5

在 SQLAlchemy 中连接同一表数据的两个查询的正确方法是什么？

即我有一个数据类定义是这样的：

class DataMeasurement(Base):
    __tablename__ = 'DataMeasurement'
    id = Column(Integer, Sequence('data_measurement_id_seq'), primary_key=True)
    data_source = Column(String)
    timestamp = Column(DateTime)
    sensor_output = Column(Float)

...并且我想加入以下两个有匹配时间戳的查询：

q1 = self.session.query(DataMeasurement).filter_by(data_source='Sensor1').order_by(DataMeasurement.timestamp)
q2 = self.session.query(DataMeasurement).filter_by(data_source='Sensor2').order_by(DataMeasurement.timestamp)
# ...and now what?

有没有办法简单地做到这一点？...或者我是否以一种根本有缺陷的方式来解决这个问题（我对 SQLAlchemy 很陌生）？

4

2 回答 2

9

使用子查询：

subq = self.session.query(DataMeasurement).\
    filter_by(data_source='Sensor1').subquery()
q = self.session.query(
    DataMeasurement.timestamp,
    # Use labels to distinguish between identically named columns.
    # This is optional.
    subq.c.sensor_output.label('output1'),
    DataMeasurement.sensor_output.label('output2')
).filter(
    (DataMeasurement.data_source == 'Sensor2') &
    (DataMeasurement.timestamp == subq.c.timestamp)
)

# Simply get a list of named tuples.
print q.all()
# Or access each column using properties.
for row in q:
    print row.timestamp, row.output1, row.output2

您还可以将结果作为DataMeasurement对象获取：

subq = self.session.query(DataMeasurement).\
    filter_by(data_source='Sensor1').subquery()
# Use alias to associate mapped class to a subquery.
dmalias = aliased(DataMeasurement, subq)
q = self.session.query(dmalias, DataMeasurement).filter(
    (DataMeasurement.data_source == 'Sensor2') &
    (DataMeasurement.timestamp == dmalias.timestamp)
)

# For each row you get a tuple containing two DataMeasurement objects.
for dm1, dm2 in q:
    print dm1.timestamp, dm1.sensor_output, dm2.sensor_output
于 2013-01-03T15:09:28.030 回答
4

您可以使用别名来建立相同表之间的关系。

您的查询可能类似于

adalias1 = aliased(DataMeasurement)
adalias2 = aliased(DataMeasurement)
q1 = self.session.query(DataMeasurement).\
        filter(
               and_(
                    adalias1.data_source in ('Sensor1', 'Sensor2'), 
                    adalias1.timestamp == adalias2.timestamp
                   )
              )
于 2013-01-03T03:54:54.500 回答