我有一个python算法,它返回给定用户的数据库行的排名顺序。此算法输出主键ID列表(这些键可以与post.id连接)。它看起来像下面,除了可能有数千个匹配:
result_rank = [1286, 1956, 6759, 3485, 2309, 3409, 9023, 912, 13098, 23489, 19023, 1239]
我想指示sqlalchemy选择这些行,并按照它们在列表中的排序顺序排序。问题是,我希望在此
上执行paginationresults = posts.query().filter(posts.id.in_(
resultIds)).order_by(??? how can I order by post.id = result_rank ???).paginate(page_num, posts_per_page)
我使用Postgresql作为数据库。
答案 0 :(得分:1)
除非有一个好的解决方案,否则我将破解我自己的paginate对象:
class paginate_obj:
""" Pagination dummy object. Takes a list and paginates it similar to sqlalchemy paginate() """
def __init__(self, paginatable, page, per_page):
self.has_next = (len(paginatable)/per_page) > page
self.has_prev = bool(page - 1)
self.next = page + self.has_next
self.prev = page - self.has_prev
self.items = paginatable[(page-1)*(per_page):(page)*(per_page)]
我认为排序的唯一方法是创建所有结果的列表,并根据一些lambda函数在python中对其进行排序:
results = my_table.query().all()
results.sort(key=lamba x: distance(x.lat, x.long, user_lat, user_long)
paginated_results = paginate_obj(results, 1, 10) #returns the first page of 10 elements
答案 1 :(得分:1)
我认为排序更重要,因为没有它,数据库级别的分页就完全没用了。注意到它后,我的回答根本没有涉及分页方面,但我认为即使是@mgoldwasser提供的答案也可用于此。
这是我想出的,以便能够根据初始过滤器列表选择一些对象并保留它们的顺序。代码是自我解释的:
# input
post_ids = [3, 4, 1]
# create helper (temporary in-query table with two columns: post_id, sort_order)
# this table looks like this:
# key | sort_order
# 3 | 0
# 4 | 1
# 1 | 2
q_subq = "\nUNION ALL\n".join(
"SELECT {} AS key, {} AS sort_order".format(_id, i)
for i, _id in enumerate(post_ids)
)
# wrap it in a `Selectable` so that we can use JOINs
s = (select([literal_column("key", Integer),
literal_column("sort_order", Integer)])
.select_from(text("({}) AS helper".format(text(q_subq))))
).alias("helper")
# actual query which is both the filter and sorter
q = (session.query(Post)
.join(s, Post.id == s.c.key) # INNER JOIN will filter implicitly
.order_by(s.c.sort_order) # apply sort order
)
适用于postgresql
和sqlite
。