
queryset = Transaction.objects.raw("your long query here where user_id=%s", [user])
添加回答
举报
0/150
提交
取消