在SQLAlchemy中如何查询符合batch=2018条件的全部匹配记录？

Hey there! Let's work through your two SQLAlchemy questions clearly:

1. Querying all records by field conditions in SQLAlchemy #

To fetch all records that match specific field conditions, you'll use SQLAlchemy's Query API with the filter() (or filter_by()) method paired with all()—this method returns a list of all matching objects instead of just a single one.

Here's a quick example with a hypothetical model:

# Assuming you've already set up your engine and session
from sqlalchemy.orm import sessionmaker
Session = sessionmaker(bind=your_engine)
session = Session()

# Example: Fetch all users where age > 18
all_eligible_users = session.query(User).filter(User.age > 18).all()

For simple equality checks, you can use filter_by() which is more concise (it takes keyword arguments instead of expression syntax):

# Fetch all users with age exactly 18
all_young_users = session.query(User).filter_by(age=18).all()

2. Getting all records where batch=2018 (instead of just the first) #

The issue with your current code is that first() only returns the very first matching record from the query result. To get every entry that matches batch=2018, just replace first() with all().

Here's how to adjust your code:

# Fetch all NewEntry records where batch equals 2018
all_matching_entries = session.query(NewEntry).filter(NewEntry.batch == 2018).all()

Or using the cleaner filter_by() syntax:

all_matching_entries = session.query(NewEntry).filter_by(batch=2018).all()

A quick note: If you're dealing with a huge number of records, all() will load everything into memory at once. For better performance in that case, use yield_per() to process records in batches:

# Process 100 records at a time to avoid memory overload
for entry in session.query(NewEntry).filter(NewEntry.batch == 2018).yield_per(100):
    # Do something with each entry
    print(entry.id, entry.some_field)

内容的提问来源于stack exchange，提问作者P Pariventhan