TypeError: issubclass() arg 1 must be a class when inheriting BaseModel of a class with a complex type

[ValiaIsNotProgrammer]

First Check

• I added a very descriptive title here.
• I used the GitHub search to find a similar question and didn't find it.
• I searched the SQLModel documentation, with the integrated search.
• I already searched in Google "How to X in SQLModel" and didn't find any information.
• I already read and followed all the tutorial in the docs and didn't find an answer.
• I already checked if it is not related to SQLModel but to Pydantic.
• I already checked if it is not related to SQLModel but to SQLAlchemy.

Commit to Help

• I commit to help with one of those options 👆

Example Code

class File(BaseModel):
    name: str
    url: str

class Producer(BaseModel):
    . . .
    files: Optional[List[File]]


class FileSQL(SQLModel, File, table=True):
    . . . 
    producer: Optional["ProducerSQL"] = Relationship(back_populates="files")


class ProducerSQL(SQLModel, Producer, table=True):
    . . .
    files: Optional[List[FileSQL]] = Relationship(back_populates="producer")

Description

When attempting to define a relationship between two models (ProducerSQL and FileSQL) using SQLModel, an error is raised if the base model (Producer) contains a field of type Optional[List[File]]. The error does not occur if this field is removed.

Actual Behavior

Traceback (most recent call last):
  File "<module>", line 83, in <module>
    class ProducerSQL(SQLModel, Producer, table=True):
  File "<virtualenv>/lib/python3.11/site-packages/sqlmodel/main.py", line 482, in __new__
    col = get_column_from_field(v)
          ^^^^^^^^^^^^^^^^^^^^^^^^
  File "<virtualenv>/lib/python3.11/site-packages/sqlmodel/main.py", line 626, in get_column_from_field
    sa_type = get_sqlalchemy_type(field)
              ^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "<virtualenv>/lib/python3.11/site-packages/sqlmodel/main.py", line 575, in get_sqlalchemy_type
    if issubclass(type_, Enum):
       ^^^^^^^^^^^^^^^^^^^^^^^
TypeError: issubclass() arg 1 must be a class

Temporary solution

I obviously took out another intermediate model (fortunately, this one fit into my pipeline) and implemented the following method:

@staticmethod
def prepare_to_db(item: ProducerLoader) -> ProducerSQL:
        files = [FileSQL(**file.dict()) for file in item.files]
        delattr(item, "files")
        sql_model = ProducerSQL(**item.dict())
        sql_model.files = files
        return sql_model

It's dirty, but working

Operating System

Linux

Operating System Details

Ubuntu 22.04.4 LTS

SQLModel Version

0.0.18

Python Version

3.11

Additional Context

pydantic = 2.7.1
sqlalchemy = 2.0.30

[imneedle]

What I would suggest is to stay away from BaseModel entirely.

Remember that SQLModel is a wrapper around Pydantic and SQLAlechemy.

So behind the scenes, the SQLModel class inherits from BaseModel.

For the most part (at least in your example above), inheriting from SQLModel with table=False is the same thing as inheriting from BaseModel.

The fix I would suggest would be to always inherit from SQLModel, and specify table=True when you want it to be an SQL table.

This would also remove your multiple inheritance pattern like ProducerSQL(SQLModel, Producer, table=True), and keep it as ProducerSQL(Producer, table=True) (assuming that Producer itself inherits from SQLModel)

from typing import List, Optional

from sqlmodel import SQLModel

class File(SQLModel):  # table=False by default
    name: str
    url: str

class Producer(SQLModel):
    ...
    files: Optional[List[File]]


class FileSQL(File, table=True):  # `File` itself inherits from SQLModel
    ...
    producer: Optional["ProducerSQL"] = Relationship(back_populates="files")


class ProducerSQL(Producer, table=True):  # `Producer` itself inherits from SQLModel
    ...
    files: Optional[List[FileSQL]] = Relationship(back_populates="producer")

[ValiaIsNotProgrammer]

Indeed, in my case it costs nothing to completely replace BaseModel with SQLModel. However, this did not solve my problem when inheriting complex type

class File(SQLModel):  # table=False by default
    name: str
    url: str

class Producer(SQLModel):
    files: Optional[List[File]]


class FileSQL(File, table=True):  # `File` itself inherits from SQLModel
    __tablename__ = "file"
    id: Optional[int] = Field(default=None, primary_key=True)
    producer_id: Optional[int] = Field(default=None, foreign_key="producer.id")
    producer: Optional["ProducerSQL"] = Relationship(back_populates="files")


class ProducerSQL(Producer, table=True):  # `Producer` itself inherits from SQLModel
    __tablename__ = "producer"
    id: Optional[int] = Field(default=None, primary_key=True)
    url: str = Field(unique=True)
    files: Optional[List['FileSQL']] = Relationship(back_populates="producer")

Traceback

  File "/home/valiaisnotprogrammer/PycharmProjects/sql_model/inherit_complex_type", line 22, in <module>
    class ProducerSQL(Producer, table=True):  # `Producer` itself inherits from SQLModel
  File "/home/valiaisnotprogrammer/.cache/pypoetry/virtualenvs/flow-engineering-requests-3i-2j0Pi-py3.11/lib/python3.11/site-packages/sqlmodel/main.py", line 482, in __new__
    col = get_column_from_field(v)
          ^^^^^^^^^^^^^^^^^^^^^^^^
  File "/home/valiaisnotprogrammer/.cache/pypoetry/virtualenvs/sql_model-3i-2j0Pi-py3.11/lib/python3.11/site-packages/sqlmodel/main.py", line 626, in get_column_from_field
    sa_type = get_sqlalchemy_type(field)
              ^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "/home/valiaisnotprogrammer/.cache/pypoetry/virtualenvs/sql_model-3i-2j0Pi-py3.11/lib/python3.11/site-packages/sqlmodel/main.py", line 575, in get_sqlalchemy_type
    if issubclass(type_, Enum):
       ^^^^^^^^^^^^^^^^^^^^^^^
TypeError: issubclass() arg 1 must be a class

[imneedle]

Ah! That problem is linked to an error in your model definitions.

More specifically, it's because in your Producer class you try to set a type as List[File]. But List has no type in sqlalchemy.

In SQL databases, when you want to use lists, you typically have to use a foreign_key, or sometimes a link_table.

What I usually like to do is define Create, Read and Update models for my data. That way, it's easy later if you want to get Read data with relationship attributes for example.

I would think the best way to get what you want in your case is the following:

from typing import Annotated, Optional

from sqlmodel import SQLModel, Field, Relationship


class FileCreate(SQLModel):
    name: str
    url: str
    # depends on how your API works, but typically, the file was created by a
    # producer, so you would have `producer_id` in the CreateModel` rather
    # than in the SQLModel... but that is up to you.


class FileSQL(FileCreate, table=True):
    __tablename__ = "file"
    id: Annotated[int | None, Field(primary_key=True)] = None
    producer_id: Annotated[int | None, Field(foreign_key="producer.id")] = None
    producer: Optional["ProducerSQL"] = Relationship(back_populates="files")


class FileRead(FileCreate):
    id: int
    producer_id: int


class ProducerCreate(SQLModel):
    url: Annotated[str, Field(unique=True)]


class ProducerSQL(ProducerCreate, table=True):
    __tablename__ = "producer"
    id: Annotated[int | None, Field(primary_key=True)] = None
    files: list["FileSQL"] = Relationship(back_populates="producer")


class ProducerRead(ProducerCreate):
    id: int


# Here, after you have defined all the Create, Read, Update and SQL models,
# you can define the read models with Relationships


class FileReadWithProducer(FileRead):
    producer: ProducerRead | None = None  # use the read model for Producer here


class ProducerReadWithFiles(ProducerRead):
    files: list[FileRead] = []  # use the read model for File here

And in your case, when you want to get the files for a producer, you make your code return the ProducerReadWithFiles Model.
Or you can use FileReadWithProducer if you want to get the producer of a file.

You can find the documentation in SQLModel for this sort of problem.

[joachimhuet]

Hey!

In fact the problem is not with the List but Rather with your Producer class that has a File type.

By inheriting Producer the typing of files uses the File type that is not castable by sqlalchemy.
So you should just avoid to declare relationship fields in base models as well as in table models:

class File(SQLModel):  # table=False by default
    name: str
    url: str

# That would be the base for your producer object
class ProducerBase(SQLModel): ...

# the files declaration here does not conflict with the declaration in SQL model
class Producer(ProducerBase):
    files: Optional[List[File]]


class FileSQL(File, table=True):  # `File` itself inherits from SQLModel
    __tablename__ = "file"
    id: Optional[int] = Field(default=None, primary_key=True)
    producer_id: Optional[int] = Field(default=None, foreign_key="producer.id")
    producer: Optional["ProducerSQL"] = Relationship(back_populates="files")


class ProducerSQL(ProducerBase, table=True):  # `Producer` itself inherits from SQLModel
    __tablename__ = "producer"
    id: Optional[int] = Field(default=None, primary_key=True)
    url: str = Field(unique=True)
    files: Optional[List['FileSQL']] = Relationship(back_populates="producer")