How to construct a slightly more complex filter using “or_” or “and_” in SQLAlchemy

Question

I\'m trying to do a very simple search from a list of terms

terms = [\'term1\', \'term2\', \'term3\']

How do I programmatically go through

Answer 1

Assuming that your terms variable contains valid SQL statement fragments, you can simply pass terms preceded by an asterisk to or_ or and_:

>>> from sqlalchemy.sql import and_, or_
>>> terms = ["name='spam'", "email='spam@eggs.com'"]
>>> print or_(*terms)
name='spam' OR email='spam@eggs.com'
>>> print and_(*terms)
name='spam' AND email='spam@eggs.com'

Note that this assumes that terms contains only valid and properly escaped SQL fragments, so this is potentially unsafe if a malicious user can access terms somehow.

Instead of building SQL fragments yourself, you should let SQLAlchemy build parameterised SQL queries using other methods from sqlalchemy.sql. I don't know whether you have prepared Table objects for your tables or not; if so, assume that you have a variable called users which is an instance of Table and it describes your users table in the database. Then you can do the following:

from sqlalchemy.sql import select, or_, and_
terms = [users.c.name == 'spam', users.c.email == 'spam@eggs.com']
query = select([users], and_(*terms))
for row in conn.execute(query):
    # do whatever you want here

Here, users.c.name == 'spam' will create an sqlalchemy.sql.expression._BinaryExpression object that records that this is a binary equality relation between the name column of the users table and a string literal that contains spam. When you convert this object to a string, you will get an SQL fragment like users.name = :1, where :1 is a placeholder for the parameter. The _BinaryExpression object also remembers the binding of :1 to 'spam', but it won't insert it until the SQL query is executed. When it is inserted, the database engine will make sure that it is properly escaped. Suggested reading: SQLAlchemy's operator paradigm

If you only have the database table but you don't have a users variable that describes the table, you can create it yourself:

from sqlalchemy import Table, MetaData, Column, String, Boolean
metadata = MetaData()
users = Table('users', metadata,
    Column('id', Integer, primary_key=True),
    Column('name', String),
    Column('email', String),
    Column('active', Integer)
)

Alternatively, you can use autoloading which queries the database engine for the structure of the database and builds users automatically; obviously this is more time-consuming:

users = Table('users', metadata, autoload=True)