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 ta         


        

Answer 1

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 2

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

Answer 3

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 4

try this code

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

Answer 5

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()