How to get last record

Question

I need to get last record from db. I'm using sqlalchemy. At the moment, I'm doing like that:

obj = ObjectRes.query.all()
return str(obj[-1].id)

But it's too heavy query. How can I get last record better?

"Last" could have done with a clear definition here. In a system with concurrent updates the most recently allocated id and the most recently committed id may not be the same, for example transaction A allocates id 42, transaction B allocates id 43, transaction B commits, then transaction A commits. Which is "last"? — snakecharmerb, Aug 28 '23 at 14:47

score 214 · Accepted Answer · answered Dec 18 '11 at 13:19

214

Take a look at Query.first(). If you specify a sort on the right column, the first will be your last. An example could look like this:

obj = session.query(ObjectRes).order_by(ObjectRes.id.desc()).first()

answered Dec 18 '11 at 13:19

miku

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@dimazubrik I get a warning using by using it `SAWarning: Can't resolve label reference '-id'; converting to text() (this warning may be suppressed after 10 occurrences)` – Shift 'n Tab Nov 07 '18 at 14:44
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^ This has now become an error; I used `Model.query.order_by(-Model.field_name).first()` to retrieve the desired object – Pratik K. Oct 09 '19 at 20:40
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Does this actually return all values of the query and then just pop the first one (ie. super inefficient) or is it smart enough to actually to something like `ORDER BY desc LIMIT 1` under the hood? – Adam Hughes Aug 24 '20 at 17:25
@AdamHughes you could specify this with a .limit() modifier such as session.query(ObjectRes).order_by(ObjectRes.id.desc()).limit(1) – Kyle Meador Dec 28 '22 at 03:43

score 9 · Answer 2 · edited Dec 02 '14 at 14:22

9

Sometimes it is difficult to reformulate simple things:

SELECT * FROM ObjectRes WHERE id IN (SELECT MAX(id) FROM ObjectRes)

but this worked for me:

session.query(ObjectRes).filter(ObjectRes.id == session.query(func.max(ObjectRes.id)))

edited Dec 02 '14 at 14:22

empiric

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answered Dec 02 '14 at 13:50

Robert

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Alex Kosh · Answer 3 · 2021-03-11T09:02:43.900

Don't forget to disable existing ordering if needed

In my case I have dynamic ordered relationships:

class Match:
  ...
  records = relationship("Record", backref="match", lazy="dynamic", order_by="Record.id")

And when I tried accepted answer I got first record, not the last, cause ORDER BY were applied twice and spoiled the results.

According to documentation:

All existing ORDER BY settings can be suppressed by passing None

So the solution will be:

match = db_session.query(Match).first()
last_record = match.records.order_by(None).order_by(Record.id.desc()).first()

score 0 · Answer 4 · answered Jan 27 '22 at 09:25

This answer modifies the others to allow for cases where you don't know what the primary key is called.

from sqlalchemy.inspection import inspect
# ...
def last_row(Table: type, *, session): # -> Table
    primary_key = inspect(Table).primary_key[0].name # must be an arithmetic type
    primary_key_row = getattr(Table, primary_key)
    # get first, sorted by negative ID (primary key)
    return session.query(Table).order_by(-primary_key_row).first()

How to get last record

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