how to use JSON type in sqlmodel , Problem with openapi.json, File "pydantic\json.py", line 90, in pydantic.json.pydantic_encoder TypeError: Object of type 'Column' is not JSON serializable

[muscodev]

First Check

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• 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.

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Example Code

from typing import Optional
from sqlmodel import Field, SQLModel, Column, JSON


class EmployeeMasterBase(SQLModel):
    additional_info: Optional[dict] = Field(sa_column=Column(JSON))

Description

i was creating a Sqlmodel to conncect with postgres.
i want a field which of postgres JSON type.
The above code is able to create the table.
But while query "/openapi.json" for swagger, It gives following error .

 \site-packages\fastapi\_compat.py", line 370, in get_model_definitions
    m_schema, m_definitions, m_nested_models = model_process_schema(
  File "pydantic\schema.py", line 582, in pydantic.schema.model_process_schema
  File "pydantic\schema.py", line 623, in pydantic.schema.model_type_schema
  File "pydantic\schema.py", line 249, in pydantic.schema.field_schema
  File "pydantic\schema.py", line 217, in pydantic.schema.get_field_info_schema
  File "pydantic\schema.py", line 996, in pydantic.schema.encode_default
  File "pydantic\json.py", line 90, in pydantic.json.pydantic_encoder
TypeError: Object of type 'Column' is not JSON serializable




### Operating System

Windows

### Operating System Details

Windows 10, 64 bit

### SQLModel Version

 sqlmodel = "^0.0.8"

### Python Version

Python 3.10.7

### Additional Context

_No response_

[muscodev]

How we approach when using json type in sqlmodel

[muscodev]

iam upgraded to sqlmodel version 0.0.11, still the issue exist.
i don't know , why pydantic taking Column for JSON serialization

additional_info: Optional[dict] = Field(default={}, sa_column=Column(JSON))

additional_info: Optional[dict] = Field(default_factory=dict, sa_column=Column(JSON))

i have used both above ways, no change

[muscodev]

@tiangolo @dhiraj any suggestions

[muscodev]

iam closing it , i have to check on it again.
Thankyou

[muscodev]

Iam posting this to avoid the same mistake by other
The issue was in another line of code, where i missed the Field

    created_on: datetime =Column(
        DateTime(timezone=True),
        server_default=func.now()
        )
    updated_on: datetime = Column(
        DateTime(timezone=True),
        server_default=func.now(),
        server_onupdate=func.now()
        )

corrected code is as follows

    created_on: datetime = Field(sa_column=Column(
        DateTime(timezone=True),
        server_default=func.now()
        ))
    updated_on: datetime = Field(sa_column=Column(
        DateTime(timezone=True),
        server_default=func.now(),
        server_onupdate=func.now()
        ))

[ad-code1993]

use this instead "Here's how to define a JSON field in SQLModel:
Import JSON from SQLAlchemy: You need to import the JSON type from sqlalchemy.
Python

from sqlalchemy import JSON

Define the field with sa_column: In your SQLModel class, define the field that will store JSON data and specify its SQLAlchemy column type using sa_column=Column(JSON).
Python

from typing import Optional, Dict, Any
from sqlmodel import Field, SQLModel, Column

class MyModel(SQLModel, table=True):
    id: Optional[int] = Field(default=None, primary_key=True)
    data: Dict[str, Any] = Field(sa_column=Column(JSON))"