SQLAlchemy and going through a large result set [duplicate]

Answer 1

If you aren't using Flask, you can use SqlAlchemy function 'slice' or a combo of 'limit' & 'offset', as mentioned here. E.g.:

some_query = Query([TableBlaa])
query = some_query.limit(number_of_rows_per_page).offset(page_number*number_of_rows_per_page)
#  -- OR --
query = some_query.slice(page_number*number_of_rows_per_page, (page_number*number_of_rows_per_page)+number_of_rows_per_page)
current_pages_rows = session.execute(query).fetchall()

Answer 2

If you are building an api to use with ReactJs, vueJs or other frontEnd framework, you can process like:

Notice:

page: current page that you need

error_out: Not display errors

max_per_page or per_page : the limit

Documentaion: SqlAchemy pagination

    record_query = Record.query.paginate(page=*Number*, error_out=False, max_per_page=15)

    result = dict(datas=record_query.items, 
                   total=record_query.total, 
                   current_page=record_query.page,
                   per_page=record_query.per_page)

On record_query you can use :

next(error_out=False)

Returns a Pagination object for the next page.

next_num

Number of the next page

page = None

the current page number (1 indexed)

pages

The total number of pages

per_page = None

the number of items to be displayed on a page.

prev(error_out=False)

Returns a Pagination object for the previous page.

prev_num

Number of the previous page.

query = None

the unlimited query object that was used to create this pagination object.

total = None

the total number of items matching the query

Hope it help you!

Answer 3

If you are using Flask-SqlAlchemy, see the paginate method of query. paginate offers several method to simplify pagination.

record_query = Record.query.paginate(page, per_page, False)
total = record_query.total
record_items = record_query.items

First page should be 1 otherwise the .total returns exception divided by zero