Strange paginate behaviour when paginating a subquery in flask

问题

I am working in a flask project (my very first project in flask) and I am having a problem when trying to paginate a query that contains a subquery. Just to put you in the picture I have three models: User, Magazine and Article. Each User is subscribed to several Magazines, each Article belongs to a Magazine. I am using queries and paginating the results in several parts of the code without any problem except when I am using a subquery. Below is the code where I am retrieving all the users subscribed to magazines a specific user (with id==2) is subscribed to:

sub = db.session.query(Magazine).join(Magazine.subscribers).filter(User.id == 2).subquery()
subscribers = db.session.query(User).select_from(sub).join(Magazine.subscribers)

this code works nice and in subscribers.all() there are the users that there should be. I have created a mockup where there are 5 users subscribed to a magazine the current user is also subscribed to. So the following code

print(len(subscribers.all()), ' friends:')
for s in subscribers.all(): print(s.name)

outputs "5 friends" and their names correctly. However, the problem arises when I try to paginate subscribers:

page=request.args.get('page', 1, type=int)
friends=subscribers.paginate(page, 3, False)
print('len(friends.items) ', len(friends.items))
for f in friends.items: print(f.name)

To start with, I am asking for 3 elements and len(friends.items) is 2. I have tryed asking for differents number of results and it always returns one less. Secondly, friends.has_next is False when it should be True: there are 5 elements in subscribers.all() and I have only asked for 3. For clarity let me put the code with some printing statements and its output:

from app import app, db
from app.models import *
from flask import render_template, url_for, request

@app.route('/')
@app.route('/index')
def index():

    page=request.args.get('page', 1, type=int)
    print('page: ', page)

    sub = db.session.query(Magazine).join(Magazine.subscribers).filter(User.id == 2).subquery()
    subscribers = db.session.query(User).select_from(sub).join(Magazine.subscribers)

    print(len(subscribers.all()), ' friends:')
    for s in subscribers.all(): print(s.name)

    friends=subscribers.paginate(page, 3, False)


    print('len(friends.items) ', len(friends.items))
    for f in friends.items: print(f.name)
    print('has_next: ', friends.has_next, ' has_previous ', friends. has_prev)
    print('next_num: ', friends.next_num, ' prev_num: ', friends.prev_num)

    next_url=url_for('index', page=friends.next_num) if friends.has_next else None
    prev_url=url_for('index', page=friends.prev_num) if friends.has_prev else None

    print(next_url)
    print(prev_url)

    return render_template("index.html", title="My friends", friends=friends.items, next_url=next_url, prev_url=prev_url)

And the output on the shell:

page:  1
5  friends:
biel
lluna
bruno
llucia
moix
len(friends.items)  2
biel
lluna
has_next:  False  has_previous  False
next_num:  None  prev_num:  None
None
None

This piece of code works perfectly if I change subscribers for some other query that does not use any subquery. What am I missing?!!

thanks a lot

来源：https://stackoverflow.com/questions/58282132/strange-paginate-behaviour-when-paginating-a-subquery-in-flask

标签

python

Flask

sqlalchemy

subquery