The sqlmodel docs gives an example of two classes
class Team(SQLModel, table=True):
id: Optional[int] = Field(default=None, primary_key=True)
name: str = Field(index=True)
headquarters: str
heroes: List["Hero"] = Relationship(back_populates="team")
class Hero(SQLModel, table=True):
id: Optional[int] = Field(default=None, primary_key=True)
name: str = Field(index=True)
secret_name: str
age: Optional[int] = Field(default=None, index=True)
team_id: Optional[int] = Field(default=None, foreign_key="team.id")
team: Optional[Team] = Relationship(back_populates="heroes")
I could get a Team object using the following code example
def get_team():
with Session(engine) as session:
statement = select(Team).where(Team.name == "avengers")
result = session.exec(statement)
avengers = result.one()
return avengers
and doing avengers.heroes
should return a list of all heroes related to that object but What if the list contains thousands of items? is there a way to paginate this without having to make a separate query to the heroes table by myself?
To do this, you have to work with the underlying SQLAlchemy library, which you can do using the sa_relationship_kwargs
argument to Relationship
.
As mentioned in this answer, if you specify the relationship as dynamic, you can then paginate it. Using SQLModel, it looks something like this:
class Team(SQLModel, table=True):
...
heroes: List["Hero"] = Relationship(
back_populates="team",
sa_relationship_kwargs={"order_by": "Hero.name", "lazy": "dynamic"},
)
# When calling it
first_ten_heroes = team.heroes[:10] # or team.heroes.limit(10)