Question

我有两个型号

class Song(db.Model):
    __tablename__ = 'song'
    id = db.Column(db.Integer, primary_key=True)
    collections = db.relationship('SongCollect', backref='song', lazy='dynamic')

class SongCollect(db.Model):
    __tablename__ = 'songcollect'
    id = db.Column(db.Integer, primary_key=True)
    song_id = db.Column(db.Integer, db.ForeignKey('song.id'))
    collect_status = db.Column(db.Boolean, default=True)

如何通过计数获取歌曲查询结果顺序（SongCollect.collect_status == True）

我使用flask-sqlalchemy。

或者我如何将以下sql翻译成sqlalchemy语法

select s.id, s.songid, s.songname,c.collect_status, sum(c.collect_status) as collect_count from song as s left join songcollect as c on s.id = c.song_id group by s.id order by collect_count desc

Answer 1

# aliases: not really required
s = db.aliased(Song, name="s")
c = db.aliased(SongCollect, name="c")

# expression for the count: simple `sum` might not work on `boolean`
# types for all databases, the used option below is more robust
# collect_count = db.func.sum(c.collect_status)
collect_count = db.func.sum(db.case([(c.collect_status, 1)], else_=0))

# actual query
q = (db.session
     .query(s, collect_count.label("collect_count"))
     .outerjoin(c)
     .group_by(s)
     .order_by(collect_count.desc())
     )
for song, song_collect_count in q:
    print(song, song_collect_count)
print("-"*80)

按关系过滤器计数的Sqlalchemy查询结果顺序

1 个答案: