python - 将结果 session.query 转换为 Object

Question

我从 SQLAlchemy 开始，我尝试获取最后的插入(一条推文)，但我只获取了 sql 查询:

conf = readConf("../utils/conf.yaml")
schema = conf['bdd']['type'] + '://' + conf['bdd']['user'] + ':' + conf['bdd']['password'] + '@' + conf['bdd']['host'] + '/' + conf['bdd']['db'] + '?charset=utf8'

engine = create_engine(schema, echo=True)

Base = getBase()
Base.metadata.create_all(engine) 

Session = sessionmaker(bind=engine)
session = Session()
tq = session.query(Tweet).group_by(Tweet.tweet_id_uniq).filter(func.max(Tweet.tweet_id_uniq) == Tweet.tweet_id_uniq)
# tweet_id_uniq is a BIGINT autoincrement, so the highest value is the last Tweet.
print tq

打印结果是 SQL 查询(为了易读性包装):

SELECT tweets.tweet_id AS tweets_tweet_id, 
       tweets.tweet_id_uniq AS tweets_tweet_id_uniq, 
       tweets.user_id AS tweets_user_id, 
       tweets.user_id_uniq AS tweets_user_id_uniq, 
       tweets.tweet_text AS tweets_tweet_text, 
       tweets.created_at AS tweets_created_at, 
       tweets.in_reply_to AS tweets_in_reply_to, 
       tweets.geo_lat AS tweets_geo_lat, 
       tweets.geo_long AS tweets_geo_long, 
       tweets.screen_name AS tweets_screen_name, 
       tweets.name AS tweets_name, 
       tweets.profile_image_url AS tweets_profile_image_url, 
       tweets.source AS tweets_source 
FROM   tweets 
WHERE  max(tweets.tweet_id_uniq) = tweets.tweet_id_uniq 
GROUP BY tweets.tweet_id_uniq 

为什么我不检索推文？

编辑:如果我添加 .one() 或 .fisrt() 或 .all()，我会收到此错误:

2012-01-31 16:05:37,644 INFO sqlalchemy.engine.base.Engine BEGIN (implicit)
2012-01-31 16:05:37,645 INFO sqlalchemy.engine.base.Engine SELECT tweets.tweet_id AS tweets_tweet_id, tweets.tweet_id_uniq AS tweets_tweet_id_uniq, tweets.user_id AS tweets_user_id, tweets.user_id_uniq AS tweets_user_id_uniq, tweets.tweet_text AS tweets_tweet_text, tweets.created_at AS tweets_created_at, tweets.in_reply_to AS tweets_in_reply_to, tweets.geo_lat AS tweets_geo_lat, tweets.geo_long AS tweets_geo_long, tweets.screen_name AS tweets_screen_name, tweets.name AS tweets_name, tweets.profile_image_url AS tweets_profile_image_url, tweets.source AS tweets_source 
FROM tweets 
WHERE max(tweets.tweet_id_uniq) = tweets.tweet_id_uniq
2012-01-31 16:05:37,645 INFO sqlalchemy.engine.base.Engine ()
...
File "/usr/local/lib/python2.6/dist-packages/SQLAlchemy-0.7.3-py2.6-linux-i686.egg/sqlalchemy/engine/default.py", line 330, in do_execute
cursor.execute(statement, parameters)
File "/usr/lib/pymodules/python2.6/MySQLdb/cursors.py", line 166, in execute
self.errorhandler(self, exc, value)
File "/usr/lib/pymodules/python2.6/MySQLdb/connections.py", line 35, in defaulterrorhandler
raise errorclass, errorvalue
sqlalchemy.exc.ProgrammingError: (ProgrammingError) (1111, 'Invalid use of group function') 'SELECT tweets.tweet_id AS tweets_tweet_id, tweets.tweet_id_uniq AS tweets_tweet_id_uniq, tweets.user_id AS tweets_user_id, tweets.user_id_uniq AS tweets_user_id_uniq, tweets.tweet_text AS tweets_tweet_text, tweets.created_at AS tweets_created_at, tweets.in_reply_to AS tweets_in_reply_to, tweets.geo_lat AS tweets_geo_lat, tweets.geo_long AS tweets_geo_long, tweets.screen_name AS tweets_screen_name, tweets.name AS tweets_name, tweets.profile_image_url AS tweets_profile_image_url, tweets.source AS tweets_source \nFROM tweets \nWHERE max(tweets.tweet_id_uniq) = tweets.tweet_id_uniq GROUP BY tweets.tweet_id_uniq' ()

Answer 1

您正在尝试匹配最大值，然后将其分组。该值是独一无二的。因此，您不妨直接订购并获得一个。

tq = session.query(Tweet).order_by(desc(Tweet.tweet_id_uniq)).first()

tq 现在是您具有最大 ID 的 Tweet 对象。

顺便说一下，问题是你得到的查询是不正确的 sql。如果不重新包含该表(这可能不是您想要做的)或使用子查询(也很尴尬)，您无法将最大值与同一表中的列进行匹配。