SQLAlchemy equivalent to SQL "LIKE" statement

Question

A tags column has values like "apple banana orange" and "strawberry banana lemon". I want to find the SQLAlchemy equivalent statement to

SELECT * FROM table WHERE tags LIKE "%banana%";

What should I pass to Class.query.filter() to do this?

Answer 1

Each column has like() method, which can be used in query.filter(). Given a search string, add a % character on either side to search as a substring in both directions.

tag = request.form["tag"]
search = "%{}%".format(tag)
posts = Post.query.filter(Post.tags.like(search)).all()

Answer 2

Adding to the above answer, whoever looks for a solution, you can also try 'match' operator instead of 'like'. Do not want to be biased but it perfectly worked for me in Postgresql.

Note.query.filter(Note.message.match("%somestr%")).all()

It inherits database functions such as CONTAINS and MATCH. However, it is not available in SQLite.

For more info go Common Filter Operators

Answer 3

Try this code:

output = dbsession.query(<model_class>).filter(
    <model_class>.email.ilike("%" + <email> + "%")
)

Answer 4

In case you want the case insensitive like clause implementation:

session.query(TableName).filter(TableName.colName.ilike(f'%{search_text}%')).all()

Answer 5

In SQLAlchemy 1.4/2.0:

q = session.query(User).filter(User.name.like('e%'))

Answer 6

If you use native sql, you can refer to my code, otherwise just ignore my answer.

SELECT * FROM table WHERE tags LIKE "%banana%";

from sqlalchemy import text

bar_tags = "banana"

# '%' attention to spaces
query_sql = """SELECT * FROM table WHERE tags LIKE '%' :bar_tags '%'"""

# db is sqlalchemy session object
tags_res_list = db.execute(text(query_sql), {"bar_tags": bar_tags}).fetchall()

Answer 7

Using PostgreSQL like (see accepted answer above) somehow didn't work for me although cases matched, but ilike (case insensisitive like) does.

Answer 8

While the accepted answer works fine, "The ORM Query object is a legacy construct as of SQLAlchemy 2.0" (ref: Legacy Query API - SQLAlchemy 2.0 Documentation).

The corresponding SQLAlchemy v2.0 equivalent to the SQL LIKE statement using the select construct is as shown in the python 3.10 code snippet below:

from typing import List

from sqlalchemy import select
from sqlalchemy.orm import Session

...

def get_multi_like_tag_substring_bidirectional(
    db: Session,
    *,
    tags_search_substring: str,
    skip: int = 0,
    limit: int = 10,
) -> List[Post]:
    return db.scalars(
        select(Post)
        .where(Post.tags.like(f"%{tags_search_substring}%"))
        .offset(skip)
        .limit(limit)
    ).all()

...

banana_tagged_posts = get_multi_like_tag_substring_bidirectional(
    db=db_session,
    tags_search_substring = "banana"
)
    

Answer 9

What i've been using all these years:

model = MyModel
col = "name_of_my_column"
value = "your-value"

query = model.query.filter(getattr(model, col).like("%{}%".format(value))).all()

You are welcome! ;) Well it seems to me bay bay that's I don't know wtf I'm doing huh

Answer 10

Doing it like this worked for me (Oracle) as native sql

"SELECT * FROM table WHERE tags LIKE '%' || :bar_tags || '%' "