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python - 将结果 session.query 转换为 Object

转载 作者:太空宇宙 更新时间:2023-11-04 01:36:42 25 4
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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' ()

最佳答案

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。

关于python - 将结果 session.query 转换为 Object,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9081383/

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