sqlalchemy ordering by descending

Asked Feb 25 '22 at 13:36

Active Feb 25 '22 at 13:36

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I am making a gym website for a project and ran into trouble with ordering a sqlalchemy field in descending order.

I am trying to give each user a unique barcode. I generate a random barcode and then check if it is already in use. If it is, another barcode is generated until one is made that isn't in use.

This is my code:

barcode = save_barcode(barcode_generator())
    barcode_list = []
    last_id = User.query.order_by(desc(id))
    for x in range(1,last_id+1):
        temp_barcode = User.query.filter_by(id=x).first()
        barcode_list.append(temp_barcode.barcode)
    while barcode in barcode_list:
        barcode = save_barcode(barcode_generator())

The save_barcode and barcode_generator functions are ones I have made earlier in the program. A random barcode is generated fine using them (I checked in DB Browser). However, as soon as I added the routine to check if the barcode is already in use, I get an error about my order_by statement.

This is the error:

sqlalchemy.exc.ArgumentError: GROUP BY / OF / etc. expression expected, got <built-in function id>

Any help would be appreciated.

asked Feb 25 '22 at 13:36

S1edgehammer500

Try putting field inside string: `last_id = User.query.order_by(desc("id"))` – sudden_appearance Feb 25 '22 at 13:40
And same for `User.query.filter_by("id"=x).first()` – sudden_appearance Feb 25 '22 at 13:41
1

Or you can try use `User.query.order_by(desc(User.id))` or even `User.query.order_by(User.id.desc())` – sudden_appearance Feb 25 '22 at 13:43
Thanks a lot, your second suggestion worked perfectly – S1edgehammer500 Feb 25 '22 at 15:02

sqlalchemy ordering by descending

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